2017-10-12 76 views
-1

這就是我現在使用的。我成功獲取第一個查詢,但是我無法從第二個查詢中檢索任何數據。第二個查詢的所有變量都是未定義的。連續獲取MySQL數據的最佳方式?

$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

你爲什麼不使用JOIN –

+1

您是否嘗試過呼應你的第二個查詢和自己運行它,以確保你應該回去數據? – RhoVisions

回答

0

只是把它拿來一次性使用連接:

$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

非常感謝!奇蹟般有效。 – Loleris54

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