IN 运算符不返回范围内的所有记录答案

【问题标题】：IN operator doesn't return all record in the rangeIN 运算符不返回范围内的所有记录
【发布时间】：2018-07-04 13:52:35
【问题描述】：

我正在尝试返回round 范围内所有可用的matches，每一轮都可以有不同的匹配。所以我写了这个查询：

SELECT m.id, m.round_id, m.gameweek
 FROM `match` m
 LEFT JOIN competition_rounds r ON m.round_id = r.id
 LEFT JOIN competition_seasons s ON r.season_id = s.id
 LEFT JOIN competition c ON c.id = s.competition_id
 WHERE 1 AND
 m.status = 5 AND
 m.round_id IN (488, 489, 490, 491) AND
 m.gameweek = (SELECT MAX(m2.gameweek)
                FROM `match` m2
                WHERE m2.round_id IN (488, 489, 490, 491))

现在的问题是结果只返回 ID 为 488 的回合的matches，为什么其他回合被忽略？谢谢。

【问题讨论】：

我猜 id 488 的回合有最大的游戏周...
@jarlh mmm 但我的目标是为每一轮获得最大游戏周我应该用 OR 替换 IN？

标签： mysql sql pdo

【解决方案1】：

我想你可能想要：

SELECT m.id, m.round_id, m.gameweek
 FROM `match` m
 LEFT JOIN competition_rounds r ON m.round_id = r.id
 LEFT JOIN competition_seasons s ON r.season_id = s.id
 LEFT JOIN competition c ON c.id = s.competition_id
 WHERE 1 AND
 m.status = 5 AND
 m.round_id IN (488, 489, 490, 491) AND
 m.gameweek = (SELECT MAX(m2.gameweek)
                FROM `match` m2
                WHERE m2.round_id = m.round_id))

【讨论】：

【解决方案2】：

我猜这个查询就足够了：

SELECT m.id, m.round_id, m.gameweek
FROM `match` m
WHERE 1 AND
      m.status = 5 AND
      m.round_id IN (488, 489, 490, 491) AND
      m.gameweek = (SELECT MAX(m2.gameweek)
                    FROM `match` m2
                    WHERE m2.round_id = m.round_id AND
                          m2.status = m.status
                   );

也就是说，您可能还想考虑状态。而且，除非您真的希望结果中有重复项，否则您不需要 JOINs。

【讨论】：

【解决方案3】：

您需要一个相关的子查询：

SELECT m.id, m.round_id, m.gameweek
 FROM `match` m
 LEFT JOIN competition_rounds r ON m.round_id = r.id
 LEFT JOIN competition_seasons s ON r.season_id = s.id
 LEFT JOIN competition c ON c.id = s.competition_id
 WHERE 1 AND
 m.status = 5 AND
 m.round_id IN (488, 489, 490, 491) AND
 m.gameweek = (SELECT MAX(m2.gameweek)
                FROM `match` m2
                WHERE m2.round_id = m.round_id)

【讨论】：