Can't get response after click on a link using jquery/ajax - php

Question

I've a link in my php page. Which I want to send a id request to server. But it doesn't response correctly. Actually, I've got two pages.
a. users.php
b. user_update.php
So I'm trying a send a request to the server using jquery/ajax by user id from users.php page to user_update.php for update the details and get back the successful result to user.php page.
The code is look like this:
<div id="edit_user_details">Edit Details</div>
<script>
$('#edit_user_details').click(function() {
id = <?php echo $uid; ?>
$.ajax({
url: edit_user_details.php,
type: post,
data: id;
success: function(response) {
$('#edit_user_result').html(response);
}
});
});
</script>
But it's not working properly. is there anything I'm missing because of I'm new learning of Jquery/Ajax. :)
Update:
I'm requesting this id in php while loop. It's look like this..
while($search_result = mysql_fetch_array($getUser)){
$uid = (int) $search_result['user_id'];
<div id="edit_user_details">Edit Details</div>
}
<script>
$('#edit_user_details').click(function() {
id = <?php echo $uid; ?>;
$.ajax({
url: 'edit_user_details.php',
type: 'post',
data: {'id' : id},
success: function(response) {
$('#edit_user_result').html(response);
}
});
});
</script>
Is it wrong ? What I'm trying is... I want to update user details base on their id. So what need to do for this ?

Answer 1

you need to create a unique id in loop or create a function and calll it onclick like
<?php while($search_result = mysql_fetch_array($getUser)){
$uid = (int) $search_result['user_id']; ?>
<div id="edit_user_details">Edit Details</div>
<?php }?>
<script>
function myfunc(id) {
id = id;
$.ajax({
url: 'db_search.php',
type: 'post',
data: {'id' : id},
success: function(response) {
$('#edit_user_result').html(response);
}
});
}
</script>

Answer 2

you should end your id initialization with a semi-colon;
id = <?php echo $uid; ?>;
Check the correct way of using ajax in this example: Demo ajax using JQuery

Answer 3

Make changes as shown below and try again..
1: Add 'post' in type.
2: Add ''edit_user_details.php' in url
3: Add semicolon at the end here
<?php echo $uid; ?>;
4: take your id as shown in below code
Updated code :-
<script>
$('#edit_user_details').click(function() {
id = <?php echo $uid; ?>;
$.ajax({
url: 'edit_user_details.php',
type: 'post',
data: {'id' : id},
success: function(response) {
$('#edit_user_result').html(response);
}
});
});
</script>
For more information :-
https://api.jquery.com/jQuery.ajax/#entry-examples

Answer 4

i think you are not declaring var id before
try
<div id="edit_user_details">Edit Details</div>
<script>
$('#edit_user_details').click(function() {
var id = <?php echo $uid; ?>;
$.ajax({
url: edit_user_details.php,
type: post,
data: [id:id],
success: function(response) {
$('#edit_user_result').html(response);
}
});
});
</script>

Answer 5

Check your url inside $.ajax. If you are sending your data to user_update.php then your code should be as follows :
<script>
$('#edit_user_details').click(function() {
id = <?php echo $uid; ?>
$.ajax({
url: user_update.php,
type: post,
data: id,
success: function(response) {
$('#edit_user_result').html(response);
}
});
});
</script>
Just change your url to user_update.php and put a comma(,) after data:id in the place of semicolon(;).

Answer 6

This code will works fine for what you need and with less lines, assuming that the PHP var $uid will be processed on loading the page:
<div id="edit_user_details">Edit Details</div>
<script>
$('#edit_user_details').click(function() {
$.get('edit_user_details.php', 'id=<?php echo $uid; ?>', function(response){
$('#edit_user_result').html(response);
});
});
</script>