So what I want to do is like this

Code:

			var rows = $('#dg').datagrid('getSelections');
			var datastr = JSON.stringify(rows);
			var posting = $.post( url, { dg: datastr, charter_type: $('#charter_type').val() } );

Only when I try and read it on the php page called I get Undefined index: dg

Code:

$dg = $_REQUEST['dg'];
$obj = json_decode($dg);
$activity_code = $obj['activity_code'];
$description = $obj['description'];
$amount = $obj['amount'];
$unit_of_measurement = $obj['unit_of_measurement'];
$quantity = $obj['quantity'];
$hours = $obj['hours'];
$charter_type = $_REQUEST['charter_type'];

Yet the charter_type came through

Thanks, does that mean I can include a form on the datagrid and pass the name of the selected list as well?

I want to have a datagrid where I go through and tick on and off items, than give this selection a name and save it to the database. How can I add a field at the topas a name, with a button to save (insert all the ticked items as rows in a table with the field name)?

Andrew

e.g.

Selection Field Name: MySelection

A          B  C Selected
Fred      0   1 Tick
Barney  1   3 Unticked
Wilma   2   1 Tick


So rows Fred and Wilma are saved with the name "MySelection" into a table