使用附加(不同)过滤器聚合列

本文介绍了使用附加(不同)过滤器聚合列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

这段代码按预期工作，但我很长而且令人毛骨悚然.

This code works as expected, but I it's long and creepy.

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:

As you can see, it consists of 3 repetitive parts for retrieving:

• 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:

And each of those also consists of 2 parts:

• 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

推荐答案

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

The aggregate FILTER clause in Postgres 9.4 or newer is shorter and faster:

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;

• 手册
• Postgres 维基
• Depesz 博文

在 Postgres 9.3(或任何版本)中，这仍然比嵌套子选择或CASE表达式更短、更快:

In Postgres 9.3 (or any version) this is still shorter and faster than nested sub-selects or CASE expressions:

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:

• 对于绝对性能，SUM 是更快还是 COUNT?