yeah, you're right though.. i had to use explode not implode
eg, if(in_array($uname['uid'], explode(' ', trim($userinfo['buddylist']))))
so i ditched the preg_split()
cheers.
----- Original Message -----
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, December 16, 2004 3:39 PM
Subject: Re: [PHP] in_array w/statement
> On Friday 17 December 2004 02:33, Sebastian wrote:
>
> > I cannot solve this problem,. sorry if this looks confusing,.
>
> It is ...
>
> > i have a form and don't want to set the variable if the in_array is
true..
> > the code works, up until i add the last !$buddy in the statement, for
some
> > reason it seems to always be true, ... something i'm doing wrong? btw, i
> > cannot add the in_array to the statement because if the $buddylist is
empty
> > it will generate errors because of the empty implode.
> >
> > $buddylist = preg_split('/( )+/', trim($userinfo['buddylist']), -1,
> > PREG_SPLIT_NO_EMPTY);
>
> OK, it looks like $userinfo['buddylist'] is a string containing buddies
> separated by some whitespace:
>
> 'buddy1 buddy2'
>
> After the above statement $buddylist becomes an array.
>
> > if($buddylist)
> > {
> > $buddy = in_array($uname['uid'], array(implode(',', $buddylist)));
> > }
>
> Here implode() returns a string containing 'buddy1,buddy2', then you stick
> that into an array() which is effectively:
>
> array('buddy1,buddy2'); // note that there is only *1* element
>
> Now unless your $uname['uid'] really is literally 'buddy1,buddy2' then
$buddy
> will be false.
>
> Hope that's enough to get you going.
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.biz
> Open Source Software Systems Integrators
> * Web Design & Hosting * Internet & Intranet Applications Development *
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