Brian Dunning wrote:
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
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
Hello Brian,
Wednesday, May 26, 2004, 4:01:30 PM, you wrote:
BD How do I check for the presence of an optional $_GET param without
BD throwing a Notice: Undefined index when the param is not present?
BD Tried all three of these, they all produce the Notice when the param is
BD not passed:
BD
From: Brian Dunning [EMAIL PROTECTED]
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
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
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'])
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]
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