How to display error messages Jquery ajax? - javascript

I am a student and am using jquery and php to add records to database. Records are being added, but i want to display a message "Record inserted" in the if the the record has been successfully been added and an error message if an error occurs.
This is my html code:
<form id="forn-newsletter" class="form-horizontal" method="POST">
<div class="form-group">
<label id="name_label" class="control-label col-xs-2">Name</label>
<div class="col-xs-10">
<input type="text" class="form-control" id="news_name" name="news_name" placeholder="Name" onblur="checkName();"/><font color="red" id="name_error"></font>
</div>
</div>
<div class="form-group">
<label id="email_label" class="control-label col-xs-2">Email</label>
<div class="col-xs-10">
<input type="text" class="form-control" id="news_email" name="news_email" placeholder="Email" onblur="vali()"/><font color="red" id="email_error"></font>
</div>
</div>
<div class="form-group">
<div class="col-xs-offset-2 col-xs-10">
<button id="register-newsletter" type="submit" class="btn btn-primary">Register for Newsletter</button>
</div>
</div>
<div id="dialog"></div>
</form>
This is my registration-newsletter.php
<?php
include('connect.php');
$name=$_POST['name'];
$email=$_POST['email'];
$val=md5($name.$email);
$query = "INSERT INTO newsletter (Id,Name,Email,Val) VALUES('','$name','$email','$val')";
$result=mysql_query($query);
if(!$result){
echo "Some error Occured..Please try later";
}
else{
echo "Your details have been saved. Thank You ";
}
mysql_close($con);
?>
This is my JQuery code
$(document).ready(function(){
$("#register-newsletter").click(function(){
var name=$("#news_name").val();
var email=$("#news_email").val();
var dataString="name="+name+"&email="+email;
var request;
request = $.ajax({
url: "registration-newsletter.php",
type: "POST",
data: dataString
});
//return false;
});
});

Add a span to your html code for displaying error.
<span id="error"></span>
Already you are echoing the message from PHP page to ajax. You can do mysql_affected_rows() to check whether the query updated the table.
$result=mysql_query($query);
if(mysql_affected_rows()>0){ // return the number of records that are inserted
echo "Your details have been saved. Thank You "; // success
}
else{
echo "Some error Occured..Please try later"; // failure
}
exit;
Then you can simply show the echoed message in the span with id error as:
request = $.ajax({
url: "registration-newsletter.php",
type: "POST",
data: dataString,
success:function(response) // response from requested PHP page
{
$('#error').html(response); // set the message as html of span
}
});

$.ajax({
url: 'registration-newsletter.php',
type: 'post',
data: dataString ,
success: function (msg) {
alert(msg);
},
error:function(msg)
{
alert(msg);
}
});

jQuery.ajax({
url: "myfile.php",
type: "POST",
data: dataString,
success:function(response) /* this response is an array which is returning from the myfile.php */
{
$status = response['status'];
if($status == 'success')
{
$('#message').html(response['msg']);
}
else
{
$('#message').html(response['msg']);
}
}
});
The function which you have added success will handle the "text to be append" or "alert to be show". Its quite equal to if condition, If the response came successfully, It will go into the condition.

This is what worked for me for form submit (see those 2 parameters to .then function):
<span id="result">Loading...</span>
<div id="form">
<form>
<input id="personId"> <-- put personID here<br>
<input id="submit" type="submit" value="Generate">
</form>
</div>
<script type="application/javascript">
$(document).ready(function() {
var submit = $('#submit');
var res = $('#result');
$('#form').submit(function() {
submit.prop('disabled', true);
res.text("Sending message...");
$.ajax({
url: '/rest/message',
contentType: "application/json;charset=UTF-8",
method: "POST",
data: JSON.stringify({'personId': $('#personId').val()})
}).then(function(result) { // OK
res.text("Message was generated");
submit.prop('disabled', false);
}, function(reason) { // ERROR
res.css('color', 'red').css('font-weight','Bold');
res.text("An error has occurred: " + reason.responseJSON.message);
submit.prop('disabled', false);
});
return false; // prevent default action
});
});
</script>

error: function(xhr) {
alert(xhr.responseText);
}
You can also get the HTTP status code or the generic message sent by your server (such as "Internal Server Error") in this XHR object if needed. More info : https://api.jquery.com/jQuery.ajax/#jqXHR

Related

Ajax post with php-mysql is not working properly

I need a ajax call to post data to the database and fetch the data from database and update in live. I have the following codes
HTML Form
<div class="hover_bkgr_fricc">
<span class="helper"></span>
<div>
<div class="popupCloseButton">×</div>
<p>
<form>
<input type="hidden" name="check_num" value="123" />
<p>Please provide more details</p>
<input type="text" name="reason" />
<a id="submit">Mark Reviewed</a>
</form>
</p>
</div>
</div>
<b id="review_result"></b>
<a class="trigger_popup_fricc">
<button> Mark Reviewed</button>
</a>
Javascript Block
$(document).ready(function() {
$(".trigger_popup_fricc").click(function() {
$('.hover_bkgr_fricc').show();
});
$('.popupCloseButton').click(function() {
$('.hover_bkgr_fricc').hide();
});
$('#submit').click(function() {
var check_num = $('input[name=check_num]').val();
var reason = $('input[name=reason]').val();
var form_data =
'check_num=' + check_num +
'&reason=' + reason;
$.ajax({
url: "loweslinkprocess.php",
type: "POST",
data: form_data,
success: function(html) {
//if process.php returned 1/true (send mail success)
if (html == 1) {
//hide the form
$('.hover_bkgr_fricc').fadeOut('slow');
$('#review_result').html(data);
} else alert('Sorry, unexpected error. Please try again later.');
}
});
});
And the php block
$link = mysqli_connect($HOST, $USER, $PWD, $DB_NAME);
$check_num = $_POST['check_num'];
$reason = mysqli_real_escape_string($link, $_POST['reason']);
$insert = mysqli_query($link, "INSERT INTO `vloer_paylink_reason` (`id`, `check_number`, `reason`) VALUES (DEFAULT, '$check_num', '$reason')");
$update = mysqli_query($link, "UPDATE vloer_paylink SET reviewed = 1 WHERE check_number ='$check_num'");
$get_check_data = mysqli_query($link, "SELECT reviewed FROM vloer_paylink WHERE check_number = '$check_num'");
$check_data = mysqli_fetch_array($get_check_data);
if($check_data['reviewed']==1){
echo "Reviewed done";
}
else {
echo "Not Reviewed done";
}
Data is inserting and updating to the database but after that not returning to html update. Its returning false (Sorry, unexpected error. Please try again later.)
Add .error : function(e){ console.log(e)} to your ajax call, to return the error.
The function will be:
$.ajax({
url: "loweslinkprocess.php",
type: "POST",
data: form_data,
success: function(data) {
if(data == "Reviewed done"){
// code goes here
}
},
error : function(e) { console.log(e)} // this will print error
});
You are sending Reviewed done or Not Reviewed done in the php code as a response. Change the javascript code like below.
$.ajax({
url: "loweslinkprocess.php",
type: "POST",
data: form_data,
success: function(response) {
//if process.php returned 1/true (send mail success)
if (response === "Reviewed done") {
//hide the form
$(".hover_bkgr_fricc").fadeOut("slow");
$("#review_result").html(response);
} else alert("Sorry, unexpected error. Please try again later.");
},
error: function(error) {
console.log(e);
} // To catch any network errors
});

