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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.}

PHP validation not working after implementing AJAX

I have code that validates the input inside the form and displays an error message in the div class="error" if condition is met. However, upon using AJAX to prevent page reload, the PHP validation no longer displays the message but still functions in the background. Does anyone why this is happening? How can I make the PHP validation work again? Thank you for help.
As you can see the validation error is supposed to show in $email_error
<form class="form_pro" action="index.php" method="post" role="form">
<input type="text" class="form" name="E-mail" placeholder="E-mail" value="<?= $email ?>">
<div class="error"><?= $email_error ?></div>
<button name="submit" type="submit" class="action_button"></button>
</form>
PHP
The error message here displays inside the "div" prior to adding AJAX code
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["E-mail"])) {
$email_error = "E-mail is required";
} else {
$email = test_input($_POST["E-mail"]);
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
$email_error = "Invalid email format";
}
}
}
jQuery
After adding this to prevent page reload on submit, the PHP error message no longer shows
$( document ).ready(function() {
$(".form_pro").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "php/interuni_process.php",
data: $(this).serialize(),
success: function(data) {
},
error: function() {
}
});
});
});

1) you need to echo the error message in server side and get the response from server and append it to that div
HTML :
htm_page.php
<form class="form_pro" action="index.php" method="post" role="form">
<input type="text" class="form" name="E-mail" placeholder="E-mail" value="<?= $email ?>">
<div class="error"></div>
<button name="submit" type="submit" class="action_button"></button>
</form>
PHP :destination.php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$email_error='';
if (empty($_POST["E-mail"])) {
$email_error = "E-mail is required";
} else {
$email = test_input($_POST["E-mail"]);
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
$email_error = "Invalid email format";
}
}
echo $email_error;
}
AJAX : html_page.php
$( document ).ready(function() {
$(".form_pro").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "destination.php",
data: $(this).serialize(),
success: function(data) {
if(data.trim() !='')
{
$('.error').html('<h1>'+data+'</h1>');
}
},
error: function() {
}
});
});
});

Simple PHP error :
<div class="error"><?= $email_error ?></div>
When you simply submit the form and according to your condition the validation fails then $email_error is set and when the page is loaded , server interprets the error and print it.
When you submit through AJAX
It checks the error validate and if fail $email_error is set but as the page s not reloaded is not interpreted . SO when you need to submit through ajax instead of setting the error echo it and it will work fine

You need to modify your PHP to return a response with either success response or the error details and then inside your Javascript "success" function you need to parse the response and if it's an error response you need to update your DOM accordingly.

Simply Ajax doesn't work with PHP validation, because the Ajax error part means that the ajax side failed not the PHP validation !!! so it won't go to the error part if the PHP validation failed, how ever you can still make the submit return false and do something like this in JS&Ajax in the success part:
success: function(data) {
if(data.error)
{
alert(data.error);
return false;
}

As you have prevented the page load, you can not see the error which were rendreing on server and display in frontend.
To achieve this you need edit your jquery code
$( document ).ready(function() {
$(".form_pro").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "php/interuni_process.php",
data: $(this).serialize(),
success: function(data) {
},
error: function(data, errorThrown)
{
$('.error').html(errorThrown);
}
});
});
});

Popuating form fields from MySQL using AJAX and Jquery

I followed a tutorial to adapt the code. Here I am trying trying to auto-populate my form fields with AJAX when an 'ID' value is provided. I am new to Jquery and can't get to work this code.
Edit 1 : While testing the code, Jquery isn't preventing the form to submit and sending the AJAX request.
HTML form
<form id="form-ajax" action="form-ajax.php">
<label>ID:</label><input type="text" name="ID" /><br />
<label>Name:</label><input type="text" name="Name" /><br />
<label>Address:</label><input type="text" name="Address" /><br />
<label>Phone:</label><input type="text" name="Phone" /><br />
<label>Email:</label><input type="email" name="Email" /><br />
<input type="submit" value="fill from db" />
</form>
I tried changing Jquery code but still I couldn't get it to work. I think Jquery is creating a problem here. But I am unable to find the error or buggy code. Please it would be be very helpful if you put me in right direction.
Edit 2 : I tried using
return false;
instead of
event.preventDefault();
to prevent the form from submitting but still it isn't working. Any idea what I am doing wrong here ?
Jquery
jQuery(function($) {
// hook the submit action on the form
$("#form-ajax").submit(function(event) {
// stop the form submitting
event.preventDefault();
// grab the ID and send AJAX request if not (empty / only whitespace)
var IDval = this.elements.ID.value;
if (/\S/.test(IDval)) {
// using the ajax() method directly
$.ajax({
type : "GET",
url : ajax.php,
cache : false,
dataType : "json",
data : { ID : IDval },
success : process_response,
error: function(xhr) { alert("AJAX request failed: " + xhr.status); }
});
}
else {
alert("No ID supplied");
}
};
function process_response(response) {
var frm = $("#form-ajax");
var i;
console.dir(response); // for debug
for (i in response) {
frm.find('[name="' + i + '"]').val(response[i]);
}
}
});
Ajax.php
if (isset($_GET['action'])) {
if ($_GET['action'] == 'fetch') {
// tell the browser what's coming
header('Content-type: application/json');
// open database connection
$db = new PDO('mysql:dbname=test;host:localhost;', 'xyz', 'xyz');
// use prepared statements!
$query = $db->prepare('select * from form_ajax where ID = ?');
$query->execute(array($_GET['ID']));
$row = $query->fetch(PDO::FETCH_OBJ);
// send the data encoded as JSON
echo json_encode($row);
exit;
}
}

