Multiple values not working in js ajax - javascript

I am trying to insert more than one data in db using js ajax but it is not working and when i am trying to inserting only one data it is successfully working
Here is my indexa.php
<html>
<head>
<script type="text/javascript">
function insert() {
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else{
xmlhttp = new ActionXObject('Microsoft.XMLHTTP');
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('success_failed_msg').innerHTML = xmlhttp.responseText;
} else {
console.log("faliled");
}
}
parameters = 'first_name='+document.getElementById('firstName').value;
console.log(parameters);
xmlhttp.open('POST','insert.inc.php',true);
xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
xmlhttp.send(parameters);
}
</script>
</head>
<body>
First Name : <input type="text" id="firstName"><br/><br/>
Last Name : <input type="text" id="lastName"><br/><br/>
Username : <input type="text" id="userName"><br/><br/>
Password : <input type="password" id="password"><br/><br/>
Re-type Password : <input type="password" id="retypePassword"><br/><br/>
<input type="button" value="Submit" onclick="insert()">
<div id="success_failed_msg"></div>
</body>
</html>
My include.inc.php
if (isset($_POST['first_name'])) {
$firstname = $_POST['first_name'];
if (!empty($firstname)) {
$insert_select = "INSERT INTO ajax_member_data(`first_name`) VALUES('".mysqli_real_escape_string($db_connect,$firstname)."')";
if ($insert_query_run = mysqli_query($db_connect,$insert_select)) {
echo 'Data inserted successfully';
} else {
echo 'Failed';
}
} else {
echo 'Please enter the value';
}
}
and when I am trying this script it
<script type="text/javascript">
function insert() {
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else{
xmlhttp = new ActionXObject('Microsoft.XMLHTTP');
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('success_failed_msg').innerHTML = xmlhttp.responseText;
} else {
console.log("faliled");
}
}
parameters = 'first_name='+document.getElementById('firstName').value'&last_name='+document.getElementById('lastName').value'&username='+document.getElementById('userName').value'&password='+document.getElementById('password').value'&retype_password='+document.getElementById('retypePassword').value;
console.log(parameters);
xmlhttp.open('POST','insert.inc.php',true);
xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
xmlhttp.send(parameters);
}
</script>
my include.inc.php
if (isset($_POST['first_name']) && isset($_POST['last_name']) && isset($_POST['username']) && isset($_POST['password']) && isset($_POST['retype_password'])) {
$firstname = $_POST['first_name'];
$lastname = $_POST['last_name'];
$usrname = $_POST['username'];
$password = $_POST['password'];
$retype_password = $_POST['retype_password'];
if (!empty($firstname) && !empty($lastname) && !empty($usrname) && !empty($password) && !empty($retype_password)) {
$insert_select = "INSERT INTO ajax_member_data(`first_name`,`last_name`,`user_name`,`password`) VALUES('".mysqli_real_escape_string($db_connect,$firstname)."', '".mysqli_real_escape_string($db_connect,$lastname)."', '".mysqli_real_escape_string($db_connect,$usrname)."', '".mysqli_real_escape_string($db_connect,$password)."')";
if ($insert_query_run = mysqli_query($db_connect,$insert_select)) {
echo 'Data inserted successfully';
} else {
echo 'Failed';
}
} else {
echo 'Please enter the value';
}
}

You haven't done concatenation properly. See here,
parameters = 'first_name='+document.getElementById('firstName').value'&last_name='+document.getElementById('lastName').value'&username='+document.getElementById('userName').value'&password='+document.getElementById('password').value'&retype_password='+document.getElementById('retypePassword').value;
^ missing + ^ missing + ^ missing + ^ missing +
It should be,
parameters = 'first_name='+document.getElementById('firstName').value+'&last_name='+document.getElementById('lastName').value+'&username='+document.getElementById('userName').value+'&password='+document.getElementById('password').value+'&retype_password='+document.getElementById('retypePassword').value;
Sidenote: Learn about prepared statements because right now your query is susceptible to SQL injection. Also see how you can prevent SQL injection in PHP.

You are missing plus(+) sign while passing parameter. Please change your code as below:
Old Code:
parameters = 'first_name='+document.getElementById('firstName').value'&last_name='+document.getElementById('lastName').value'&username='+document.getElementById('userName').value'&password='+document.getElementById('password').value'&retype_password='+document.getElementById('retypePassword').value;
New Code:(Added plus(+) sign wherever it was required)
parameters = 'first_name='+document.getElementById('firstName').value + '&last_name='+document.getElementById('lastName').value + '&username='+document.getElementById('userName').value + '&password='+document.getElementById('password').value + '&retype_password='+document.getElementById('retypePassword').value;

If you wrap the input fields with a form and use jQuery serialize could be easier.
Example HTML:
<form>
First Name : <input type="text" id="firstName"><br/><br/>
Last Name : <input type="text" id="lastName"><br/><br/>
Username : <input type="text" id="userName"><br/><br/>
Password : <input type="password" id="password"><br/><br/>
Re-type Password : <input type="password" id="retypePassword"><br/><br/>
<input type="button" value="Submit">
<div id="success_failed_msg"></div>
</form>
Example JS:
//you can use the var sending to avoid
// more than one request at the same time
var sending = false;
$('form').on('submit', function(ev){
ev.preventDefault();
if (!sending) {
$.ajax({
url: 'insert.inc.php',
method: 'post',
dataType: 'json',
data: $('form').serialize(),
cache: false,
beforeSend: function() {
//here you can show an ajax loading icon
sending = true;
},
success: function(response){
//here you can show a success message and check if the
//response is correct
//response object depends on the server side response
if (response.success || response.success === 'true') {
//show success message...
} else {
//show error message...
}
},
error: function(err){
//here you can show an error message
},
complete: function(){
//here you can hide the ajax loading icon
sending = false;
}
});
}
});
Documentation from jQuery:
Ajax http://api.jquery.com/jquery.ajax/
Serialize https://api.jquery.com/serialize/
And you can format a json response from the server side
json_encode (with this you "transform" a php array to a js object) http://php.net/manual/en/function.json-encode.php
header('Content-Type: application/json'); with this the app will know what type of response is given
You can read more about json and ajax here:
http://www.json.org/
https://developer.mozilla.org/en-US/docs/AJAX
And a tutorial about how to see the request on Firefox: https://developer.mozilla.org/en-US/docs/Tools/Network_Monitor
And on Chrome: https://developers.google.com/web/tools/chrome-devtools/network-performance/resource-loading

var sending = false;
$('form').on('submit', function(ev){
ev.preventDefault();
if (!sending) {
$.ajax({
url: 'insert.inc.php',
method: 'post',
dataType: 'json',
data: $('form').serialize(),
cache: false,
beforeSend: function() {
console.log("processing");
sending = true;
},
success: function(response){
if (response.success || response.success === 'true') {
('#success_failed_msg').text(response);
} else {
('#success_failed_msg').text('response failed');
}
},
error: function(err){
('#success_failed_msg').text(err);
},
complete: function(){
console.log('process complete');
sending = false;
}
});
}
});

