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Hi,

seems that I dont understand -> frustrated.
This is my main code:

[code]

<script src="codebase/dhtmlxcommon.js"></script>
<script src="codebase/dhtmlxform.js"></script>
<script src="codebase/dhtmlxdataprocessor.js"></script>
<script src="codebase/connector/connector.js"></script>
[/code] and this is the PHP [code]<?php $_POST['drawingNumber'] = 1695484;

// common settings
$hostname_dblink = “localhost”;
$username_dblink = “root”;
$password_dblink = “”;
$database_dblink = “ppap”;

// Connect database
$dblink = mysql_connect($hostname_dblink, $username_dblink, $password_dblink);
mysql_select_db($database_dblink, $dblink);

require_once(’…/…/codebase/connector/form_connector.php’);

$sql = ‘SELECT part.id AS partid, part_number, drawing_number, part_name, part_level.level AS partlevel, engineering_level.level AS engineeringlevel FROM part
LEFT JOIN part_relation ON part_relation.id_part = part.id
LEFT JOIN part_level ON part_level.id_part = part.id
LEFT JOIN engineering_level ON engineering_level.id_drawing = part.id
WHERE part_relation.id_program = 1
AND drawing_number = ‘’.$_POST[‘drawingNumber’].’’’;

$form = new FormConnector($dblink);
$form->render_sql($sql, ‘partid’, ‘part_number’);
?>[/code]
It always results in a fatal error
Uncaught exception ‘Exception’ with message ‘ID parameter is missed’ in J:\xampp\htdocs\Webserver\ppap_db\codebase\connector\form_connector.php:48

Where is my fault?

Try to use

$form->render_sql($sql, ‘part.id(partid)’, ‘part_number’);

instead of

$form->render_sql($sql, ‘partid’, ‘part_number’);

Here is the documentation: docs.dhtmlx.com/doku.php?id=dhtm … ta#aliases