将AJAX变量传递给PHP

Question

Hi wondering how to send a AJAX variable to php, I thought I had it but apparently not. In my console I get the error "Uncaught TypeError: Illegal invocation line 6"

Im taking it there is something wrong with my code straight after the alert? (NOTE where it say "jquery" is in replace of $ simply because joomla does not like $ in scripts for some reason) UPDATED, Pay attention to

Script to click and get rowID

<script language="javascript" type="text/javascript">
  jQuery(document).ready(function()
  {
     jQuery("tr.getRow").click(function()
     {
       rowID = jQuery(this).find("td.idCell");
       alert(jQuery(rowID).text());
       //Send the row ID to ajaxupdate.php
       jQuery.post("ajaxupdate.php", { submit: "update", ID_ID: rowID})
       .done( function(data) {
       var results = jQuery.parseJSON(data);
       console.log( results );
       })
       .fail( function() {
       console.log("AJAX POST failed."); 
       });
     });
  });
</script>

Load first table(the one thats being clicked)

<table border="",th,td, width="500", align="center">
<tr>
    <th>TP ID</th> <th>Permit Deny</th> <th>Level</th> <th>Session</th> <th>Information Specialist</th>
</tr>

<?php foreach ($results as $row): ?>

<tr class="getRow">
<td class="idCell"><?php echo $row->TP_ID ?></td>
<td><?php echo $row->Permit_or_Deny ?></td>   
<td><?php echo $row->Level ?></td>
<td><?php echo $row->Session ?></td>
<td><?php echo $row->Information_specialist ?></td>
</tr>
<?php endforeach ?>
<br>
</table>

Second table, the one that im trying to get to load

<?php
// In ajaxupdate.php file

if( (isset($_POST['ID_ID'])) || (isset($_POST['submit']))) //im Sure this part is wrong
{
$ID_ID =($_POST['ID_ID']); // pass JS var as a PHP var

$db = JFactory::getDbo();
$query = $db->getQuery(true);
$query
->select($db->quoteName(array('CV_ID', 'Classifier', 'Value', 'TP_ID')))
->from($db->quoteName('sessionta'))
->where($db->quoteName('TP_ID') . ' LIKE '. $db->quote('".$ID_ID."'));

$db->setQuery($query);
$results = $db->loadObjectList();


}
?>

Answer 1

3425742,

I rewrote your script and tested it with this JSfiddle . Try it out.

I see that you are using Joomla. Diving into Joomla as a novice is daunting. Within ajaxupdate.php the script is expecting to see a $_POST['submit'] variable. Either remove that requirement or add it like I did below. At the bottom of ajaxupdate.php add this line so that jQuery has something to test.

echo $results ? json_encode("succeeded") : json_encode("failed"); die();

Here is the jQuery ajax code:

//Send the row ID to ajaxupdate.php
$.post("ajaxupdate.php", { submit: "update", TP_ID: rowID})
.done( function(data) {
   var results = $.parseJSON(data);
   console.log( results );
})
.fail( function() {
   console.log("AJAX POST failed.");                
});

Edit "ajaxupdate.php" to the correct location of that file. If ajaxupdate.php is in a different directory you have to tell jQuery to look there. For example, if your $.post is in index.php in the root of your webserver and ajaxupdate is in the /js directory change "ajaxupdate.php" to "js/ajaxupdate.php".

Answer 2

you are passing rowID as an object, not as a single text-variable. You'd need

$.post("ajaxupdate.php", ({ TP_ID: rowID.attr('id') }), function( data )
...