在mysql中使用DISTINCT SELECT更新表

假设我有一个表，其中包含关于

game (PRIMARY INT id, TINYINT offline)

还有第二张表，其中包含该游戏的详细信息：

gamedetail (PRIMARY INT id, INT game_id (fk to game table), TINYINT offline) 

详细信息会从各种程序中频繁更新。在这里，我设置了细节的脱机标志。我不可能设置游戏本身的离线标志。（但是，如果我找到在线详细信息，我会将游戏的离线标志设置为0）。但我想通过更新查询在数据库中设置此信息。其理念是：

SELECT DISTINCT game.id FROM game 
    LEFT JOIN gamedetail AS gdon 
           ON (gdon.game_id = game.id AND gdon.offline = 0)
    LEFT JOIN gamedetail AS gdoff 
           ON (gdoff.game_id = game.id AND gdoff.offline = 1)
WHERE gdoff.id IS NOT NULL AND gdon.id IS NULL;

这给了我很好的所有游戏，我只有离线的游戏细节。因此，我想将此作为UPDATE语句的输入，如下所示：

UPDATE game SET game.offline=1 WHERE game id IN (
    SELECT DISTINCT game.id FROM game 
        LEFT JOIN gamedetail AS gdon 
               ON (gdon.game_id = game.id AND gdon.offline = 0)
        LEFT JOIN gamedetail AS gdoff 
               ON (gdoff.game_id = game.id AND gdoff.offline = 1)
    WHERE gdoff.id IS NOT NULL AND gdon.id IS NULL;

)

UPDATE game SET game.offline=1 WHERE game.id IN (
  SELECT * FROM (
    SELECT DISTINCT game.id FROM game 
        LEFT JOIN gamedetail AS gdon 
               ON (gdon.game_id = game.id AND gdon.offline = 0)
        LEFT JOIN gamedetail AS gdoff 
               ON (gdoff.game_id = game.id AND gdoff.offline = 1)
    WHERE gdon.id IS NOT NULL AND gdon.id IS NULL;
  ) t
)

不幸的是，这在mysql中失败了，因为

错误1093（HY000）：表'game'被指定了两次，既作为'UPDATE'的目标，也作为数据的独立源

我的问题是如何将我的update语句更改为在mysql中工作的语句

编辑：更正WHERE条件
将查询包装在子查询中，如下所示：

UPDATE game SET game.offline=1 WHERE game id IN (
    SELECT DISTINCT game.id FROM game 
        LEFT JOIN gamedetail AS gdon 
               ON (gdon.game_id = game.id AND gdon.offline = 0)
        LEFT JOIN gamedetail AS gdoff 
               ON (gdoff.game_id = game.id AND gdoff.offline = 1)
    WHERE gdoff.id IS NOT NULL AND gdon.id IS NULL;

)

UPDATE game SET game.offline=1 WHERE game.id IN (
  SELECT * FROM (
    SELECT DISTINCT game.id FROM game 
        LEFT JOIN gamedetail AS gdon 
               ON (gdon.game_id = game.id AND gdon.offline = 0)
        LEFT JOIN gamedetail AS gdoff 
               ON (gdoff.game_id = game.id AND gdoff.offline = 1)
    WHERE gdon.id IS NOT NULL AND gdon.id IS NULL;
  ) t
)

只需使用一个简单的连接，不需要子查询

UPDATE game 
LEFT JOIN gamedetail AS gdon 
               ON (gdon.game_id = game.id AND gdon.offline = 0)
        LEFT JOIN gamedetail AS gdoff 
               ON (gdoff.game_id = game.id AND gdoff.offline = 1)
SET game.offline=1
    WHERE gdoff.id IS NOT NULL AND gdon .id IS NULL;

或者，如果您正在使用子查询，您需要提供一个新的别名，因为在更新查询中，您可以在where子句中指定相同的表，我想where条件应该是
where gdoff.id不为NULL，而gdon.id为NULL
thx，我更正了where条件。这是对实际表的简化，因此存在错误。我现在尝试一下你的建议，与外部连接相比，这个查询可能具有巨大的优势，因为它使用标准SQL，并且也可以用于其他数据库。