Related

Mailchimp Wordpress Plugin

I have built this Mailchimp Newsletter sign up plugin. PHP its not my strength so I am not being able to fix the error it has.
I used This Mailchimp wrapper: https://github.com/drewm/mailchimp-api.
My PHP file is:
<?php
include('MailChimp.php'); // path to API wrapper downloaded from GitHub
use \DrewM\MailChimp\MailChimp;
function sm_mailchimp_subscribe()
{
// Settings:
$mcAPIKey = '70000000000000000007e-us00';
$mcListID = '323231214212';
$status = 'subscribed';
// Settings End
$MailChimp = new MailChimp($mcAPIKey);
$email = $_POST['email'];
$fname = $_POST['fname'];
$sname = $_POST['sname'];
$merge_vars = array(
'FNAME' => $fname,
'LNAME' => $sname
);
if (isset($email) and isset($fname) and isset($sname)) {
$result = $MailChimp->post("lists/" . $mcListID . "/members", [
'email_address' => $email,
'merge_fields' => $merge_vars,
'status' => $status,
]);
return json_encode($result);
}
}
if ($_POST['ajax']) {
echo sm_mailchimp_subscribe();
} elseif ($_POST['action'] == 'setCookies') {
echo setCookiesTemp();
}
function setCookiesTemp()
{
setcookie('sm_mailchimp-temp', true, time() + (86400 * 28), '/');
}
function setCookiesPerm()
{
setcookie('sm_mailchimp-perm', true, time() + (86400 * 28 * 24), '/');
}
And here is the JS:
jQuery(function($) {
// From wp_localized
const ajaxURL = ajax_obj.ajax_url;
$(document).ready(function() {
var body = $("body");
var modal = `<div class="sm_mailchimp-modal">
<div class="sm_mailchimp-box">
<button class="sm_mailchimp-close">X</button>
<h3>Join Our Mailing List</h3>
<form name="sm_mailchimp-form" method="post" class="sm_mailchimp-form light">
<p>Subscribe to our newsletter and receive tips for a successful business.</p>
<input type="email" name="email" placeholder="Type Your Email" class="sm_sname">
<div class="sm_mailchimp-message"></div>
<button type="submit">SIGN UP</button>
</form>
</div>
</div>`;
if (!document.cookie.includes("sm_mailchimp-temp=1")) {
$(modal)
.hide()
.appendTo(body);
setTimeout(function() {
$(".sm_mailchimp-modal").fadeIn();
}, 5000);
}
$(".sm_mailchimp-close").click(function() {
closeModal();
});
$(".sm_mailchimp-modal").click(function(e) {
if (e.target == this) {
closeModal();
}
});
function closeModal() {
$(".sm_mailchimp-modal").fadeOut();
$.ajax({
url: ajaxURL,
method: "POST",
data: { action: "setCookies" },
success: function(data) {
console.log("Cookie added as sm_mailchimp-temp=1");
}
});
}
$(".sm_mailchimp-form").submit(function() {
var emailVal, fnameVal, snameVal;
emailVal = $(".sm_email", this).val();
fnameVal = $(".sm_fname", this).val();
snameVal = $(".sm_sname", this).val();
if (emailVal === "" || fnameVal === "" || snameVal === "") {
$(".sm_mailchimp-message").html("Please fill out all the fields!");
return false;
} else {
$(".sm_mailchimp-message").html("Adding your email address...");
var form = $(this);
$.ajax({
url: ajaxURL,
type: "POST",
data: form.serialize() + "&ajax=true",
success: function(msg) {
var message = $.parseJSON(msg),
result = "";
if (message.status === "pending") {
result =
"Success! Please click the confirmation link that will be emailed to you shortly.";
} else if (message.status === "subscribed") {
result =
"Success! You have been subscribed to our mailing list.";
} else {
result = "Oops: " + message.title;
console.log("Error: ", message);
}
$(".sm_mailchimp-message").html(result);
}
});
return false;
}
});
});
});
The error is the following:
Once the Sign Up button is clicked the message "Adding your email address..." shows as it should but then it freezes there. The console shows the following error:
Weirdly, it used to work fine. And without a change it stopped. I built it a couple of months ago.
UPDATE
The error is actually coming from the success: function(msg) { line. The msg is empty. So, when it gets to message.status it cannot read the property of status null.
Question: Why is the success callback returning msg empty?

Prevent PHP form from redirecting on success and show HTML message [duplicate]

