Question

我正在使用PostgreSQL 9.4。

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

给定一个workout_ids列表和两个user_id，我想为每个用户返回每个锻炼的最佳分数。如果用户没有该锻炼的结果，我想返回填充/ null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)
)

在此查询中，左连接充当内连接;如果用户没有得到锻炼结果，我没有得到任何填充。无论存在多少结果，此查询应始终返回六行。

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

where过滤掉你的NULL值，这就是为什么结果不符合你的期望。

Joinint 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以获取正确的列。

Answer 2

如何使用distinct on或row_number()？

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。在重复列名称的情况下，子查询可能会返回错误。

编辑：

您需要先生成行，然后生成每个行的结果。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

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;

解释

形成给定锻炼和用户的完整笛卡尔积。
假设给定的训练始终存在假设并非所有给定用户都有针对所有给定锻炼的结果。
然后 LEFT JOIN到results。所有条件都进入ON的{{1}}子句，而不是LEFT JOIN子句，这将排除没有结果的WHERE组合。详细说明：
- Rails includes query with conditions not returning all results from left table
最后使用(workout_id, user_id)为(user_id, workout_id)选择最佳结果。在此期间，产生所需的排序顺序 DISTINCT ON的详细信息：
- 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：

Optimize GROUP BY query to retrieve latest record per user

简易版

如果您想要的是每个DISTINCT ON组合的最大score，则有简单的版本：

(user_id, workout_id)

SQL Fiddle.

左边加入Group By

3 个答案:

解释

简易版