Question

我的JSON响应有问题。在我的mysql数据库中，我有两个表：用户和游戏。当我尝试从1个用户获得所有游戏时，我会在我的响应中获得这些游戏，但*我在Users表中拥有的不同用户数量。因此，如果我想要来自id为8的用户的所有游戏，例如3个游戏，我会得到那些游戏，但是我拥有的独特用户数量，让我们说9个。所以查询返回27个游戏（9 * 3）同样的游戏）。我在这里做错了什么。

 $pm_id = $_POST["pm_id"];
 $pm_pass = $_POST["pm_pass"];
 $pm_timestamp = $_POST["pm_timestamp"];

try {
     $dbconn = 'mysql:host=' . DBHOST . ';dbname=' . DBDATA;
     $db = new PDO($dbconn, DBUSER, DBPASS);
 } catch (PDOException $e) {
     echo 'Connection failed: ' . $e->getMessage();
 }

 $statement = $db->prepare('SELECT * FROM users WHERE user_id = :id AND password =      :pass');
 $statement->execute(array(':id' => $pm_id, ':pass' => $pm_pass));
 $statement->setFetchMode(PDO::FETCH_ASSOC);

 $row = $statement->fetch();
     if($row['timestamp'] == $pm_timestamp){
         #no sync necessary
         echo json_encode("no sync needed");
     }else{
         #start sync
         $games = array();
         $statement_getAllGames = $db->prepare('SELECT game.game_id, game.user_id,       game.name, game.buyin, game.result, game.startDate, game.endDate, game.location,      game.isTournament, game.participants, game.endposition, game.comment, game.blinds,      game.pause,
         game.visibility, users.timestamp FROM game, users WHERE game.user_id = :id AND    game.timestamp > :timestamp');
         $statement_getAllGames->execute(array(':id' => $pm_id, ':timestamp' =>   $pm_timestamp));

         while($row = $statement_getAllGames->fetch(PDO::FETCH_ASSOC)){
             $games[] = array('game'=>$row);
         }
         echo json_encode(array('games'=>$games));

 };

如果它有用，我可以添加表格的结构。

Answer 1

FROM game, users

您缺少加入条件game.user_id = user.user_id。您可以将其添加到WHERE子句中，也可以使用较新的SQL-92连接语法（请参阅INNER JOIN ON vs WHERE clause）：

FROM game INNER JOIN users ON game.user_id = user.user_id

PDO查询返回PHP的结果很多

1 个答案: