if ( ! ISSET($img_url) )
-----Original Message-----
From: ROBERT MCPEAK [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 21, 2002 8:18 AM
To: [EMAIL PROTECTED]
Subject: [PHP] testing for blank var
if $img_url has a value, then I'd like to show the image, if it doesn't,
then I'd like to show a message. What's wrong with my code? Am I
incorrectly testing for the value? The else works fine, but not the if.
Thanks!
if (!$img_url)
{
echo "<b>No Image URL Entered"</b><br>";
}
else
{
echo "<img src=\"$img_url\">";
}
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