PHP Blocked Out Code?

[Guest]

I would thoroughly enjoy if someone solved this.  The problem is, is that I want the forum name of any created forum I make displayed by PHP.  All of my other codes work except this one...

 

<!-- forum info-->
				<?php
					#get forum infos
					$forum = fetch("SELECT fid, name FROM forums WHERE fid = ".$fid);
					if(!$forum) $forum = array();
				?>
				<table class="foruminfo" cellspacing="0">
					<tbody>
						<tr>
							<td class="foruminfoleft">
								<div class="forumdescwrap">
									<img class="forumicon" src="/images/cornerpic.jpg">
									<h3 class="forumname"><a href="?fid=<?php echo $forum[0] ;?>"><?php echo $forum[1];?></a></h3>

 

Now, that code above seems to work and show the forum name when I take out this code on another PHP file I have:

 

function fetch($query) { $r = mysql_query($query); if($f) {return mysql_fetch_row($r);}}

 

Any suggestions?

[PaulRyan]

Change this:

function fetch($query) { $r = mysql_query($query); if($f) {return mysql_fetch_row($r);}}

 

To this:

function fetch($query) { $r = mysql_query($query); if($r) {return mysql_fetch_row($r);}}

 

In the IF statement you were checking $f when you should have been checking $r

[Guest]

Oh, wow, thanks, man!  Fixed both my issues. 

 

If you are reading this again, would you know how to do this?:

 

http://trigamer.com/forums/newest.php?fid=1

 

If you go to that link, how could I fix the blue bar at the top?

 

 

But on this page it is perfect and I don't even think any of the coding is different.

 

http://trigamer.com/forums/replythread.php?tid=11