2017-10-12 75 views
-1

这就是我现在使用的。我成功获取第一个查询,但是我无法从第二个查询中检索任何数据。第二个查询的所有变量都是未定义的。连续获取MySQL数据的最佳方式?

$sql = "SELECT * FROM history WHERE ended=1 ORDER BY chance ASC LIMIT 1"; 
$result = $conn->query($sql); 
$row = mysqli_fetch_assoc($result); 
$luck_steamid = $row["winner"]; 
$luck_total = $row["total"]; 
$luck_chance = $row["chance"]; 
$luck_round = $row["id"]; 
$sql = "SELECT * FROM users WHERE steamid=".$luck_steamid; 
$result = $conn->query($sql); 
$row = mysqli_fetch_assoc($result); 
$luck_avatar = $row["avatar"]; 
$luck_name = $row["name"]; 
$luck_profile = $row["profile"]; 
+1

你为什么不使用JOIN –

+1

您是否尝试过呼应你的第二个查询和自己运行它,以确保你应该回去数据? – RhoVisions

回答

0

只是把它拿来一次性使用连接:

$sql = "SELECT history.winner, history.total, history.chance, history.id, users.avatar, users.name, users.profile FROM history JOIN users ON history.winner = users.steamid WHERE ended=1 ORDER BY chance ASC LIMIT 1"; 
$result = $conn->query($sql); 
$row = mysqli_fetch_assoc($result); 
$luck_steamid = $row["winner"]; 
$luck_total = $row["total"]; 
$luck_chance = $row["chance"]; 
$luck_round = $row["id"]; 
$luck_avatar = $row["avatar"]; 
$luck_name = $row["name"]; 
$luck_profile = $row["profile"]; 
+0

非常感谢!奇迹般有效。 – Loleris54

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