I don't see where you're parsing your json response into a javascript object (hash). This jQuery method should help. It also looks like you're not posting your form using jquery, but rather trying to make a get request. To properly submit the form using jquery, use something like this:
$.post( "form-ajax.php", $( "#form-ajax" ).serialize() );
Also, have you tried adding id attributes to your form elements?
<input type="text" id="name" name="name"/>
It would be easier to later reach them with
var element = $('#'+element_id);
If this is not a solution, can you post the json that is coming back from your request?

Replace the submit input with button:
<button type="button" id="submit">
Note the type="button".
It's mandatory to prevent form submition
Javascript:
$(document).ready(function() {
$("#submit").on("click", function(e) {
$.ajax({type:"get",
url: "ajax.php",
data: $("#form-ajax").serialize(),
dataType: "json",
success: process_response,
error: function(xhr) { alert("AJAX request failed: " + xhr.status); }
});
});
});

Pass view as data to js in symfony

I am tryng to implement a search function in my index page using java script. I hav got a form to enter the name and when apply serach, the index page will get updated and load the new index page with the search results
Here is the form in my index page
<div id="content">
<form id="myForm" action="{{path('index_search')}}" method="POST" >
Write your name here:
<input type="text" name="name" id="name_id" value="" /><br />
<input type="submit" value="Send" />
</form>
</div>
<div id="output">#current index</div>
Here is the action exexcuted
public function searchAction()
{
$request = $this->get('request');
$name=$request->request->get('formName');
$em = $this->getDoctrine()->getManager();
$entities = $em->getRepository('SystemVmsBundle:VisitorsDetails')->findOneByfirstname($name);
$view = $this->render('SystemVmsBundle:VisitorsDetails:index.html.twig',array(
'entities' => $entities,
));
$return=array("responseCode"=>200, "view"=>$view);
$return=json_encode($return);//jscon encode the array
return new Response($return,200,array('Content-Type'=>'application/json'));
}
Here is the js
$(document).ready(function() {
//listen for the form beeing submitted
$("#myForm").submit(function(){
//get the url for the form
var url=$("#myForm").attr("action");
$.post(url,{
formName:$("#name_id").val(),
other:"attributes"
},function(data){
//the response is in the data variable
if(data.responseCode==200 ){
$('#output').html(data.view);
$('#output').css("color","red");
}
else{
alert("An unexpeded error occured.");
}
});
return false;
});
});
However my js is working,but can not pass data as view to the js.How to pass the view 'index.html.twig' to the js?
When inspects with firebug,i got like
{"responseCode":200,"view":{"headers":{}}}
Any ideas?Thanks in advance!

Try to specify the dataType on your $.post function, like this:
$.post(url, {formName: $("#name_id").val(), other: "attributes"},
function(data) {
if(data.responseCode == 200){
$('#output')
.html(data.view)
.css("color","red");
} else {
alert("An unexpeded error occured.");
}
}, 'json');
But If you only need the html, not other data inside the json, you should change the logic, render a normal template and use dataType 'html'. goes like this:
// ...Controller.php
$view = $this->render('SystemVmsBundle:VisitorsDetails:index.html.twig',array(
'entities' => $entities,
));
return new Response($view);
// index.html.twig
$.ajax({
url: url,
dataType: 'html',
type: 'POST',
data: {formName: $("#name_id").val(), other:"attributes"},
success: function (data) {
$('#output').html(data).css("color","red");
}
});