我需要查询一个数据库，其中location = $location。这将返回一些user_id s。
然后我需要查询另一个数据库，其中user_id =以前的user_id结果。这将找到game_id s。
然后我需要用这些game_id查询另一个表来获取每个游戏的名称(每个游戏名称都分配了一个id)
这就是我拥有的:
if (isset($_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>";
}
听起来复杂。有更简单的方法吗?
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，以便您能够轻松地通过代码处理它:
$query = mysql_query("SELECT game_name FROM another_table '
JOIN users '
JOIN games '
WHERE users.location = '$location'")
or die(mysql_error());
while ($row = mysql_fetch_array($query)) {
echo $row['game_name']; //Here is the game name
}