Question

我需要查询location = $location的数据库。这将返回一些user_id。

然后我需要查询另一个数据库user_id =之前的user_id结果。这将找到game_id s。

然后，我需要使用game_id查询另一个表来获取每个游戏的名称（每个游戏名称都分配了一个ID。）

这就是我所拥有的：

 if (isset($_POST['location'])) {
        $location = $_POST['location'];
        $query = mysql_query("SELECT * FROM users WHERE location = '$location'") or die(mysql_error());

        echo "<ul>";
        while ($row = mysql_fetch_array($query)) {
            foreach (explode(", ", $row['user_id']) as $users) {
                echo "<li>" . $row['user_name'] . "<ul><li>";
               //this where I need I need to query another table where user_Id = $user (each user found in previous query)
//this will find multiple game id's (as $game) , I then need to query another table to find game_id = $game
// output name of game
                echo "</li></ul></li>";
            }  
        }
        echo "</ul>";
      }

听起来很复杂。有更简单的方法吗？

Answer 1

这样的查询将生成location, user_id, game_id表。在SQL查询中投入尽可能多的精力，以便您能够通过代码轻松处理它：

$query = mysql_query("SELECT game_name FROM another_table \
                      JOIN users \
                      JOIN games \
                      WHERE users.location = '$location'")
         or die(mysql_error());
while ($row = mysql_fetch_array($query)) {
    echo $row['game_name']; //Here is the game name
}

查询表，然后查询每个结果的另一个表，然后...等

1 个答案: