Wednesday 22 September 2004 6:34:37 am
Hi all, I have a PHP script on another server with the following content:
<?php
/* grabs the POST variables and puts them into variables that we can use */
$relatieCode=$_POST['relatieCode'];
echo("code:".$relatieCode.""); ?> I have a clickable image in ezpublish, which should open the PHP script and pass the user-id of the logged in user as 'relatieCode'. Unfortunately I don't know how to pass the user-id. In my eZ template I've got the following code:
{let user=fetch( 'user', 'current_user' )}
{$user.contentobject.name}
{$user.email}
{$user.login}
{/let}
<form method="POST" action="http://mysite.com/myform.php" name="form">
<input type="hidden" name="relatieCode" id="relatieCode" value={$user.login}>
<input type="image" name="formbutton" onClick="return validate(form)" SRC={"buttons/bt_contact.png"|ezimage} class=button alt="My alt text"> </form> In my eZ site I can see the right user id (user.login), so the fetch is OK. But my PHP script doesn't show the user ID. When I change: <input type="hidden" name="relatieCode" id="relatieCode" value={$user.login}> into <input type="hidden" name="relatieCode" id="relatieCode" value="MYVALUE"> then my PHP script does show 'MYVALUE'. Can anyone please tell me what I'm doing wrong? Thanks, Maarten UPDATE; Sorry, my bad: I closed the let too early...
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