Question

假设我有一个包含

信息的表格

game (PRIMARY INT id, TINYINT offline)

以及第二张表格，其中包含有关该游戏的详细信息：

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

详细信息会经常更新并从各种程序中更新。在那里我设置了细节的离线标志。我没有编程设置游戏本身的离线标志的可能性。（我确实将游戏的离线标志设置为0，但是，如果我找到在线详细信息）。但我想通过更新查询在数据库中设置此信息。这个想法就是这个SELECT：

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，我只有离线游戏细节。所以我想将此作为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;

)

由于ERROR 1093 (HY000): Table 'game' is specified twice, both as a target for 'UPDATE' and as a separate source for data，这很遗憾在mysql中失败。

我的问题是如何将我的更新语句更改为适用于mysql的内容？

编辑：更正了WHERE条件

Answer 1

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 * 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
)

Answer 2

只需使用简单的连接，无需使用子查询

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，如果您使用子查询，则需要提供一个新别名，因为在更新查询中您可以在where子句中指定相同的表，我猜其中where条件应为WHERE gdoff.id IS NOT NULL AND gdon .id IS NULL

在mysql中使用DISTINCT SELECT更新表

2 个答案: