syntax problems with checking for constants in an array?

[saf-t scissors]

if ($user[membergroupids] == USERGROUP_A) always returns true, but
if ($user[membergroupids] == 'USERGROUP_A') and
if ($user[membergroupids] == " . USERGROUP_A . ") always returns false

If I just go
if ($user[membergroupids] == '2'), it works as it should.

Similarly, later on,

if ($groupid != USERGROUP_A) branches properly.

WTF am I doing wrong?

[Code Monkey]

Have you tried $user['membergroupids'] ?

[saf-t scissors]

That doesn't seem to affect it.

Thanks, though.

[Dream]

$user[membergroupids] is a list of user groups separated by commas

you need to explode(',') it and get an array with a list of user group ids to work with

not sure if constants get parsed inside quotes, but this

if ($user[membergroupids] == " . USERGROUP_A . ")

should work I think O_o

[saf-t scissors]

Correct. I was actually trying to check for one (and only one) membergroup. If there were more than one, it branched and did something else.

Originally Posted by Dream

if ($user[membergroupids] == " . USERGROUP_A . ")

That always returns as false. I'm not sure why.

[Dismounted]

if ($user[membergroupids] == " . USERGROUP_A . ")

There is no need to concencate the constant....You're not including any other bits of data.

[Dream]

still he's curious, as am I

[Adrian Schneider]

PHP Code:

if (is_member_of($user, USERGROUPA)) 


Which will check primary & secondary usergroups for USERGROUP_A.

[Dismounted]

Originally Posted by Dream

still he's curious, as am I

Are you referring to my post? If you are, you're not starting the string, so it'd error out.

if ($user[membergroupids] == "" . USERGROUP_A . "")

That would work.

[Adrian Schneider]

Why even add the empty strings around it though?

PHP Code:

if ($user['membergroupids'] == USERGROUP_A) 


is correct, but membergroupids is a comma separated list, so you need to either explode it, or use is_member_of().