如何一次从三个表中 return 行？

Question

如何将这三个查询合并为一个？

1.

SELECT "Skills"."name", "Skills"."id", "TrainerScores"."fellow_uid", MIN("TrainerScores"."score") AS "score"
FROM "TrainerScores"
INNER JOIN "Skills" ON "TrainerScores"."skill_id" = "Skills"."id"
WHERE "TrainerScores"."fellow_uid" = 'google:105697533513134511631'
AND DATE("TrainerScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id", "TrainerScores"."fellow_uid"

2.

Select "Skills"."name", "Skills"."id", MIN("PeerScores"."score") AS "score"
FROM "PeerScores"
LEFT OUTER JOIN "Skills" ON "PeerScores"."skill_id" = "Skills"."id"
WHERE "PeerScores"."evaluatee_uid" = 'google:105697533513134511631'
AND DATE("PeerScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id"

3.

Select "Skills"."name", "Skills"."id", MIN("SelfScores"."score") AS "score"
FROM "SelfScores"
LEFT OUTER JOIN "Skills" ON "SelfScores"."skill_id" = "Skills"."id"
WHERE "SelfScores"."fellow_uid" = 'google:105697533513134511631'
AND DATE("SelfScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id"

我想将其用作报告，我不想在任何时候想要获取数据时调用每个查询。

Answer 1

我提出了一个完全符合您要求的解决方案，但它确实有效。使用 raw 查询，您可以运行并获得多个查询的结果，如下所示：

var sequelize = require('./libs/pg_db_connect');

var query = "SELECT Skills.name, Skills.id, TrainerScores.fellow_uid, MIN(TrainerScores.score) AS score
FROM TrainerScores
INNER JOIN Skills ON TrainerScores.skill_id = Skills.id
WHERE TrainerScores.fellow_uid = 'google:105697533513134511631' AND DATE(TrainerScores.created_at) BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY Skills.name, Skills.id, TrainerScores.fellow_uid";


sequelize.query(query, {
   type: sequelize.QueryTypes.SELECT
}).success(function (query1) {
   done = _.after(query1.length, function () {
      callback(query1)
   })

   query = "Select Skills.name, Skills.id, MIN(PeerScores.score) AS score
        FROM PeerScores
        LEFT OUTER JOIN Skills ON PeerScores.skill_id = Skills.id
        WHERE PeerScores.evaluatee_uid = 'google:105697533513134511631' AND DATE(PeerScores.created_at) BETWEEN '2015-10-01' AND '2015-10-30'
        GROUP BY Skills.name, Skills.id";

   sequelize.query(query, {
      type: sequelize.QueryTypes.SELECT
   }).success(function (query2) {

      query = "Select Skills.name, Skills.id, MIN(SelfScores.score) AS score
        FROM SelfScores
        LEFT OUTER JOIN Skills ON SelfScores.skill_id = Skills.id
        WHERE SelfScores.fellow_uid = 'google:105697533513134511631' AND DATE(SelfScores.created_at) BETWEEN '2015-10-01' AND '2015-10-30'
        GROUP BY Skills.name, Skills.id";

      sequelize.query(query, {
         type: sequelize.QueryTypes.SELECT
      }).success(function (query3) {
         console.log(query1); // show the returns of query 1
         console.log(query2); // show the returns of query 2
         console.log(query3); // show the returns of query 3
      });

success函数的结果sequelize.query也可以存储在json变量中。

Answer 2

备选方案 1，只是一个巨大的 UNION ALL:

SELECT "Skills"."name", "Skills"."id", "TrainerScores"."fellow_uid", MIN("TrainerScores"."score") AS "score"
FROM "TrainerScores"
INNER JOIN "Skills" ON "TrainerScores"."skill_id" = "Skills"."id"
WHERE "TrainerScores"."fellow_uid" = 'google:105697533513134511631'
AND DATE("TrainerScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id", "TrainerScores"."fellow_uid"

UNION ALL

Select "Skills"."name", "Skills"."id", NULL, MIN("PeerScores"."score") AS "score"
FROM "PeerScores"
LEFT OUTER JOIN "Skills" ON "PeerScores"."skill_id" = "Skills"."id"
WHERE "PeerScores"."evaluatee_uid" = 'google:105697533513134511631'
AND DATE("PeerScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id"

UNION ALL

Select "Skills"."name", "Skills"."id", NULL, MIN("SelfScores"."score") AS "score"
FROM "SelfScores"
LEFT OUTER JOIN "Skills" ON "SelfScores"."skill_id" = "Skills"."id"
WHERE "SelfScores"."fellow_uid" = 'google:105697533513134511631'
AND DATE("SelfScores"."created_at") BETWEEN '2015-10-01' AND '2015-10-30'
GROUP BY "Skills"."name", "Skills"."id"

Answer 3

基本上，像一样使用 UNION ALL。
"Combining Queries".

章节中手册中的详细信息

但是还有很多。我有根据的猜测，你真的想要这个：

WITH vals AS (SELECT timestamp '2015-10-01 00:00' AS ts_low  -- incl. lower bound
                   , timestamp '2015-10-31 00:00' AS ts_hi   -- excl. upper bound
                   , text 'google:105697533513134511631' AS uid)
SELECT s.name, sub.*
FROM  (
   SELECT skill_id AS id, min(score) AS score, 'T' AS source
   FROM   "TrainerScores", vals v
   WHERE  fellow_uid =  v.uid
   AND    created_at >= v.ts_low
   AND    created_at <  v.ts_hi
   GROUP  BY 1

   UNION ALL
   SELECT skill_id, min(score), 'P'
   FROM   "PeerScores", vals v
   WHERE  evaluatee_uid = v.uid
   AND    created_at >= v.ts_low
   AND    created_at <  v.ts_hi
   GROUP  BY 1

   UNION ALL
   SELECT skill_id, min(score), 'S'
   FROM   "SelfScores", vals v
   WHERE  fellow_uid =  v.uid
   AND    created_at >= v.ts_low
   AND    created_at <  v.ts_hi
   GROUP  BY 1
   ) sub
JOIN   "Skills" s USING (id);

要点

首先，我trim从你的语法（可能由你的 ORM 产生）中消除噪音，使其易于阅读：删除多余的双引号，添加 table 别名,trim 杂词...
您对 LEFT [OUTER] JOIN 的使用被破坏了，因为您过滤了左侧 table 的列，这抵消了 LEFT JOIN。替换为 [INNER] JOIN.
在 WHERE 子句中使用 sargable 表达式，否则您的查询不能使用普通索引，并且对于大 tables 会非常慢。相关：
- Get difference of another field between first and last timestamps of grouping
在 CTE（WITH 子句）中一次提供您的参数 - 这是不需要的在准备好的语句中，您将 uid、ts_low 和 ts_hi 作为参数传递。
我从您的第一个查询的输出中删除了 "TrainerScores"."fellow_uid" 以简化查询。无论如何，这只是您的输入参数。
您可以在加入[=21=之前聚合您各自的主要table ] 一次.

我添加了一列 source 来表示每一行的来源。

旁白：您似乎想匹配整个 2015 年 10 月，但随后排除了 10 月 31 日。这是故意的吗？

如何一次从三个表中 return 行？

How to return rows from three tables at once?

sql

postgresql

node.js

sequelize.js

要点