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.}

JavaScript Ajax and Mail/Ticket

I'm working on some project which there is a Bootstrap modal it sending tickets but I want also send email to the same person at the same time. But I'm missing something I guess.
<div class="modal-body">
<form onsubmit="return false" id="new_message_form" autocomplete="off">
<input type="hidden" name="form_type" value="new_message">
<input type="hidden" name="ticket_id" id="ticket_id" value="<?=$id;?>">
<div class="row">
<div class="col-md-12">
<div class="form-group no-margin">
<textarea class="ckeditor" id="message_new" name="message_new"></textarea>
</div>
</div>
</div>
</div>
<div class="modal-footer">
Close
<button id="ticket_message_add" class="btn <?=$this->Configs->get_info("theme_buttons");?>">Send</button>
</form>
and it sends it to admin.js
$("#ticket_message_update").click(function (){
CKEDITOR.instances['message_update'].updateElement();
$("#UpdateMessageModal").modal("hide");
$.post(base_admin+"support", $("#update_message_form").serialize() ,function(data){
var data= JSON.parse(data);
if(data.update== 1){
$('#update_message_form')[0].reset();
location.reload();
}
});
});
So here is the thing. I have fully working ajax form with php mail system.(PHPMailer on github) When I click send button it automatically sending email to target mail addresses. But when I try to merge the ticket and email system seems not working but in logic it should be.
I'm adding function on button/input fields
onclick="AjaxFunction();"
<button id="ticket_message_add" onclick="AjaxFunction();" class="btn <?=$this->Configs->get_info("theme_buttons");?>">Send</button>
and adding my script also;
<script>
function AjaxFunction() {
var bilgi = {
userid: $('#user').val(),
ad: $('#ticket_id').val(),
soyad: $('#message_new').val()
}
$.ajax({
type: 'post',
url: 'gonder.php',
data: {query: bilgi},
success: function(result) {
}
});
}
</script>
And it needs to send the variables to "gonder.php" right? But it doesn't. Apparently I'm missing something.
Hi try to send the data like this:
function AjaxFunction() {
var bilgi = [{
userid: $('#user').val(),
ad: $('#ticket_id').val(),
soyad: $('#message_new').val()
}];
$.ajax({
type: 'post',
url: 'gonder.php',
data: {query: JSON.stringify(bilgi)},
success: function(result) {
}
});
}
then on the server side:
<?php
$data = json_decode($_POST["query"]);
// will echo the JSON.stringified - string:
echo $_POST["query"];
// will echo the json_decode'd object
var_dump($data);
//traversing the whole object and accessing properties:
foreach($data as $cityObject){
echo "userid: " . $cityObject->userid. ", ad: " . $cityObject->ad. "<br/>";
}
?>
Hope it Helps

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.}

AJAX pass data to PHP

I've been searching this forum, but I had no good result. I have this HTML:
<form method="post" id="postform">
<input type="text" name="message" id="message" placeholder='Say "Hello!"' class="mymessage" maxlength="255"/>
<br />
<input id="submit" type="button" value="Send"/>
</form>
PHP:
<?php
include ('connect.php');
session_start();
if (isset($_POST['message'])){
mysql_query("INSERT INTO messages VALUES ('','".$_POST['message']."','".$_SESSION['user']."')");
}
?>
And jQuery:
$('#submit').click(
function(){
var message = $('#message').val();
$.ajax({
url: post.php,
type:'POST',
data: {"message":message}
});
});
I need to pass #message content to PHP without refreshig the page. This is what I've made, but it's not working.
Check your syntax:
Wrap url with quotes '
Remove single quotes around 'message' variable
$.ajax({
url: 'post.php',
type: 'POST',
data: {message : message}
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
See the first example of $.ajax() at the bottom of its documentation page.
I need to pass #message content to PHP without refreshig the page.
Handle form submit instead:
$("#postform").submit(function(){
var message = $('#message').val();
$.ajax({
url: 'post.php',
type:'POST',
data: {message:message}
}).done(function( data) {
// handle response from the server in case of success.
});
return false; //to prevent submitting and refreshing the page.
});

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