I am trying to send data from a form to a database. Here is the form I am using:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
The typical approach would be to submit the form, but this causes the browser to redirect. Using jQuery and Ajax, is it possible to capture all of the form's data and submit it to a PHP script (an example, form.php)?
Basic usage of .ajax would look something like this:
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
jQuery:
// Variable to hold request
var request;
// Bind to the submit event of our form
$("#foo").submit(function(event){
// Prevent default posting of form - put here to work in case of errors
event.preventDefault();
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "/form.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
});
Note: Since jQuery 1.8, .success(), .error() and .complete() are deprecated in favor of .done(), .fail() and .always().
Note: Remember that the above snippet has to be done after DOM ready, so you should put it inside a $(document).ready() handler (or use the $() shorthand).
Tip: You can chain the callback handlers like this: $.ajax().done().fail().always();
PHP (that is, form.php):
// You can access the values posted by jQuery.ajax
// through the global variable $_POST, like this:
$bar = isset($_POST['bar']) ? $_POST['bar'] : null;
Note: Always sanitize posted data, to prevent injections and other malicious code.
You could also use the shorthand .post in place of .ajax in the above JavaScript code:
$.post('/form.php', serializedData, function(response) {
// Log the response to the console
console.log("Response: "+response);
});
Note: The above JavaScript code is made to work with jQuery 1.8 and later, but it should work with previous versions down to jQuery 1.5.
To make an Ajax request using jQuery you can do this by the following code.
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>
JavaScript:
Method 1
/* Get from elements values */
var values = $(this).serialize();
$.ajax({
url: "test.php",
type: "post",
data: values ,
success: function (response) {
// You will get response from your PHP page (what you echo or print)
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus, errorThrown);
}
});
Method 2
/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
var ajaxRequest;
/* Stop form from submitting normally */
event.preventDefault();
/* Clear result div*/
$("#result").html('');
/* Get from elements values */
var values = $(this).serialize();
/* Send the data using post and put the results in a div. */
/* I am not aborting the previous request, because it's an
asynchronous request, meaning once it's sent it's out
there. But in case you want to abort it you can do it
by abort(). jQuery Ajax methods return an XMLHttpRequest
object, so you can just use abort(). */
ajaxRequest= $.ajax({
url: "test.php",
type: "post",
data: values
});
/* Request can be aborted by ajaxRequest.abort() */
ajaxRequest.done(function (response, textStatus, jqXHR){
// Show successfully for submit message
$("#result").html('Submitted successfully');
});
/* On failure of request this function will be called */
ajaxRequest.fail(function (){
// Show error
$("#result").html('There is error while submit');
});
The .success(), .error(), and .complete() callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use .done(), .fail(), and .always() instead.
MDN: abort() . If the request has been sent already, this method will abort the request.
So we have successfully send an Ajax request, and now it's time to grab data to server.
PHP
As we make a POST request in an Ajax call (type: "post"), we can now grab data using either $_REQUEST or $_POST:
$bar = $_POST['bar']
You can also see what you get in the POST request by simply either. BTW, make sure that $_POST is set. Otherwise you will get an error.
var_dump($_POST);
// Or
print_r($_POST);
And you are inserting a value into the database. Make sure you are sensitizing or escaping All requests (whether you made a GET or POST) properly before making the query. The best would be using prepared statements.
And if you want to return any data back to the page, you can do it by just echoing that data like below.
// 1. Without JSON
echo "Hello, this is one"
// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));
And then you can get it like:
ajaxRequest.done(function (response){
alert(response);
});
There are a couple of shorthand methods. You can use the below code. It does the same work.
var ajaxRequest= $.post("test.php", values, function(data) {
alert(data);
})
.fail(function() {
alert("error");
})
.always(function() {
alert("finished");
});
I would like to share a detailed way of how to post with PHP + Ajax along with errors thrown back on failure.
First of all, create two files, for example form.php and process.php.
We will first create a form which will be then submitted using the jQuery .ajax() method. The rest will be explained in the comments.
form.php
<form method="post" name="postForm">
<ul>
<li>
<label>Name</label>
<input type="text" name="name" id="name" placeholder="Bruce Wayne">
<span class="throw_error"></span>
<span id="success"></span>
</li>
</ul>
<input type="submit" value="Send" />
</form>
Validate the form using jQuery client-side validation and pass the data to process.php.
$(document).ready(function() {
$('form').submit(function(event) { //Trigger on form submit
$('#name + .throw_error').empty(); //Clear the messages first
$('#success').empty();
//Validate fields if required using jQuery
var postForm = { //Fetch form data
'name' : $('input[name=name]').val() //Store name fields value
};
$.ajax({ //Process the form using $.ajax()
type : 'POST', //Method type
url : 'process.php', //Your form processing file URL
data : postForm, //Forms name
dataType : 'json',
success : function(data) {
if (!data.success) { //If fails
if (data.errors.name) { //Returned if any error from process.php
$('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
}
}
else {
$('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
}
}
});
event.preventDefault(); //Prevent the default submit
});
});
Now we will take a look at process.php
$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`
/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
$errors['name'] = 'Name cannot be blank';
}
if (!empty($errors)) { //If errors in validation
$form_data['success'] = false;
$form_data['errors'] = $errors;
}
else { //If not, process the form, and return true on success
$form_data['success'] = true;
$form_data['posted'] = 'Data Was Posted Successfully';
}
//Return the data back to form.php
echo json_encode($form_data);
The project files can be downloaded from http://projects.decodingweb.com/simple_ajax_form.zip.
You can use serialize. Below is an example.
$("#submit_btn").click(function(){
$('.error_status').html();
if($("form#frm_message_board").valid())
{
$.ajax({
type: "POST",
url: "<?php echo site_url('message_board/add');?>",
data: $('#frm_message_board').serialize(),
success: function(msg) {
var msg = $.parseJSON(msg);
if(msg.success=='yes')
{
return true;
}
else
{
alert('Server error');
return false;
}
}
});
}
return false;
});
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
I use the way shown below. It submits everything like files.
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url = $(this).attr("action");
$.ajax({
url: url,
type: 'POST',
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
console.log("error");
}
});
});
If you want to send data using jQuery Ajax then there is no need of form tag and submit button
Example:
<script>
$(document).ready(function () {
$("#btnSend").click(function () {
$.ajax({
url: 'process.php',
type: 'POST',
data: {bar: $("#bar").val()},
success: function (result) {
alert('success');
}
});
});
});
</script>
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input id="btnSend" type="button" value="Send" />
<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<form method="post" id="form_content" action="Javascript:void(0);">
<button id="desc" name="desc" value="desc" style="display:none;">desc</button>
<button id="asc" name="asc" value="asc">asc</button>
<input type='hidden' id='check' value=''/>
</form>
<div id="demoajax"></div>
<script>
numbers = '';
$('#form_content button').click(function(){
$('#form_content button').toggle();
numbers = this.id;
function_two(numbers);
});
function function_two(numbers){
if (numbers === '')
{
$('#check').val("asc");
}
else
{
$('#check').val(numbers);
}
//alert(sort_var);
$.ajax({
url: 'test.php',
type: 'POST',
data: $('#form_content').serialize(),
success: function(data){
$('#demoajax').show();
$('#demoajax').html(data);
}
});
return false;
}
$(document).ready(function_two());
</script>
In your php file enter:
$content_raw = file_get_contents("php://input"); // THIS IS WHAT YOU NEED
$decoded_data = json_decode($content_raw, true); // THIS IS WHAT YOU NEED
$bar = $decoded_data['bar']; // THIS IS WHAT YOU NEED
