从两个不同的表中创建对两个列求和的 pgsql 视图

Posted 2023-02-26

【中文标题】从两个不同的表中创建对两个列求和的 pgsql 视图

【英文标题】：Create pgsql view who SUM two column from two different table

【发布时间】：2021-04-23 10:07:58

【问题描述】：

我有 2 个视图：

预后评分：

user_id	isGoodPrognosis	pts_won	good_gap	good_score
1	true	1	0	0
1	true	1	0	0
2	true	3	1	1
2	false	0	0	0

question_scores：

user_id	isGoodPrognosis	pts_won
1	false	0
2	true	10

并且想创建第三个视图来计算用户的总分： （我还需要 users 表中的其他数据）

score_calculations：

user_id	pts__won	good_gap	good_score	good_winner	company_id	team_id	name
1	2	0	0	2	1	1	John
2	13	1	1	1	1	1	Sam

为此，我这样做了：

 CREATE VIEW score_calculations
    AS SELECT 
    users.id as user_id,
    users.name as name,
    users.company_id as company_id,
    users.team_id as team_id,
    users.email_verified AS email_verified,
    users.banned AS banned,
    -- users.email as email,
    SUM(COALESCE(prognosis_scores."pts_won", 0) + COALESCE (question_scores."pts_won", 0) ) as pts_won,
    SUM(prognosis_scores."good_gap") as good_gap,
    SUM(prognosis_scores."good_score") as good_score,
    SUM(prognosis_scores."isGoodPrognosis"::INT) as good_winner
    FROM users
    
    LEFT JOIN prognosis_scores
    ON prognosis_scores.user_id=users.id
    LEFT JOIN question_scores
    ON question_scores.user_id=users.id
    GROUP BY users.id , users.name, users.company_id,team_id,email_verified,banned;

但SUM(COALESCE(prognosis_scores."pts_won", 0) + COALESCE (question_scores."pts_won", 0) ) as pts_won, 效果不佳：SUM with multiple LEFT JOINS with VIEWS

所以我最终得到了这个：

CREATE VIEW score_calculations
AS SELECT u.id as user_id, u.name, u.company_id, u.team_id, u.email_verified,u.banned,
-- users.email as email,
   COALESCE(ps.pts_won, 0) + COALESCE (qs.pts_won, 0) as pts_won,
   ps.good_gap, ps.good_score, ps.good_winner
FROM users u LEFT JOIN LATERAL
     (SELECT SUM(ps."pts_won") as pts_won,
             SUM(ps.good_gap) as good_gap,
             SUM(ps.good_score) as good_score,
             SUM(ps."isGoodPrognosis"::INT) as good_winner
      FROM prognosis_scores ps
      WHERE ps.user_id = u.id
     ) ps
     ON 1=1 LEFT JOIN LATERAL
     (SELECT SUM(qs."pts_won") as pts_won
      FROM question_scores qs
      WHERE qs.user_id = u.id
     ) qs
     ON 1=1;

问题是第二块代码很慢，当我尝试运行SELECT * FROM score_calculations时，执行时间大约是16s，而第一块代码很快，执行时间大约是400ms。

对于这个测试，我有大约 1000 个用户和大约 30000 个预测分数

问题是：如何优化或更改第二段代码（score_calculations 视图）？

【问题讨论】：

要诊断性能问题，请使用 EXPLAIN ANALYZE。将分析查询的结果包含在您的问题中以获得最佳答案。

【参考方案1】：

答案试图优化查询：

SELECT * FROM score_calculations 

如果您需要多个用户的信息，使用关键字LATERAL 可以提高性能，但对于所有不使用 LATERAL 的用户而言，性能通常会更好。

CREATE VIEW score_calculations
AS 
SELECT 
  u.id as user_id, u.name, u.company_id, u.team_id, u.email_verified,u.banned,
-- users.email as email,
   COALESCE(ps.pts_won, 0) + COALESCE (qs.pts_won, 0) as pts_won,
   ps.good_gap, ps.good_score, ps.good_winner
FROM users u 
  LEFT JOIN (
    SELECT 
      ps.user_id,
      SUM(ps.pts_won) as pts_won,
      SUM(ps.good_gap) as good_gap,
      SUM(ps.good_score) as good_score,
      SUM(ps."isGoodPrognosis"::INT) as good_winner
    FROM 
      prognosis_scores ps
    GROUP BY
      ps.user_id
    ) ps ON (ps.user_id = u.id)
  LEFT JOIN (
    SELECT
      qs.user_id,
      SUM(qs.pts_won) as pts_won
    FROM 
      question_scores qs
    GROUP BY 
      qs.user_id 
    ) qs ON (qs.user_id = u.id);

如果您将 WHERE 条件与 users 表中的属性一起使用，则您的原始视图很好。您应该检查您是否在prognosis_scores.user_id 和question_scores.user_id 上存在缺陷。

【参考方案2】：

对于横向连接，您需要以下索引：

prognosis_scores(user_id) question_scores(user_id)

需要索引来在这些表中查找用户的特定行——没有它们，数据库需要求助于其他机制，例如嵌套循环。

我想如果使用正确的索引，这个版本实际上可能比第一个版本更快，因为它避免了外部聚合（对此没有任何承诺）。

【参考方案3】：

试试这个方法（你可以添加与用户的 JOIN）：

SELECT 
    ps.user_id,
    SUM(COALESCE(ps."pts_won", 0) + COALESCE (qs."pts_won", 0) ) as pts_won,
    SUM(ps."good_gap") as good_gap,
    SUM(ps."good_score") as good_score,
    SUM(ps.good_winner) as good_winner

FROM (SELECT ps.user_id,
             SUM(ps."pts_won") AS pts_won,
             SUM(ps."good_gap") as good_gap,
             SUM(ps."good_score") as good_score,
             SUM(ps."isGoodPrognosis"::INT) as good_winner
      FROM prognosis_scores ps
      GROUP BY ps.user_id) AS ps
      
LEFT JOIN question_scores qs ON qs.user_id = ps.user_id
  
GROUP BY ps.user_id 
ORDER BY ps.user_id;