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我的JSON響應有問題。在我的mysql數據庫中有兩個表:用戶和遊戲。當我嘗試從1位用戶那裏獲得所有遊戲時,我會在我的用戶表中獲得這些遊戲,但*我的用戶表中有不同用戶的數量。因此,如果我想要8位用戶的所有遊戲,例如3款遊戲,那麼我只能獲得這些遊戲的數量,而不是我擁有的唯一用戶數量,比方說9。因此,查詢會返回27個遊戲(9 * 3相同的遊戲)。我在這裏做錯了什麼?PDO查詢返回很多結果PHP
$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));
};
如果有幫助,我可以添加我的表的結構。
謝謝!這是我的一天。 – nostradamus