vb.org Archive - $ismod function - I want something similar?

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- vB3 General Discussions (https://vborg.vbsupport.ru/forumdisplay.php?f=111)

- - $ismod function - I want something similar? (https://vborg.vbsupport.ru/showthread.php?t=48776)

N9ne	02-10-2003 08:22 PM

$ismod function - I want something similar?

I have 5 separate usergroups for banned users, now what I want to do is instead of having to write out long if statements, just say $isbanned...how can I do this? I looked in functions.php but didn't find much to help me.

Xenon

02-11-2003 11:10 AM

create a function in functions.php

PHP Code:


		
			
function isbanned($userid=-1) {

      if ($userid==-1) {

         $usergroupid=$bbuserinfo[usergroupid];

} else {

    $user=$DB_site->query_first("SELECT usergroupid FROM user where userid=$userid");

  $usergroupid=$user[usergroupid];

}

return in_array($usergroupid, array(8,9,12,13));

}

replace the 8,9,12,13 with the usergroupids of your banned usergrous..

Dean C

02-11-2003 03:21 PM

Geez your too clever with that in_array function xenon ;)

- miSt

N9ne	02-11-2003 03:29 PM

And that will make $isbanned work globally right? It's just that I don't see any 'global' in there :p

Xenon

02-11-2003 04:03 PM

it's a function, so it would always work globaly.

it's like ismoderator.

so your $ in front is incorrect!

also, please change your sig to fit with the rules!

N9ne	02-11-2003 04:30 PM

So if it's a function, I can't use it as a variable? I'd have to use isbanned() wouldn't I? Is there a way I could use it like you can use $ismod...I want to use $isbanned instead.

edited my sig, completely forgot about the rules :o.

Xenon

02-11-2003 04:39 PM

thx

i don't see why you want to use it as a variable, it has no sense normally...
if use use a var or the function it's equal..

but you can do this:

Code:

$isbanned=isbanned();

that's the way it's done with $ismod, it's just because of a caching feature...

N9ne	02-11-2003 05:13 PM

In an if statement, which of the following can I do then?

PHP Code:


		
			
if ($isbanned=isbanned();) {

echo("blah");

}



if($isbanned) {

echo("blah");

}



if(isbanned()) {

echo("blah");

}

N9ne	02-11-2003 05:14 PM

Grr, it keeps parsing a smilie, the first one eshould be:

if ($isbanned=isbanned()__)__{
echo("Blah");
}

Just remove the underscores.

Xenon

02-11-2003 05:21 PM

use this:

PHP Code:


		
			
if(isbanned()) {

echo("blah");

}

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