piehead Posted February 15, 2013 Share Posted February 15, 2013 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") , $(this).form("id") etc and nothing gets it in.. Any advice would be great MY PHP Loop while($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=\"$row[comments]\"</td>"; echo"<td>"; echo "<td class=\"delete\" align=\"center\" id=".$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"</tr>"; echo "</form>"; echo ' </table>'; } My AJAX <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { <!--$('#load').hide();--> }); $(function() { $(".approve").click(function() { var commentContainer = $(this).parent('tr:first'); var id = $(this).attr("id"); var string = 'id='+ id; var formData = $(this).attr("id") $.ajax({ type: "POST", url: "approve.php", data: $(formData).serialize(), cache: false, success: function(){ commentContainer.slideUp('slow', function() {$(this).remove();}); } }); return false; }); }); </script> Quote Link to comment Share on other sites More sharing options...
Jessica Posted February 15, 2013 Share Posted February 15, 2013 Your unique id should be a hidden input field. Quote Link to comment Share on other sites More sharing options...
piehead Posted February 15, 2013 Author Share Posted February 15, 2013 How do I pull that in my ajax? Quote Link to comment Share on other sites More sharing options...
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