左加入 Group By

Question

我使用的是 PostgreSQL 9.4。

我有一张workouts 的表格。用户可以为每个workout创建多个results，一个result有一个score。

给定一个锻炼 ID 列表和两个用户 ID，我想返回每个用户每次锻炼的最佳分数。如果用户没有该锻炼的结果，我想返回填充/空结果。

SELECT "results".*, "workouts".* 
FROM "results" LEFT JOIN "workouts" ON "workouts"."id" = "results"."workout_id" 
WHERE (
  (user_id, workout_id, score) IN 
  (SELECT user_id, workout_id, MAX(score) 
    FROM results WHERE user_id IN (1, 2) AND workout_id IN (1, 2, 3) 
    GROUP BY user_id, workout_id)
) 

在这个查询中，左连接充当内连接；如果用户没有得到锻炼结果，我不会得到任何填充。无论存在多少结果，此查询都应始终返回六行。

示例数据：

results
user_id | workout_id | score 
-----------------------------
      1 |          1 |     10
      1 |          3 |     10
      1 |          3 |     15
      2 |          1 |      5

Desired result:

results.user_id | results.workout_id | max(results.score) | workouts.name
-------------------------------------------------------------------------
              1 |                  1 |                 10 | Squat
              1 |                  2 |               null | Bench
              1 |                  3 |                 15 | Deadlift
              2 |                  1 |                  5 | Squat
              2 |                  2 |               null | Bench
              2 |                  3 |               null | Deadlift

Answer 1

SELECT DISTINCT ON (1, 2)
       u.user_id
     , w.id AS workout_id
     , r.score
     , w.name AS workout_name
FROM   workouts w
CROSS  JOIN (VALUES (1), (2)) u(user_id)
LEFT   JOIN  results r ON r.workout_id = w.id
                      AND r.user_id = u.user_id
WHERE  w.id IN (1, 2, 3)
ORDER  BY 1, 2, r.score DESC NULLS LAST;

分步说明

1. 形成给定锻炼和用户的完整笛卡尔积。
   假设给定的锻炼始终存在。
   假设并非所有给定用户都有所有给定锻炼的结果。

2. LEFT JOIN 到 results。所有条件都进入LEFT JOIN 的ON 子句，而不是WHERE 子句，这将排除没有结果的(workout_id, user_id) 组合。见：

   • Rails includes query with conditions not returning all results from left table

3. 最后选择(user_id, workout_id) 和DISTINCT ON 的最佳结果。在此过程中，生成所需的排序顺序。见：

   • Select first row in each GROUP BY group?

根据表的大小和数据分布，可能会有更快的解决方案。见：

• Optimize GROUP BY query to retrieve latest row per user

简单版

如果你想要的只是每个(user_id, workout_id) 组合的最大score，那么有一个简单的版本：

SELECT user_id, workout_id, max(r.score) AS score
FROM        unnest('{1,2}'::int[])   u(user_id)
CROSS  JOIN unnest('{1,2,3}'::int[]) w(workout_id)
LEFT   JOIN results r USING (user_id, workout_id)
GROUP  BY 1, 2
ORDER  BY 1, 2;

db小提琴here
_{旧sqlfiddle.}

Answer 2

使用distinct on 或row_number() 怎么样？

SELECT DISTINCT ON (r.user_id, r.workout_id) r.*, w.* 
FROM "results" r LEFT JOIN
     "workouts" w
     ON "w."id" = r."workout_id" 
WHERE r.user_id IN (1, 2) AND r.workout_id IN (1, 2, 3) 
ORDER BY r.user_id, r.workout_id, score desc;

row_number() 等效项需要子查询：

SELECT rw.*
FROM (SELECT r.*, w.*,
             row_number() over (partition by user_id, workout_id order by score desc) as seqnum 
      FROM "results" r LEFT JOIN
           "workouts" w
           ON "w."id" = r."workout_id" 
      WHERE r.user_id IN (1, 2) AND r.workout_id IN (1, 2, 3) 
     ) rw
WHERE seqnum = 1;

您应该比使用* 更明智地选择列。如果列名重复，子查询可能会返回错误。

编辑：

您需要先生成行，然后再生成每个行的结果。这是一种基于第二个查询的方法：

SELECT u.user_id, w.workout_id, rw.score, rw.name
FROM (SELECT 1 as user_id UNION ALL SELECT 2) u CROSS JOIN
     (SELECT 1 as workout_id UNION ALL SELECT 2 UNION ALL SELECT 3) w LEFT JOIN
     (SELECT r.*, w.*,
             row_number() over (partition by user_id, workout_id order by score desc) as seqnum 
      FROM "results" r LEFT JOIN
           "workouts" w
           ON "w."id" = r."workout_id" 
      WHERE r.user_id IN (1, 2) AND r.workout_id IN (1, 2, 3) 
     ) rw
     ON rw.user_id = u.user_id and rw.workout_id = w.workout_id and
        rw.seqnum = 1;

Answer 3

where 过滤掉了你的 NULL 值，这就是为什么结果不是你所期望的。

加入 WHERE 子句结果，而不是过滤 where 子句结果。

SELECT "results".*, "workouts".*,"max_score".*
FROM "results" 
LEFT JOIN "workouts" ON "workouts"."id" = "results"."workout_id"
LEFT JOIN (SELECT user_id, workout_id, MAX(score) 
    FROM results WHERE user_id IN (1, 2) AND workout_id IN (1, 2, 3) 
    GROUP BY user_id, workout_id) max_score ON workouts.workout_id=max_score.workout_id;

您需要更改 SELECT 以获得正确的列。