$time = $decoded_data['time'];
$hash = $decoded_data['hash'];
echo "You have sent a POST request containing the bar variable with the value $bar";
and in your js file send an ajax with the data object
var data = {
bar : 'bar value',
time: calculatedTimeStamp,
hash: calculatedHash,
uid: userID,
sid: sessionID,
iid: itemID
};
$.ajax({
method: 'POST',
crossDomain: true,
dataType: 'json',
crossOrigin: true,
async: true,
contentType: 'application/json',
data: data,
headers: {
'Access-Control-Allow-Methods': '*',
"Access-Control-Allow-Credentials": true,
"Access-Control-Allow-Headers" : "Access-Control-Allow-Headers, Origin, X-Requested-With, Content-Type, Accept, Authorization",
"Access-Control-Allow-Origin": "*",
"Control-Allow-Origin": "*",
"cache-control": "no-cache",
'Content-Type': 'application/json'
},
url: 'https://yoururl.com/somephpfile.php',
success: function(response){
console.log("Respond was: ", response);
},
error: function (request, status, error) {
console.log("There was an error: ", request.responseText);
}
})
or keep it as is with the form-submit. You need this only, if you want to send a modified request with calculated additional content and not only some form-data, which is entered by the client. For example a hash, a timestamp, a userid, a sessionid and the like.
Handling Ajax errors and loader before submit and after submitting success shows an alert boot box with an example:
var formData = formData;
$.ajax({
type: "POST",
url: url,
async: false,
data: formData, // Only input
processData: false,
contentType: false,
xhr: function ()
{
$("#load_consulting").show();
var xhr = new window.XMLHttpRequest();
// Upload progress
xhr.upload.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = (evt.loaded / evt.total) * 100;
$('#addLoad .progress-bar').css('width', percentComplete + '%');
}
}, false);
// Download progress
xhr.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = evt.loaded / evt.total;
}
}, false);
return xhr;
},
beforeSend: function (xhr) {
qyuraLoader.startLoader();
},
success: function (response, textStatus, jqXHR) {
qyuraLoader.stopLoader();
try {
$("#load_consulting").hide();
var data = $.parseJSON(response);
if (data.status == 0)
{
if (data.isAlive)
{
$('#addLoad .progress-bar').css('width', '00%');
console.log(data.errors);
$.each(data.errors, function (index, value) {
if (typeof data.custom == 'undefined') {
$('#err_' + index).html(value);
}
else
{
$('#err_' + index).addClass('error');
if (index == 'TopError')
{
$('#er_' + index).html(value);
}
else {
$('#er_TopError').append('<p>' + value + '</p>');
}
}
});
if (data.errors.TopError) {
$('#er_TopError').show();
$('#er_TopError').html(data.errors.TopError);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
else
{
$('#headLogin').html(data.loginMod);
}
} else {
//document.getElementById("setData").reset();
$('#myModal').modal('hide');
$('#successTop').show();
$('#successTop').html(data.msg);
if (data.msg != '' && data.msg != "undefined") {
bootbox.alert({closeButton: false, message: data.msg, callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
} else {
bootbox.alert({closeButton: false, message: "Success", callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
}
}
}
catch (e) {
if (e) {
$('#er_TopError').show();
$('#er_TopError').html(e);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
}
});
I am using this simple one line code for years without a problem (it requires jQuery):
<script src="http://malsup.github.com/jquery.form.js"></script>
<script type="text/javascript">
function ap(x,y) {$("#" + y).load(x);};
function af(x,y) {$("#" + x ).ajaxSubmit({target: '#' + y});return false;};
</script>
Here ap() means an Ajax page and af() means an Ajax form. In a form, simply calling af() function will post the form to the URL and load the response on the desired HTML element.
<form id="form_id">
...
<input type="button" onclick="af('form_id','load_response_id')"/>
</form>
<div id="load_response_id">this is where response will be loaded</div>
Since the introduction of the Fetch API there really is no reason any more to do this with jQuery Ajax or XMLHttpRequests. To POST form data to a PHP-script in vanilla JavaScript you can do the following:
async function postData() {
try {
const res = await fetch('../php/contact.php', {
method: 'POST',
body: new FormData(document.getElementById('form'))
})
if (!res.ok) throw new Error('Network response was not ok.');
} catch (err) {
console.log(err)
}
}
<form id="form" action="javascript:postData()">
<input id="name" name="name" placeholder="Name" type="text" required>
<input type="submit" value="Submit">
</form>
Here is a very basic example of a PHP-script that takes the data and sends an email:
<?php
header('Content-type: text/html; charset=utf-8');
if (isset($_POST['name'])) {
$name = $_POST['name'];
}
$to = "test#example.com";
$subject = "New name submitted";
$body = "You received the following name: $name";
mail($to, $subject, $body);
Please check this. It is the complete Ajax request code.
$('#foo').submit(function(event) {
// Get the form data
// There are many ways to get this data using jQuery (you
// can use the class or id also)
var formData = $('#foo').serialize();
var url = 'URL of the request';
// Process the form.
$.ajax({
type : 'POST', // Define the type of HTTP verb we want to use
url : 'url/', // The URL where we want to POST
data : formData, // Our data object
dataType : 'json', // What type of data do we expect back.
beforeSend : function() {
// This will run before sending an Ajax request.
// Do whatever activity you want, like show loaded.
},
success:function(response){
var obj = eval(response);
if(obj)
{
if(obj.error==0){
alert('success');
}
else{
alert('error');
}
}
},
complete : function() {
// This will run after sending an Ajax complete
},
error:function (xhr, ajaxOptions, thrownError){
alert('error occured');
// If any error occurs in request
}
});
// Stop the form from submitting the normal way
// and refreshing the page
event.preventDefault();
});
Pure JS
In pure JS it will be much simpler
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
This is a very good article that contains everything that you need to know about jQuery form submission.
Article summary:
Simple HTML Form Submit
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get the form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = $(this).serialize(); // Encode form elements for submission
$.ajax({
url : post_url,
type: request_method,
data : form_data
}).done(function(response){ //
$("#server-results").html(response);
});
});
HTML Multipart/form-data Form Submit
To upload files to the server, we can use FormData interface available to XMLHttpRequest2, which constructs a FormData object and can be sent to server easily using the jQuery Ajax.
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="file" name="my_file[]" /> <!-- File Field Added -->
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = new FormData(this); // Creates new FormData object
$.ajax({
url : post_url,
type: request_method,
data : form_data,
contentType: false,
cache: false,
processData: false
}).done(function(response){ //
$("#server-results").html(response);
});
});
I hope this helps.
That's the code that fills a select option tag in HTML using ajax and XMLHttpRequest with the API is written in PHP and PDO
conn.php
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "db_event";
try {
$conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
?>
category.php
<?php
include 'conn.php';
try {
$data = json_decode(file_get_contents("php://input"));
$stmt = $conn->prepare("SELECT * FROM events ");
http_response_code(200);
$stmt->execute();
header('Content-Type: application/json');
$arr=[];
while($value=$stmt->fetch(PDO::FETCH_ASSOC)){
array_push($arr,$value);
}
echo json_encode($arr);
} catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
script.js
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
data = JSON.parse(this.responseText);
for (let i in data) {
$("#cars").append(
'<option value="' + data[i].category + '">' + data[i].category + '</option>'
)
}
}
};
xhttp.open("GET", "http://127.0.0.1:8000/category.php", true);
xhttp.send();
index.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link rel="stylesheet" href="style.css">
<script src="https://code.jquery.com/jquery-3.6.0.min.js"
integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
<title>Document</title>
</head>
<body>
<label for="cars">Choose a Category:</label>
<select name="option" id="option">
</select>
<script src="script.js"></script>
</body>
</html>
I have one other idea.
Which the URL that of PHP files which provided the download file.
Then you have to fire the same URL via ajax and I checked this second request only gives the response after your first request complete the download file. So you can get the event of it.
It is working via ajax with the same second request.}

