What is wrong with this?

[Scrub]

I've tried several diff ways to get this to work, but no luck with any. :ermm:

PHP Code:

if ($userinfo['usergroupid'] == 1 && 3 && 4 && 8 && 9)
{
    $show['link'] = false;
}
else
{
    $show['link'] = true;
} 


[assassingod]

PHP Code:

if (in_array($bbuserinfo['usergroupid'], 1,3,4,8,9)
{
$show['link'] = false;
}
else
{
$show['link'] = true;
} 


That'll do it

[Scrub]

Parse error: parse error, unexpected '{' in ***** on line 487

On line 487 is the { below the if in the code you posted.

[Dean C]

Steve just forgot a trailing bracket and to actualy have an array

Use this it's more efficient:

PHP Code:

if (in_array($bbuserinfo['usergroupid'], array(1,3,4,8,9)))
{
$show['link'] = false;
}
else
{
$show['link'] = true;
} 


[filburt1]

If you are using vB3, always prefer is_member_of() over a check of the usergroupid key. This will, among other things, take additional usergroups into effect (not just the primary usergroup).

[assassingod]

Originally Posted by Dean C

Steve just forgot a trailing bracket and to actualy have an array

Whoops

[Scrub]

Thanks guys!

Filburt1, where would that go? Kind of a newbie in vB3 code :speechless:

[Gary King]

Originally Posted by Scrub

Thanks guys!

Filburt1, where would that go? Kind of a newbie in vB3 code :speechless:

PHP Code:

if (is_member_of($bbuserinfo, 1, 3, 4, 8, 9));
{
    // etc.
} 


[Scrub]

Thank you!

[Scrub]

Instead of making another thread, I'll post in this one. . Anyways, I am stuck yet again and on something that I thought would've been easy to do. :x.

Anyways, I am trying to achive something like member.php does, but no luck in making it work. I've scanned countless files trying almsot everything. LOL. But I need to show a username according to the userid, hence like the member.php file. Thanks in advanced.