JQUERY AJAX loading content w/ PHP loop

Question

I have PHP loop that loops through a bunch of user IDs.

<?php while ($row = mysql_fetch_assoc($result)) : ?>
<?php $id = $row['id']; 

 <td width="90px" class="resultsDisplay"><a href=userdetail.php?id=<?php echo $id ?>> <?php echo $row['Username']; ?></a></td>

I would like to replace the HTML link there with an AJAX call to load that page in a particular DIV. The problem I'm having with the code below is that after a link is clicked if continues passing the rest of the user IDs for the loop. So, I click on a link, and it begins cycling through all of the user IDs in the database, loading about 70 pages in a row.

jQuery(document).ready(function(){
$(".resultsDisplay").click(function() {
  $.ajax({
   url: "userdetail.php?id=<?php echo $id ?>",
   success: function(msg){
     $("#results").html(msg);
   }
 });
});

Can you provide a little more context. How is this being written to the page? I'm guessing that you are looping in your php and outputting this and that would not work. — Hemlock
– Hemlock, Commented Jan 1, 2011 at 17:30

ifaour · Accepted Answer · 2011-01-01 17:27:30Z

3

I really didn't get your question but I suppose you are outputting more than one ID? so you have more than one TD..etc?
The way you should do this (what ever you are doing) is like this:

jQuery(document).ready(function(){
$(".resultsDisplay a").click(function() {
  $.ajax({
   url: $(this).attr("href"),
   success: function(msg){
     $("#results").html(msg);
   }
 });
 return false;
});

Also you are missing the quotes for your href attribute.

answered Jan 1, 2011 at 17:27

ifaour

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7 Comments

user547794 Over a year ago

Yes, that's right - Each <td> has it's own user ID which is gathered by looping through the database. With your code do I change anything with my <td> code (besides adding quotes to href?) I still need to pass in the userID somehow.

ifaour Over a year ago

no the idea is to find a way to pass the user ID to your ajax call which can be achieved by my code above.

ifaour Over a year ago

please note that the jQuery(document).ready is not closed..assuming you have more code under the click event..!

user547794 Over a year ago

Thanks, that worked beautifully! I also did forget to paste the end of the document.ready, good eye.

user547794 Over a year ago

I have found that the return false; in this code is breaking an upload script I have further up on the page. Any idea why that's happening?

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jeroen · Accepted Answer · 2011-01-01 17:27:43Z

0

You seem to be mixing the initial page-load with an ajax request. How are you generating your document.ready function?

What seems definitely wrong is your use of php in the document.ready function; for the ajax call to work, the ID you are sending to userdetail.php needs to be grabbed from the specific link / div that was clicked, not a static ID that was generated at page-load.

answered Jan 1, 2011 at 17:27

jeroen

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