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

[rija]

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";



wrote in message news:20021104022815.4043.qmail@;pb1.pair.com...
> 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?
> 
>  if(!isset($HTTP_SESSION_VARS['svUserAccess'])){
>   echo " style='cursor:hand' onMouseover='this.style.backgroundColor='#C0E0FF''
> onMouseout='this.style.backgroundColor='''
> onClick='window.location.href='login.php''> href='login.php'>Login";
> } else {
>   echo " style='cursor:hand' onMouseover='this.style.backgroundColor='#C0E0FF''
> onMouseout='this.style.backgroundColor='''
> onClick='window.location.href='logout.php''> href='login.php'>Logout";
>   }
> ?>
> 
> Thanks
> 
> 
> 
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> 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

[vernon]

That's the thing, everything works but the error message keeps coming up?

Runtime Error: Syntax Error



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

[John W. Holmes]

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 an error, which
I
> don't understand because the code actually works (other than the
error).
> Any
> ideas?
> 
>  if(!isset($HTTP_SESSION_VARS['svUserAccess'])){
>   echo " style='cursor:hand' onMouseover='this.style.backgroundColor='#C0E0FF''
> onMouseout='this.style.backgroundColor='''
> onClick='window.location.href='login.php''> href='login.php'>Login";
> } else {
>   echo " style='cursor:hand' onMouseover='this.style.backgroundColor='#C0E0FF''
> onMouseout='this.style.backgroundColor='''
> onClick='window.location.href='logout.php''> href='login.php'>Logout";
>   }
> ?>
> 
> Thanks
> 
> 
> 
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php




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[PHP] Error in code - Seems simple enough

[vernon]

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



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