简单的sql多表查询

Question

I am programming a chat room on my site but I am really new to php. I want users to be able to chat with the users that played in the same teams of a game (knowing that users can have participated together to differents team) and who work in the same area.

Assume there are three tables : the account user's table, the area's t able, games'table

I have a function that returns my query that looks like

function myfunction($userid){

$games_user=mysql_query('select theme from games where games.userid="'.$userid.'"');  
$games_theme = mysql_fetch_array($games_user);

$sql = ("select  userid, username, area.userid 

    from account 
        left join area
            on account.userid = area.userid    

        left join games
            on account.userid = games.userid

    where account.userid <> '".mysql_real_escape_string($userid)."' and '".(in_array(games.theme,$games_theme))."' and area.userid=1 
 );
 return $sql;
}

Reformatted:

$sql = "
SELECT userid, username, area.userid 
FROM account 
LEFT JOIN area ON account.userid = area.userid    
LEFT JOIN games ON account.userid = games.userid
WHERE account.userid <> '".mysql_real_escape_string($userid)."'
  AND '".(in_array(games.theme,$games_theme))."'
  AND area.userid = 1 
";

But it really doesn't work, I think I have syntax problems. I don't really understand how in_array is indexed, and I don't know how to do in a simpler way that query

Can anybody help ?

Answer 1

I'm still not entirely sure what you are doing, but I think this is what you want; you can do this in a single query:

<?php
function myfunction($userid){
    $id = mysql_real_escape_string($userid);
    $sql = "SELECT  userid, username, area.userid 
        FROM account 
            LEFT JOIN area
                ON account.userid = area.userid    
            LEFT JOIN games
                ON account.userid = games.userid
        WHERE account.userid<>'$id' AND area.userid=1
            AND games.theme IN (SELECT theme FROM games WHERE games.userid='$id')
    ";
    return $sql;
}
?>