aslo you could change it to something else like this
$_GET['id'] = $id;
if (!$id) {
// whatever you want to happen put here
}From: "Daniel Clark" <[EMAIL PROTECTED]>
Reply-To: "Daniel Clark" <[EMAIL PROTECTED]>
To: "Brian Dunning" <[EMAIL PROTECTED]>,"[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
Subject: Re: [PHP] How to check for a $_GET without throwing a Notice?
Date: Wed, 26 May 2004 10:59:20 -0700
if (isset( $_GET['id']))
>>How do I check for the presence of an optional $_GET param without >>throwing a "Notice: Undefined index" when the param is not present? >> >>Tried all three of these, they all produce the Notice when the param is >>not passed: >> >>if ($_GET['id']) >>if ($_GET['id'] != "") >>if (isset $_GET['id']) >> >>-- >>PHP General Mailing List (http://www.php.net/) >>To unsubscribe, visit: http://www.php.net/unsub.php >>
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