在另一个查询的WHERE条件中使用查询中的数据

I have database consist of different tables, the table I care about are, user and module, after the user login the second query must return the max(id) from the module user. What I am trying to do is: I want to get the max(id) of the same user who is logged in, the user are stored in module table in create_user column, is possible to do that and how should the where condition be?

The table structure:
user table: (id int pri.key, username varchar, password varchar).
module table: (id int pri.key, name varchar, creat_time date, create_user int)

When inserting new module the user id is stored in create_user.

The code:

<?php
$json["status"] = "running";
$details[] = "started get_tables ";
include_once('confi.php');
$request_body = file_get_contents('php://input');
$statement = "INSERT INTO tbl_archive (content) VALUES ('$request_body' );";
mysql_query($statement);
$input = json_decode($request_body, true);
$user = $input["user"];
$password = $user["password"];      
    $mySQLstring = "SELECT username, password, id FROM tbl_user where username = '$username' ;";
    $json["statement"][] = $mySQLstring;
    $qur = mysql_query($mySQLstring);
            if ($qur){
        $max = mysql_fetch_assoc($qur);
    }
    if ($max){
        $json["max"] = $max;
        if ($max["password"] == $password){
            $json["username"] = $username;
            $json["id"] = $max["id"];               
            $json["status"] = "ok";
            $tables = array("class", "class_user", "module", "module_class", "module_user", "rating", "student", "student_class");
            foreach($tables as $table){
                $id= $max["id"];
                if ( $table == 'module' ){

                $statement ='SELECT create_user, MAX(id) AS id FROM tbl_'.$table;
                  $statement .= ' WHERE create_user =$id' ;

                $qur = mysql_query($statement);
                if ($qur){
                    while($r = mysql_fetch_array($qur, MYSQL_ASSOC)){
                        $result[$table][] = $r;
                    }
                }
                }
            }
            $json = array("status" => "ok", "data" => $result);
        }
    }
}



@mysql_close($conn);
header('Content-type: application/json');
echo json_encode($json);
?>