PHP/MySQL Sessions

**Tracer Round** · November 12th, 2007, 8:13 PM

I had this code

Code:

      	if( $kernel->session->vars['session_group_id'] <> 1 )
      	{
      		$kernel->ld['lang_f_control_panel'] = "";
      	}
			else
			{
				$kernel->tp->call( "admin_copyright_notice" );
				
				$kernel->tp->call( "admin_footer" );
			}

I want to include session_group_id 23 so with the help of DewKnight I now have

Code:

      		if(($kernel->session->vars['session_group_id'] != 1 )|| ($kernel->session->vars['session_group_id'] != 23))
      		{
      			$kernel->ld['lang_f_control_panel'] = "";
			}		
			else
			{
				$kernel->tp->call( "admin_copyright_notice" );
				
				$kernel->tp->call( "admin_footer" );
			}

Now I get

Quote:

Parse error: syntax error, unexpected ';', expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or '}' in /<path>/include/function_class_page.php on line 1029

Line 1029 is

Quote:

?>

Any suggestions?

**MarkRH** · November 12th, 2007, 8:41 PM

Darned if I can find anything wrong with that code. Right number of parenthesis and curly braces and no misplaced semi-colons. Might make sure the file was transferred back to the server in ASCII and not binary or possible copy that section of code to Notepad, delete in the original and paste it back in case there's some weird hidden whitespace character.

That or something was mistakenly modified or changed somewhere else.

Or, I'm an idiot and don't see the problem...

**Tracer Round** · November 12th, 2007, 8:57 PM

Quote:

Originally Posted by MarkRH

Darned if I can find anything wrong with that code. Right number of parenthesis and curly braces and no misplaced semi-colons. Might make sure the file was transferred back to the server in ASCII and not binary or possible copy that section of code to Notepad, delete in the original and paste it back in case there's some weird hidden whitespace character.

That or something was mistakenly modified or changed somewhere else.

Or, I'm an idiot and don't see the problem...

Ok, I took your advice and reverted back to the original code and entered the changes without using copy/paste. I am not getting an error anymore.

Bad thing is it didnt do what I want. Neither session_group_id's can see the footer info now. Before group id 1 could.

wgm · November 12th, 2007, 9:23 PM

Seems out of place, and I am no expert.

Code:

if(($kernel->session->vars['session_group_id'] != 1 )|| ($kernel->session->vars['session_group_id'] != 23))

Possibly

Code:

if(($kernel->session->vars['session_group_id'] != 1) || ($kernel->session->vars['session_group_id'] != 23))

or

Code:

if(( $kernel->session->vars['session_group_id'] != 1 ) || ( $kernel->session->vars['session_group_id'] != 23 ))

**MarkRH** · November 13th, 2007, 12:19 AM

Yeah, I was looking at the logic of that and didn't think it would work how you wanted, change it to:

PHP Code:


			
    if (($kernel->session->vars['session_group_id'] != 1 ) && ($kernel->session->vars['session_group_id'] != 23))

**DewKnight** · November 13th, 2007, 12:38 AM

but he wants it if it's either 1 or 23... since it can't be both 1 and 23.

I couldn't find anything wrong, but I am sure there has to be something!

wgm · November 13th, 2007, 12:55 AM

I'll just leave this to the experts (or gurus) to solve, mine was a guess only.

**MarkRH** · November 13th, 2007, 12:57 AM

Correct, so if it's 1, the first variation will result in true and not show the footer.

if true or true = true (id = 5)
if true or false = true (id = 23)
if false or true = true (id = 1)
if false or false = false (an impossible case since the value can't be 1 and 23 at the same time)

Using and:

if true and true = true (id = 5)
if true and false = false (id = 23)
if false and true = false (id = 1)
if false and false = false (impossible case, id can't be both 1 and 23)

**Tracer Round** · November 13th, 2007, 9:56 PM

Quote:

Originally Posted by MarkRH

Yeah, I was looking at the logic of that and didn't think it would work how you wanted, change it to:

PHP Code:


			
    if (($kernel->session->vars['session_group_id'] != 1 ) && ($kernel->session->vars['session_group_id'] != 23))

That worked Mark! Thanks

But I havent the slightest clue why AND works and OR doesnt.

November 12th, 2007, 8:41 PM	#2 (permalink)
MarkRH Race Surpass Super #1 Joined in Jul 2006 Lives in Oklahoma City, OK Hosted on sh102 1,218 posts Gave thanks: 18 Thanked 86 times	Darned if I can find anything wrong with that code. Right number of parenthesis and curly braces and no misplaced semi-colons. Might make sure the file was transferred back to the server in ASCII and not binary or possible copy that section of code to Notepad, delete in the original and paste it back in case there's some weird hidden whitespace character. That or something was mistakenly modified or changed somewhere else. Or, I'm an idiot and don't see the problem... __________________ SH102 : Mark Headrick - Blog - Gallery Pelicar \| Company of Valor \| Mysstie Wolfestar

November 13th, 2007, 12:19 AM	#5 (permalink)
MarkRH Race Surpass Super #1 Joined in Jul 2006 Lives in Oklahoma City, OK Hosted on sh102 1,218 posts Gave thanks: 18 Thanked 86 times	Yeah, I was looking at the logic of that and didn't think it would work how you wanted, change it to: PHP Code: `if (($kernel->session->vars['session_group_id'] != 1 ) && ($kernel->session->vars['session_group_id'] != 23))` __________________ SH102 : Mark Headrick - Blog - Gallery Pelicar \| Company of Valor \| Mysstie Wolfestar

November 13th, 2007, 12:38 AM	#6 (permalink)
DewKnight Skittles Super #1 Joined in Aug 2004 Lives in a space ship Hosted on dedi 6,665 posts Gave thanks: 96 Thanked 190 times	but he wants it if it's either 1 or 23... since it can't be both 1 and 23. I couldn't find anything wrong, but I am sure there has to be something! __________________ Mountain Dew Knight People should not be afraid of their governments. Governments should be afraid of their people.

November 13th, 2007, 12:57 AM	#8 (permalink)
MarkRH Race Surpass Super #1 Joined in Jul 2006 Lives in Oklahoma City, OK Hosted on sh102 1,218 posts Gave thanks: 18 Thanked 86 times	Correct, so if it's 1, the first variation will result in true and not show the footer. if true or true = true (id = 5) if true or false = true (id = 23) if false or true = true (id = 1) if false or false = false (an impossible case since the value can't be 1 and 23 at the same time) Using and: if true and true = true (id = 5) if true and false = false (id = 23) if false and true = false (id = 1) if false and false = false (impossible case, id can't be both 1 and 23) __________________ SH102 : Mark Headrick - Blog - Gallery Pelicar \| Company of Valor \| Mysstie Wolfestar

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November 13th, 2007, 12:55 AM	#7 (permalink)
wgm Registered User Super #1 Joined in Mar 2005 Lives in Washougal, WA 1,302 posts Gave thanks: 23 Thanked 39 times	I'll just leave this to the experts (or gurus) to solve, mine was a guess only.

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