Question

您好我一直在寻找这个，我找到了其他示例代码的解决方案，但我无法弄清楚如何为我的实现。

SELECT `gameDBGames`, `game_id`, MIN(`gamePrice`) AS `gamePrice`
FROM `games` 
LEFT JOIN `platforms` ON `gamePlatform` = `platform_id` 
LEFT JOIN `bundles` ON `gameBundle` = `bundle_id` 
LEFT JOIN `currency` ON `bundleCurrency` = `currency_id` 
WHERE `bundleEnd` > CURDATE() AND `bundleType` = "1" AND `gameDBGames` != "0"
GROUP BY `gameDBGames`

这是我的实际查询。这会返回最低价格，但不会与game_id对应。我怎么能这样做？我相信正在做这样的内部联接：

 SELECT `gameDBGames`, `game_id`, MIN(`gamePrice`) AS `gamePrice`
 FROM `games` 
 LEFT JOIN `platforms` ON `gamePlatform` = `platform_id`
 LEFT JOIN `bundles` ON `gameBundle` = `bundle_id`
 INNER JOIN (....)
 LEFT JOIN `currency` ON `bundleCurrency` = `currency_id` 
 WHERE b.`bundleEnd` > CURRDATE() AND b.`bundleType` = "1" AND g.`gameDBGames` != "0"

谢谢。

编辑：抱歉，我不知道我在考虑不发布表格结构。

game_id是唯一ID（例如同一游戏，但价格不同），gameDBGames是游戏的ID（例如gameDBGames = 1，价格可能是40或30），这就是我使用gameDBGames进行分组的原因。目标是以最低价格获得独特的gameDBGames。

我有什么。

+--------------+---------+----+---------+
| table games  |         |    |         |
+--------------+---------+----+---------+
| game_id      | int     | AI | PRIMARY |
| gameDBGames  | int     |    |         |
| gamePrice    | float   |    |         |
| gamePlatform | tinyint |    |         |
| gameBundle   | int     |    |         |
+--------------+---------+----+---------+

+---------+-------------+-----------+--------------+------------+
| game_id | gameDBGames | gamePrice | gamePlatform | gameBundle |
+---------+-------------+-----------+--------------+------------+
|       1 |           1 |        20 |            1 |          1 |
|       2 |           2 |        20 |            2 |          1 |
|       3 |           2 |        15 |            2 |          1 |
|       4 |           3 |        17 |            1 |          1 |
|       5 |           3 |        20 |            1 |          1 |
|       6 |           3 |        15 |            1 |          1 |
|       7 |           4 |        16 |            2 |          2 |
|       8 |           5 |        18 |            2 |          2 |
|       9 |           5 |        14 |            2 |          2 |
|      10 |           6 |        15 |            1 |          2 |
+---------+-------------+-----------+--------------+------------+

我得到了什么。

+---------+-------------+-----------+--------------+------------+
| game_id | gameDBGames | gamePrice | gamePlatform | gameBundle |
+---------+-------------+-----------+--------------+------------+
|       1 |           1 |        20 |            1 |          1 |
|       2 |           2 |        15 |            2 |          1 |
|       4 |           3 |        15 |            1 |          1 |
|       7 |           4 |        16 |            2 |          2 |
|       8 |           5 |        14 |            2 |          2 |
|      10 |           6 |        15 |            1 |          2 |
+---------+-------------+-----------+--------------+------------+

正如您所看到的，game_id与gamePrice不对应。它应该是这样的。

+---------+-------------+-----------+--------------+------------+
| game_id | gameDBGames | gamePrice | gamePlatform | gameBundle |
+---------+-------------+-----------+--------------+------------+
|       1 |           1 |        20 |            1 |          1 |
|       3 |           2 |        15 |            2 |          1 |
|       6 |           3 |        15 |            1 |          1 |
|       7 |           4 |        16 |            2 |          2 |
|       9 |           5 |        14 |            2 |          2 |
|      10 |           6 |        15 |            1 |          2 |
+---------+-------------+-----------+--------------+------------+

希望你现在明白，对不起之前没有得到更多的解释。如果你需要别的东西请问。谢谢。

编辑2（标记） Mark提出的解决方案给了我这个错误。

有重复的gameDBGames。
有些游戏没有显示。

我已经更新了WHERE子句以防万一。

Answer 1

尝试：

SELECT g.`gameDBGames`, g.`game_id`, g.`gamePrice`
FROM (SELECT `gameDBGames`, MIN(`gamePrice`) AS `minPrice`
      FROM `games` 
      GROUP BY `gameDBGames`) mn
JOIN `games` g 
  ON mn.`gameDBGames`=g.`gameDBGames` and mn.`minPrice`=g.`gamePrice`
LEFT JOIN `platforms` p ON g.`gamePlatform` = p.`platform_id` 
LEFT JOIN `bundles` b ON g.`gameBundle` = b.`bundle_id` 
LEFT JOIN `currency` c ON b.`bundleCurrency` = c.`currency_id` 
WHERE ....

由于子查询中的分组，不应要求主查询中的GROUP BY - 但是，对于给定的game_id ，将返回多个gameDBGames如果有多个game_id具有相同gameDBGames的最低价格。

MYSQL组具有最小值

1 个答案: