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 个用户那里获取所有游戏时,我会在我的响应中得到这些游戏,但 * 我的用户表中的不同用户数量。因此,如果我想要一个 id 为 8 的用户的所有游戏,例如 3 个游戏,我会得到这些游戏,但乘以我拥有的唯一用户数量,比如说 9。所以查询返回 27 个游戏(9 * 3相同的游戏)。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));
};
如果有帮助,我可以添加表格的结构。