Re: [PHP] Error in code - Seems simple enough

This is rather problem about javascript than PHP - But I think that you need to use double quotes (escaped) to distinguish query and variables in your javascript query: Try out this, It work fine for me: if(!isset($HTTP_SESSION_VARS['svUserAccess'])){ echo "Login"; } else { echo "Logout"; w

Re: [PHP] Error in code - Seems simple enough

That's the thing, everything works but the error message keeps coming up? Runtime Error: Syntax Error -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

RE: [PHP] Error in code - Seems simple enough

And the error is?? > -Original Message- > From: vernon [mailto:vernon@;comp-wiz.com] > Sent: Sunday, November 03, 2002 9:28 PM > To: [EMAIL PROTECTED] > Subject: [PHP] Error in code - Seems simple enough > > I'm having trouble with the following code dispalying

[PHP] Error in code - Seems simple enough

I'm having trouble with the following code dispalying an error, which I don't understand because the code actually works (other than the error). Any ideas? Login"; } else { echo "Logout"; } ?> Thanks -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.n