內部聯接查詢以獲取每種武器的結果

Question

我有一個帶有內部聯接的查詢，但是WEAPON是未知的。 我想為每種武器爭取最好的殺手。 如何使用此查詢執行此操作？ 我對高級查詢不是很了解。

    $q = $mysql->query("SELECT `killerID`
            , COUNT(`killerID`) AS tot_kills
            , MIN(`Username`) AS username
            FROM `kills`
            INNER JOIN `players`
            ON `players`.`id` = `kills`.`killerid`
            WHERE `killText` LIKE '%###WEAPON###%'
            GROUP BY `killerID`
            ORDER BY `tot_kills` DESC") or die($mysql->error);

Answer 1

由於使用join ，因此需要在select子句中指定表名。 嘗試類似：

SELECT `kills`.`killerID`
            , COUNT(`kills`.`killerID`) AS tot_kills
            , MIN(`players`.`Username`) AS username
            FROM `kills`
            INNER JOIN `players`
            ON `players`.`id` = `kills`.`killerid`
            WHERE `kills`.`killText` LIKE '%###WEAPON###%'
            GROUP BY `kills`.`killerID`
            ORDER BY `kills`.`tot_kills` DESC

Answer 2

我想你應該按武器，而不是殺手ID，只要你想的數killerID的weapon 。

請注意，您不需要所有這些反引號。 僅當查詢中的表名或列名是保留字之一時，才需要反引號。

SELECT killerID
        , COUNT(killerID) AS tot_kills
        , MIN(Username) AS username
FROM kills
INNER JOIN players
  ON players.id = kills.killerid
WHERE killText LIKE '%###WEAPON###%'
GROUP BY kills.killText
ORDER BY tot_kills DESC