Driving Me Crazy (MYQL Help Please)

[noonespecial]

Problem with this is that I need to be able to specify the userids ...

IE: there are 4 matches in LINKID between userid-A and userid-B

Where would I specify those in this example?

--------------- Added [DATE]1208335849[/DATE] at [TIME]1208335849[/TIME] ---------------

   $sql="SELECT linkid, userid FROM links_favorites WHERE userid=1083";
   $result = mysql_query($sql);  
   $number1 = mysql_fetch_array($result);
   
   $sql2="SELECT linkid, userid FROM links_favorites WHERE userid=1";
   $result2 = mysql_query($sql2);  
   $number2 = mysql_fetch_array($result2);
   
   $result3 = array_intersect($number1, $number2);
   
   $joined = count($result3);
   
   echo $joined;

Also doesn't work.

--------------- Added [DATE]1208335899[/DATE] at [TIME]1208335899[/TIME] ---------------

     $sql="SELECT linkid FROM links_favorites WHERE userid=1083";
   $result = mysql_query($sql);  
   $number1 = mysql_fetch_array($result);
   
   $myids=implode(",", $number1);
   
   $sql2="SELECT COUNT(linkid) as count FROM links_favorites WHERE userid=1 AND linkid IN (".$myids.")";
   $result2 = mysql_query($sql2);  
   $number2 = mysql_fetch_array($result2);

Also doesn't work.

--------------- Added [DATE]1208337252[/DATE] at [TIME]1208337252[/TIME] ---------------

$fandom = $vbulletin->db->query_read("SELECT *
FROM links_favorites AS links_favorites
LEFT JOIN links_favorites AS favs2 ON (links_favorites.linkid=favs2.linkid)
WHERE links_favorites.userid=".$vbulletin->userinfo['userid']."
AND favs2.userid=".$friend[userid]."
");

$joined= $db->num_rows($fandom);

I think this works ..... ugly.