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我有一个三表设置：孩子、玩具和游戏，每个都有唯一的主键：id_kid、id_toy 和 id_game。每个孩子可以拥有多个玩具和游戏，但每个玩具或游戏只有一个孩子拥有。

玩具和游戏有一个带有 3 个状态的购买列：-1,0,1 表结构是这样的：

kids

id_kid
kid_name
etc

games

id_game
id_kid_games --> links with id_kid in kids_table (maybe not the best name, I know)
game_name
bought --> can be -1,0,1

toys

id_toy
id_kid_toys --> links with id_kid in kids_table
toy_name
bought --> can be -1,0,1

对于每个孩子，我尝试使用下面的查询获得购买和未购买的玩具和游戏的总数，但是结果翻了一番：

SELECT kids.*, 

COUNT(DISTINCT toys.id_toy) AS total_toys, 
SUM(CASE toys.bought WHEN 1 THEN 1 ELSE 0 END) AS toys_bought, 
SUM(CASE toys.bought WHEN -1 THEN 1 ELSE 0 END) AS toys_not_bought, 

COUNT(DISTINCT games.id_game) AS total_games, 
SUM(CASE games.bought WHEN 1 THEN 1 ELSE 0 END) AS games_bought, 
SUM(CASE games.bought WHEN -1 THEN 1 ELSE 0 END) AS games_not_bought 

FROM kids as k 
LEFT JOIN toys t ON k.id_kid = t.id_kid_toys
LEFT JOIN games g ON k.id_kid = g.id_kid_games
GROUP BY k.id_kid
ORDER BY k.name ASC

一个孩子有 2 个玩具和 4 个游戏，都买了，结果是总共 2 个玩具（正确），总共 4 个游戏（正确），买了 8 个玩具，买了 8 个游戏。（都错了）

请帮助回答 - 如果可能的话 - 不使用子选择。谢谢你。

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1 回答 1

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当您从两个不相关的关系（孩子加入玩具，孩子加入游戏）中选择数据时，子查询是执行此操作的自然方式。由于可能会使用不相关的子查询，因此这应该不会特别慢。

试试这个查询是否足够有效：

与您的原始查询相比，它基本上只是颠倒了加入和分组的顺序。

SELECT kids.*, t.total_toys, t.toys_bought, t.toys_not_bought,
               g.total_games, g.games_bought, g.games_not_bought
FROM kids
LEFT JOIN (SELECT id_kids_toys,
                  COUNT(*) AS total_toys,
                  SUM(CASE bought WHEN 1 THEN 1 ELSE 0 END) as toys_bought,
                  SUM(CASE bought WHEN -1 THEN 1 ELSE 0 END) as toys_not_bought
           FROM toys
           GROUP BY id_kids_toys) AS t
ON t.id_kids_toys = kids.id_kid
LEFT JOIN (SELECT id_kids_games,
                  COUNT(*) AS total_games,
                  SUM(CASE bought WHEN 1 THEN 1 ELSE 0 END) as games_bought,
                  SUM(CASE bought WHEN -1 THEN 1 ELSE 0 END) as games_not_bought
           FROM games
           GROUP BY id_kids_games) AS g
ON g.id_kids_games = kids.id_kid
ORDER by kids.name;

如果您坚持避免使用子查询，那么这个查询可能会效率低得多：

SELECT kids.*, 
COUNT(DISTINCT toys.id_toy) AS total_toys, 

-- sum only toys joined to first game
SUM(IF(g2.id_game IS NULL AND bought = 1, 1, 0)) AS toys_bought, 
SUM(IF(g2.id_game IS NULL AND bought = -1, 1, 0)) AS toys_not_bought, 

-- sum only games joined to first toy
COUNT(DISTINCT games.id_game) AS total_games, 
SUM(IF(t2.id_toy IS NULL AND bought = 1, 1, 0)) AS games_bought, 
SUM(IF(t2.id_toy IS NULL AND bought = -1, 1, 0)) AS games_not_bought 

FROM kids as k 
LEFT JOIN toys t ON k.id_kid = t.id_kid_toys
LEFT JOIN games g ON k.id_kid = g.id_kid_games

-- select only rows where either game or toy is the first one for this kid
LEFT JOIN toys t2 on k.id_kid = t.id_kid_toys AND t2.id_toy < t.id_toy
LEFT JOIN games g2 ON k.id_kid = g.id_kid_games AND g2.id_game < g.id_game 
WHERE t2.id_toy IS NULL OR g2.id_game IS NULL

GROUP BY k.id_kid
ORDER BY k.name ASC

它的工作原理是确保对于每个孩子，只计算加入第一个玩具的游戏，并且只计算加入第一个游戏的玩具。

于 2012-11-28T17:45:37.303 回答