使用其他（不同的）过滤器聚合列答案

【问题标题】：Aggregate columns with additional (distinct) filters使用其他（不同的）过滤器聚合列
【发布时间】：2015-01-24 00:28:08
【问题描述】：

此代码按预期工作，但我很长而且令人毛骨悚然。

select p.name, p.played, w.won, l.lost from

(select users.name, count(games.name) as played
from users
inner join games on games.player_1_id = users.id
where games.winner_id > 0
group by users.name
union
select users.name, count(games.name) as played
from users
inner join games on games.player_2_id = users.id
where games.winner_id > 0
group by users.name) as p

inner join

(select users.name, count(games.name) as won
from users
inner join games on games.player_1_id = users.id
where games.winner_id = users.id
group by users.name
union
select users.name, count(games.name) as won
from users
inner join games on games.player_2_id = users.id
where games.winner_id = users.id
group by users.name) as w on p.name = w.name

inner join

(select users.name, count(games.name) as lost
from users
inner join games on games.player_1_id = users.id
where games.winner_id != users.id
group by users.name
union
select users.name, count(games.name) as lost
from users
inner join games on games.player_2_id = users.id
where games.winner_id != users.id
group by users.name) as l on l.name = p.name

如您所见，它由 3 个用于检索的重复部分组成：

玩家姓名和他们玩的游戏数量
玩家姓名和他们赢得的游戏数量
玩家姓名和他们输掉的游戏数量

每一个也由两部分组成：

玩家姓名和他们作为玩家_1参与的游戏数量
玩家姓名和他们作为玩家_2参与的游戏数量

如何简化？

结果如下：

           name            | played | won | lost 
---------------------------+--------+-----+------
 player_a                  |      5 |   2 |    3
 player_b                  |      3 |   2 |    1
 player_c                  |      2 |   1 |    1

【问题讨论】：

你还没有运行 postgres 9.4 是吗？
@JoeLove，还没有，但感谢您提到聚合过滤器，我以后一定会考虑升级。

标签： sql postgresql aggregate-functions aggregate-filter

【解决方案1】：

这是一种关联子查询可以简化逻辑的情况：

select u.*, (played - won) as lost
from (select u.*,
             (select count(*)
              from games g
              where g.player_1_id = u.id or g.player_2_id = u.id
             ) as played,
             (select count(*)
              from games g
              where g.winner_id = u.id
             ) as won
      from users u
     ) u;

这假设没有关系。

【讨论】：

【解决方案2】：

select users.name, 
       count(case when games.winner_id > 0 
                  then games.name 
                  else null end) as played,
       count(case when games.winner_id = users.id 
                  then games.name 
                  else null end) as won,
       count(case when games.winner_id != users.id 
                  then games.name 
                  else null end) as lost
from users inner join games 
     on games.player_1_id = users.id or games.player_2_id = users.id
group by users.name;

【讨论】：

这在 9.4 之后实施聚合过滤器时将完全不重要。这些类型的“案例”陈述将成为过去。

【解决方案3】：

Postgres 9.4 或更新版本中的 aggregate FILTER 子句 更短更快：

SELECT u.name
     , count(*) FILTER (WHERE g.winner_id  > 0)    AS played
     , count(*) FILTER (WHERE g.winner_id  = u.id) AS won
     , count(*) FILTER (WHERE g.winner_id <> u.id) AS lost
FROM   games g
JOIN   users u ON u.id IN (g.player_1_id, g.player_2_id)
GROUP  BY u.name;

在 Postgres 9.3（或 any 版本）中，这仍然比嵌套子选择或 CASE 表达式更短且更快：

SELECT u.name
     , count(g.winner_id  > 0 OR NULL)    AS played
     , count(g.winner_id  = u.id OR NULL) AS won
     , count(g.winner_id <> u.id OR NULL) AS lost
FROM   games g
JOIN   users u ON u.id IN (g.player_1_id, g.player_2_id)
GROUP  BY u.name;

详情：

For absolute performance, is SUM faster or COUNT?

【讨论】：