PHP: submitting checkbox value through onChange event without page redirect [duplicate]

I am trying to send data from a form to a database. Here is the form I am using:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
The typical approach would be to submit the form, but this causes the browser to redirect. Using jQuery and Ajax, is it possible to capture all of the form's data and submit it to a PHP script (an example, form.php)?
Basic usage of .ajax would look something like this:
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
jQuery:
// Variable to hold request
var request;
// Bind to the submit event of our form
$("#foo").submit(function(event){
// Prevent default posting of form - put here to work in case of errors
event.preventDefault();
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "/form.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
});
Note: Since jQuery 1.8, .success(), .error() and .complete() are deprecated in favor of .done(), .fail() and .always().
Note: Remember that the above snippet has to be done after DOM ready, so you should put it inside a $(document).ready() handler (or use the $() shorthand).
Tip: You can chain the callback handlers like this: $.ajax().done().fail().always();
PHP (that is, form.php):
// You can access the values posted by jQuery.ajax
// through the global variable $_POST, like this:
$bar = isset($_POST['bar']) ? $_POST['bar'] : null;
Note: Always sanitize posted data, to prevent injections and other malicious code.
You could also use the shorthand .post in place of .ajax in the above JavaScript code:
$.post('/form.php', serializedData, function(response) {
// Log the response to the console
console.log("Response: "+response);
});
Note: The above JavaScript code is made to work with jQuery 1.8 and later, but it should work with previous versions down to jQuery 1.5.
To make an Ajax request using jQuery you can do this by the following code.
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>
JavaScript:
Method 1
/* Get from elements values */
var values = $(this).serialize();
$.ajax({
url: "test.php",
type: "post",
data: values ,
success: function (response) {
// You will get response from your PHP page (what you echo or print)
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus, errorThrown);
}
});
Method 2
/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
var ajaxRequest;
/* Stop form from submitting normally */
event.preventDefault();
/* Clear result div*/
$("#result").html('');
/* Get from elements values */
var values = $(this).serialize();
/* Send the data using post and put the results in a div. */
/* I am not aborting the previous request, because it's an
asynchronous request, meaning once it's sent it's out
there. But in case you want to abort it you can do it
by abort(). jQuery Ajax methods return an XMLHttpRequest
object, so you can just use abort(). */
ajaxRequest= $.ajax({
url: "test.php",
type: "post",
data: values
});
/* Request can be aborted by ajaxRequest.abort() */
ajaxRequest.done(function (response, textStatus, jqXHR){
// Show successfully for submit message
$("#result").html('Submitted successfully');
});
/* On failure of request this function will be called */
ajaxRequest.fail(function (){
// Show error
$("#result").html('There is error while submit');
});
The .success(), .error(), and .complete() callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use .done(), .fail(), and .always() instead.
MDN: abort() . If the request has been sent already, this method will abort the request.
So we have successfully send an Ajax request, and now it's time to grab data to server.
PHP
As we make a POST request in an Ajax call (type: "post"), we can now grab data using either $_REQUEST or $_POST:
$bar = $_POST['bar']
You can also see what you get in the POST request by simply either. BTW, make sure that $_POST is set. Otherwise you will get an error.
var_dump($_POST);
// Or
print_r($_POST);
And you are inserting a value into the database. Make sure you are sensitizing or escaping All requests (whether you made a GET or POST) properly before making the query. The best would be using prepared statements.
And if you want to return any data back to the page, you can do it by just echoing that data like below.
// 1. Without JSON
echo "Hello, this is one"
// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));
And then you can get it like:
ajaxRequest.done(function (response){
alert(response);
});
There are a couple of shorthand methods. You can use the below code. It does the same work.
var ajaxRequest= $.post("test.php", values, function(data) {
alert(data);
})
.fail(function() {
alert("error");
})
.always(function() {
alert("finished");
});
I would like to share a detailed way of how to post with PHP + Ajax along with errors thrown back on failure.
First of all, create two files, for example form.php and process.php.
We will first create a form which will be then submitted using the jQuery .ajax() method. The rest will be explained in the comments.
form.php
<form method="post" name="postForm">
<ul>
<li>
<label>Name</label>
<input type="text" name="name" id="name" placeholder="Bruce Wayne">
<span class="throw_error"></span>
<span id="success"></span>
</li>
</ul>
<input type="submit" value="Send" />
</form>
Validate the form using jQuery client-side validation and pass the data to process.php.
$(document).ready(function() {
$('form').submit(function(event) { //Trigger on form submit
$('#name + .throw_error').empty(); //Clear the messages first
$('#success').empty();
//Validate fields if required using jQuery
var postForm = { //Fetch form data
'name' : $('input[name=name]').val() //Store name fields value
};
$.ajax({ //Process the form using $.ajax()
type : 'POST', //Method type
url : 'process.php', //Your form processing file URL
data : postForm, //Forms name
dataType : 'json',
success : function(data) {
if (!data.success) { //If fails
if (data.errors.name) { //Returned if any error from process.php
$('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
}
}
else {
$('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
}
}
});
event.preventDefault(); //Prevent the default submit
});
});
Now we will take a look at process.php
$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`
/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
$errors['name'] = 'Name cannot be blank';
}
if (!empty($errors)) { //If errors in validation
$form_data['success'] = false;
$form_data['errors'] = $errors;
}
else { //If not, process the form, and return true on success
$form_data['success'] = true;
$form_data['posted'] = 'Data Was Posted Successfully';
}
//Return the data back to form.php
echo json_encode($form_data);
The project files can be downloaded from http://projects.decodingweb.com/simple_ajax_form.zip.
You can use serialize. Below is an example.
$("#submit_btn").click(function(){
$('.error_status').html();
if($("form#frm_message_board").valid())
{
$.ajax({
type: "POST",
url: "<?php echo site_url('message_board/add');?>",
data: $('#frm_message_board').serialize(),
success: function(msg) {
var msg = $.parseJSON(msg);
if(msg.success=='yes')
{
return true;
}
else
{
alert('Server error');
return false;
}
}
});
}
return false;
});
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
I use the way shown below. It submits everything like files.
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url = $(this).attr("action");
$.ajax({
url: url,
type: 'POST',
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
console.log("error");
}
});
});
If you want to send data using jQuery Ajax then there is no need of form tag and submit button
Example:
<script>
$(document).ready(function () {
$("#btnSend").click(function () {
$.ajax({
url: 'process.php',
type: 'POST',
data: {bar: $("#bar").val()},
success: function (result) {
alert('success');
}
});
});
});
</script>
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input id="btnSend" type="button" value="Send" />
<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<form method="post" id="form_content" action="Javascript:void(0);">
<button id="desc" name="desc" value="desc" style="display:none;">desc</button>
<button id="asc" name="asc" value="asc">asc</button>
<input type='hidden' id='check' value=''/>
</form>
<div id="demoajax"></div>
<script>
numbers = '';
$('#form_content button').click(function(){
$('#form_content button').toggle();
numbers = this.id;
function_two(numbers);
});
function function_two(numbers){
if (numbers === '')
{
$('#check').val("asc");
}
else
{
$('#check').val(numbers);
}
//alert(sort_var);
$.ajax({
url: 'test.php',
type: 'POST',
data: $('#form_content').serialize(),
success: function(data){
$('#demoajax').show();
$('#demoajax').html(data);
}
});
return false;
}
$(document).ready(function_two());
</script>
In your php file enter:
$content_raw = file_get_contents("php://input"); // THIS IS WHAT YOU NEED
$decoded_data = json_decode($content_raw, true); // THIS IS WHAT YOU NEED
$bar = $decoded_data['bar']; // THIS IS WHAT YOU NEED
$time = $decoded_data['time'];
$hash = $decoded_data['hash'];
echo "You have sent a POST request containing the bar variable with the value $bar";
and in your js file send an ajax with the data object
var data = {
bar : 'bar value',
time: calculatedTimeStamp,
hash: calculatedHash,
uid: userID,
sid: sessionID,
iid: itemID
};
$.ajax({
method: 'POST',
crossDomain: true,
dataType: 'json',
crossOrigin: true,
async: true,
contentType: 'application/json',
data: data,
headers: {
'Access-Control-Allow-Methods': '*',
"Access-Control-Allow-Credentials": true,
"Access-Control-Allow-Headers" : "Access-Control-Allow-Headers, Origin, X-Requested-With, Content-Type, Accept, Authorization",
"Access-Control-Allow-Origin": "*",
"Control-Allow-Origin": "*",
"cache-control": "no-cache",
'Content-Type': 'application/json'
},
url: 'https://yoururl.com/somephpfile.php',
success: function(response){
console.log("Respond was: ", response);
},
error: function (request, status, error) {
console.log("There was an error: ", request.responseText);
}
})
or keep it as is with the form-submit. You need this only, if you want to send a modified request with calculated additional content and not only some form-data, which is entered by the client. For example a hash, a timestamp, a userid, a sessionid and the like.
Handling Ajax errors and loader before submit and after submitting success shows an alert boot box with an example:
var formData = formData;
$.ajax({
type: "POST",
url: url,
async: false,
data: formData, // Only input
processData: false,
contentType: false,
xhr: function ()
{
$("#load_consulting").show();
var xhr = new window.XMLHttpRequest();
// Upload progress
xhr.upload.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = (evt.loaded / evt.total) * 100;
$('#addLoad .progress-bar').css('width', percentComplete + '%');
}
}, false);
// Download progress
xhr.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = evt.loaded / evt.total;
}
}, false);
return xhr;
},
beforeSend: function (xhr) {
qyuraLoader.startLoader();
},
success: function (response, textStatus, jqXHR) {
qyuraLoader.stopLoader();
try {
$("#load_consulting").hide();
var data = $.parseJSON(response);
if (data.status == 0)
{
if (data.isAlive)
{
$('#addLoad .progress-bar').css('width', '00%');
console.log(data.errors);
$.each(data.errors, function (index, value) {
if (typeof data.custom == 'undefined') {
$('#err_' + index).html(value);
}
else
{
$('#err_' + index).addClass('error');
if (index == 'TopError')
{
$('#er_' + index).html(value);
}
else {
$('#er_TopError').append('<p>' + value + '</p>');
}
}
});
if (data.errors.TopError) {
$('#er_TopError').show();
$('#er_TopError').html(data.errors.TopError);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
else
{
$('#headLogin').html(data.loginMod);
}
} else {
//document.getElementById("setData").reset();
$('#myModal').modal('hide');
$('#successTop').show();
$('#successTop').html(data.msg);
if (data.msg != '' && data.msg != "undefined") {
bootbox.alert({closeButton: false, message: data.msg, callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
} else {
bootbox.alert({closeButton: false, message: "Success", callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
}
}
}
catch (e) {
if (e) {
$('#er_TopError').show();
$('#er_TopError').html(e);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
}
});
I am using this simple one line code for years without a problem (it requires jQuery):
<script src="http://malsup.github.com/jquery.form.js"></script>
<script type="text/javascript">
function ap(x,y) {$("#" + y).load(x);};
function af(x,y) {$("#" + x ).ajaxSubmit({target: '#' + y});return false;};
</script>
Here ap() means an Ajax page and af() means an Ajax form. In a form, simply calling af() function will post the form to the URL and load the response on the desired HTML element.
<form id="form_id">
...
<input type="button" onclick="af('form_id','load_response_id')"/>
</form>
<div id="load_response_id">this is where response will be loaded</div>
Since the introduction of the Fetch API there really is no reason any more to do this with jQuery Ajax or XMLHttpRequests. To POST form data to a PHP-script in vanilla JavaScript you can do the following:
async function postData() {
try {
const res = await fetch('../php/contact.php', {
method: 'POST',
body: new FormData(document.getElementById('form'))
})
if (!res.ok) throw new Error('Network response was not ok.');
} catch (err) {
console.log(err)
}
}
<form id="form" action="javascript:postData()">
<input id="name" name="name" placeholder="Name" type="text" required>
<input type="submit" value="Submit">
</form>
Here is a very basic example of a PHP-script that takes the data and sends an email:
<?php
header('Content-type: text/html; charset=utf-8');
if (isset($_POST['name'])) {
$name = $_POST['name'];
}
$to = "test#example.com";
$subject = "New name submitted";
$body = "You received the following name: $name";
mail($to, $subject, $body);
Please check this. It is the complete Ajax request code.
$('#foo').submit(function(event) {
// Get the form data
// There are many ways to get this data using jQuery (you
// can use the class or id also)
var formData = $('#foo').serialize();
var url = 'URL of the request';
// Process the form.
$.ajax({
type : 'POST', // Define the type of HTTP verb we want to use
url : 'url/', // The URL where we want to POST
data : formData, // Our data object
dataType : 'json', // What type of data do we expect back.
beforeSend : function() {
// This will run before sending an Ajax request.
// Do whatever activity you want, like show loaded.
},
success:function(response){
var obj = eval(response);
if(obj)
{
if(obj.error==0){
alert('success');
}
else{
alert('error');
}
}
},
complete : function() {
// This will run after sending an Ajax complete
},
error:function (xhr, ajaxOptions, thrownError){
alert('error occured');
// If any error occurs in request
}
});
// Stop the form from submitting the normal way
// and refreshing the page
event.preventDefault();
});
Pure JS
In pure JS it will be much simpler
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
This is a very good article that contains everything that you need to know about jQuery form submission.
Article summary:
Simple HTML Form Submit
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get the form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = $(this).serialize(); // Encode form elements for submission
$.ajax({
url : post_url,
type: request_method,
data : form_data
}).done(function(response){ //
$("#server-results").html(response);
});
});
HTML Multipart/form-data Form Submit
To upload files to the server, we can use FormData interface available to XMLHttpRequest2, which constructs a FormData object and can be sent to server easily using the jQuery Ajax.
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="file" name="my_file[]" /> <!-- File Field Added -->
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = new FormData(this); // Creates new FormData object
$.ajax({
url : post_url,
type: request_method,
data : form_data,
contentType: false,
cache: false,
processData: false
}).done(function(response){ //
$("#server-results").html(response);
});
});
I hope this helps.
That's the code that fills a select option tag in HTML using ajax and XMLHttpRequest with the API is written in PHP and PDO
conn.php
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "db_event";
try {
$conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
?>
category.php
<?php
include 'conn.php';
try {
$data = json_decode(file_get_contents("php://input"));
$stmt = $conn->prepare("SELECT * FROM events ");
http_response_code(200);
$stmt->execute();
header('Content-Type: application/json');
$arr=[];
while($value=$stmt->fetch(PDO::FETCH_ASSOC)){
array_push($arr,$value);
}
echo json_encode($arr);
} catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
script.js
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
data = JSON.parse(this.responseText);
for (let i in data) {
$("#cars").append(
'<option value="' + data[i].category + '">' + data[i].category + '</option>'
)
}
}
};
xhttp.open("GET", "http://127.0.0.1:8000/category.php", true);
xhttp.send();
index.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link rel="stylesheet" href="style.css">
<script src="https://code.jquery.com/jquery-3.6.0.min.js"
integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
<title>Document</title>
</head>
<body>
<label for="cars">Choose a Category:</label>
<select name="option" id="option">
</select>
<script src="script.js"></script>
</body>
</html>
I have one other idea.
Which the URL that of PHP files which provided the download file.
Then you have to fire the same URL via ajax and I checked this second request only gives the response after your first request complete the download file. So you can get the event of it.
It is working via ajax with the same second request.}

form field is not stored in mysql database

I have created a simple form consisting of a textarea field, so when user clicks on submit button its linked to a jquery script containing a url executing the process and store the data, but problem is every time i hit submit, ID & created_at data is stored but the data given on textarea is ignored and not stored, never faced this problem before..please help me out!
HTML
<form id="form" name="form" method="POST" action="profile_1.php" class="wizard-big" autocomplete="off" enctype="multipart/form-data" required="">
<div class="form-group col-sm-12">
<textarea type="text" name="status" id="status" placeholder="What's on your mind.." class="form-control" style="height:100px;"></textarea>
</div>
<div class="col-sm-12 form-group">
<input style="width:100%" type="submit" name="submit" id="submit" value="Post" class="btn btn-success">
</div>
</form>
Jquery
$(document).ready(function() {
$("#submit").click(function(e) {
var status = $('form')[0].checkValidity();
if (status) {
var formData = new FormData($('form')[0]);
$.ajax({
url: "form_post.php",
type: "POST",
data: formData,
processData: false,
contentType: false,
async: false,
dataType: "JSON",
success: function(json) {
if (json.error) {
alert(json.error_msg);
e.preventDefault();
} else {
alert("Post updated successfully!");
}
},
error: function(jqXHR, textStatus, errorThrown) {
alert(errorThrown);
}
});
}
});
});
php
<?php
session_start();
define('HOST','localhost');
define('USER','**');
define('PASS','**');
define('DB','**');
$response = array();
$con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect');
if(!mysqli_connect_errno()){
$error_flag = false;
/*foreach($_POST as $value){
if(empty($value)){
$error_flag = true;
break;
}
}*/
if(!$error_flag){
//receiving post parameters
$status =$_POST['status'];
// create a new user profile
$sql = "INSERT INTO status (via, status, created_at) VALUES ('".$_SESSION['vault_no']."', '$status', NOW())";
if(mysqli_query($con,$sql)){
$response["error"] = false;
$response['via'] = $via;
echo json_encode($response);
}else{
$response["error"] = true;
$response["error_msg"] = "INSERT operation failed";
echo json_encode($response);
}
}else{
$response["error"] = true;
$response["error_msg"] = "Few fields are missing";
echo json_encode($response);
}
}else{
$response["error"] = true;
$response["error_msg"] = "Database connection failed";
echo json_encode($response);
}
?>
Note: The solution is in the comments for other readers of this question
Maybe this helps you out. You need to change it to your wish offcourse
And save this function, it could be usefull for you in the future.
This function serializes the form as how it should be done.
<script>
$.fn.serializeObject = function()
{
var o = {};
var a = this.serializeArray();
$.each(a, function() {
if (o[this.name] !== undefined) {
if (!o[this.name].push) {
o[this.name] = [o[this.name]];
}
o[this.name].push(this.value || '');
} else {
o[this.name] = this.value || '';
}
});
return o;
};
$(function() {
$('form').submit(function() {
var formData = $('form').serializeObject();
$.ajax({
data: formData,
type: 'POST',
url: 'form_post.php',
success: function(result) {
$('#result').html(result);
},
error: function(jqXHR, textStatus, errorThrown) { alert(textStatus); }
});
return false;
});
});
</script>

Login on same page, check SESSION without reload

I currently have a form at the top of every page on my website that lets the user input the username and password to login.
but Once the button is clicked, I use JQuery AJAX method to submit it to login.php without page refresh where it validates credentials and returns users whether the username / password submitted was valid or invalid.
Finally, the result is returned back to the page the user tried to login on.
I would also like the page to after once there is a successful login, the form disappears and is replaced with "Welcome back, USER!"
Everything works except for I want this to happen without a page reload. Here is what I have so far:-
Display form if session is not set, otherwise say Welcome back, user:-
<div class="user">
<?php
if (isset($_SESSION['logged_user'])) {
$logged_user = $_SESSION['logged_user'];
print("<p>Welcome back, $logged_user!</p>");
}
else {
print('<div class="forms">
<form id="login">
<label>Username <input type="text" name="username" id="username"></label>
<label>Password <input type="text" name="password" id="password"></label>
<input type="submit" class="submit" value="Login" id="login">
<span class="error"></span>
</form>
</div>');
}
?>
</div>
Javascript / AJAX to handle submission:
$(document).ready(function() {
$("#login").click(function() {
var username = $("#username").val();
var password = $("#password").val();
var dataString = 'username='+username+'&password='+password;
if(username != '' && password != '') {
$.ajax({
type: "POST",
url: "login.php",
data: dataString,
success: function(responseText) {
if(responseText == 0) {
$("error").html("Invalid login.");
}
else if(responseText == 1) {
window.location = 'index.php';
}
else {
alert('Error');
}
}
});
}
return false;
});
And my login.php does everything fine.
Right now, the only way for my page to update to show the Welcome back message is if I include the line: window.location = 'index.php'; so that the page reloads. But without this, the user will have logged in successfully but will not be able to see this.
Is there a way to do this without AngularJS? This is for a class and we are not allowed to use frameworks, which has been quite frustrating! Thanks!
You can use Dynamic HTML (DHTML).
The success function you don't redirect instead using jquery to change the content.
Something like this:
success: function(responseText) {
if(responseText == 0) {
alert('Invalid login');
}
else if(responseText == 1) {
// Success
$('#login').parent().append('<p>Welcome back, '+username+'!</p>');
$('#login').remove();
}
else {
alert('Another Invalid login');
}
}
The API Login return:
0 if invalid
1 if valid
use this code,
<div class="user">
<div class="after_login" style="display:none">
<?php
$logged_user = $_SESSION['logged_user'];
print("<p>Welcome back, $logged_user!</p>");?>
</div>
<div class = "before_login">
print('<div class="forms">
<form id="login">
<label>Username <input type="text" name="username" id="username"></label>
<label>Password <input type="text" name="password" id="password"></label>
<input type="submit" class="submit" value="Login" id="login">
<span class="error"></span>
</form>
</div>');
?>
</div>
</div>
in script use like this instead of:
window.location = 'index.php';
use below lines,
$('.after_login').show();
$('.before_login').hide();
You have to change some modification in your javascript code as below :
$(document).ready(function() {
$("#login").click(function() {
var username = $("#username").val();
var password = $("#password").val();
var dataString = 'username='+username+'&password='+password;
if(username != '' && password != '') {
$.ajax({
type: "POST",
url: "login.php",
data: dataString,
dataType: "json",
success: function(responseText) {
if(responseText.status == 'invalid') {
$(".error").html("Invalid login.");
}
else if(responseText.status == 'valid') {
$('.after_login').html('Welcome back,'+responseText.username);
$('.after_login').show();
$('.before_login').hide();
}
else {
alert('Error');
}
}
});
}
return false;
});
In above example, I have added one extra line dataType: "json", So now you have need to send output in json format from your login.php and also some modification done in success.
$response = array();
// $result variable get result from your database
if(count($result) == 0)
{
$response['status'] = 'invalid';
}else{
$response['status'] = 'valid';
$response['username'] = $result['username'];
}
echo json_encode($response);
exit;
So your success become like that below
success: function(responseText) {
if(responseText.status == 'invalid') {
$(".error").html("Invalid login.");
}
else if(responseText.status == 'valid') {
$('.after_login').html('Welcome back,'+responseText.username);
$('.after_login').show();
$('.before_login').hide();
}
else {
alert('Error');
}
In your login.php, change success condition to output username instead of 1. So responseText will have the username on ajax callback.
Remove the final else condition in your ajax success, and do something like this in the success condition
$('.forms').html('<p>Welcome back, ' + responseText + '!</p>');
PHP:
if (isset($_SESSION['logged_user'])) {
$logged_user = $_SESSION['logged_user'];
//Leave out the print(Welcome...) part, it won't do anything because the page won't be refreshed.
}
else{
exit('1');
}
In the Ajax, don't worry about responseText == 0, just stick with ==1:
success: function(responseText) {
if(responseText == "1") {
alert("Nope");
}
else {
$("#successdiv").html("welcome...");
}

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