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Help with getting ID from a form in While loop to my AJAX

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#1 piehead

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Posted 15 February 2013 - 01:27 AM

I have been stuck on this for 3 days... I would love the experts to shoot me some words of advice, appreciate it.

I have a form I create from a while loop in php. I can submit it if I name the form and place that in the AJAX but it just uploads the last record in the table not the one I actually click. I know I need a unique ID from the form, which I have but I can get it in the AJAX ;( I have used $(this).attr("id") , $(this).form("id") etc and nothing gets it in.. Any advice would be great


while($row = mysql_fetch_array($pendingresult))
 $id = "myForm".$row['reg_id'];
 echo '<table width="100%" border="0" cellspacing="0" cellpadding="0" >';
  print "<form id=\"$id\" name=\"CDs\" method=\"post\" >";
    echo '<tr class="commentContainer" style="color:#FFF">';
    echo"<td><input type=\"text\" name=\"team_name\" value=\"$row[team_name]\"</td>";

    echo"<td><input type=\"text\" name=\"reg_id\" value=\"$id\"</td>";
    echo"<td><input type=\"text\" name=\"team_level\" value=\"$row[team_level]\"</td>";
    echo"<td><input type=\"text\" name=\"notes\" value=\"$row[comments]\"</td>";

echo "<td class=\"delete\" align=\"center\" id=".$row['reg_id']." width=\"10\"><a href=\"#\" id=\"$row[reg_id]\"><img src=\"admin/images/delete.png\" border=\"0\" ></a></td>";
    echo "<td class=\"approve\" align=\"center\" id=".$id." width=\"10\"><a href=\"#\" ><img src=\"admin/images/approve.png\" border=\"0\" ></a></td>";

echo "</td>";

  echo "</form>";
  echo ' </table>';

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script>

<script type="text/javascript">
$(document).ready(function() {

$(function() {
$(".approve").click(function() {
var commentContainer = $(this).parent('tr:first');
var id = $(this).attr("id");
var string = 'id='+ id;
var formData = $(this).attr("id")

   type: "POST",
   url: "approve.php",
   data: $(formData).serialize(),
   cache: false,
   success: function(){
commentContainer.slideUp('slow', function() {$(this).remove();});


return false;

#2 Jessica

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Posted 15 February 2013 - 01:31 AM

Your unique id should be a hidden input field.
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#3 piehead

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Posted 15 February 2013 - 01:34 AM

How do I pull that in my ajax?

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