多个array_agg（）调用单个查询

Question

我试图用我的查询来完成某些事情，但它并没有真正起作用。我的应用程序曾经有一个mongo db，所以应用程序用于在一个字段中获取数组，现在我们不得不改为Postgres，我不想更改我的应用程序代码以保持v1工作。

为了在Postgres中的1个字段中获取数组，我使用了array_agg()函数。到目前为止这个工作正常。但是，我需要另一个不同的表中的字段中的另一个数组。

例如：

我有我的员工。员工有多个地址，有多个工作日。

SELECT name, age, array_agg(ad.street) FROM employees e 
JOIN address ad ON e.id = ad.employeeid
GROUP BY name, age

现在这对我来说很好，这会导致例如：

| name  | age| array_agg(ad.street)
| peter | 25 | {1st street, 2nd street}|

现在我想在工作日加入另一张桌子，所以我这样做：

SELECT name, age, array_agg(ad.street), arrag_agg(wd.day) FROM employees e 
JOIN address ad ON e.id = ad.employeeid 
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY name, age

这导致：

| peter | 25 | {1st street, 1st street, 1st street, 1st street, 1st street, 2nd street, 2nd street, 2nd street, 2nd street, 2nd street}| "{Monday,Tuesday,Wednesday,Thursday,Friday,Monday,Tuesday,Wednesday,Thursday,Friday}

但我需要结果：

| peter | 25 | {1st street, 2nd street}| {Monday,Tuesday,Wednesday,Thursday,Friday}

我知道它与我的连接有关，因为多个连接行多次但我不知道如何实现这一点，任何人都可以给我正确的提示吗？

Answer 1

DISTINCT通常用于修复从内部腐烂的查询，这通常很慢和/或不正确。不要将行数乘以开头，然后您不必在最后对不需要的重复项进行排序。

一次加入多个n表（“有很多”）会使结果集中的行相乘。这就像代理商的CROSS JOIN或Cartesian product ：

• Two SQL LEFT JOINS produce incorrect result

有多种方法可以避免这种错误。

首先聚合，稍后加入

从技术上讲，只要您在聚合之前一次加入多行 一个 表，该查询就会起作用：

SELECT e.id, e.name, e.age, e.streets, arrag_agg(wd.day) AS days
FROM  (
   SELECT e.id, e.name, e.age, array_agg(ad.street) AS streets
   FROM   employees e 
   JOIN   address  ad ON ad.employeeid = e.id
   GROUP  BY e.id    -- id enough if it is defined PK
   ) e
JOIN   workingdays wd ON wd.employeeid = e.id
GROUP  BY e.id, e.name, e.age;

最好还包括主键id和GROUP BY，因为name和age不一定是唯一的。你可能错误地合并了两名员工。

但是你可以在加入之前在子查询中进行聚合，除非你在WHERE上有选择性的employees条件，否则这是优越的：

SELECT e.id, e.name, e.age, ad.streets, arrag_agg(wd.day) AS days
FROM   employees e 
JOIN  (
   SELECT employeeid, array_agg(ad.street) AS streets
   FROM   address
   GROUP  BY 1
   ) ad ON ad.employeeid = e.id
JOIN   workingdays wd ON e.id = wd.employeeid
GROUP  BY e.id, e.name, e.age, ad.streets;

或聚合两者：

SELECT name, age, ad.streets, wd.days
FROM   employees e 
JOIN  (
   SELECT employeeid, array_agg(ad.street) AS streets
   FROM   address
   GROUP  BY 1
   ) ad ON ad.employeeid = e.id
JOIN  (
   SELECT employeeid, arrag_agg(wd.day) AS days
   FROM   workingdays
   GROUP  BY 1
   ) wd ON wd.employeeid = e.id;

如果您检索基表中的所有或大多数行，则最后一个通常更快。

请注意，使用JOIN而非LEFT JOIN会从结果中删除没有地址或没有工作日的员工。这可能是也可能不是。切换到LEFT JOIN以在结果中保留 所有 员工。

相关子查询/ LATERAL join

对于小选择，我会考虑相关的子查询：

SELECT name, age
    , (SELECT array_agg(street) FROM address WHERE employeeid = e.id) AS streets
    , (SELECT arrag_agg(day) FROM workingdays WHERE employeeid = e.id) AS days
FROM   employees e
WHERE  e.namer = 'peter';  -- very selective

或者，使用Postgres 9.3或更高版本，您可以使用LATERAL联接：

SELECT e.name, e.age, a.streets, w.days
FROM   employees e
LEFT   JOIN LATERAL (
   SELECT array_agg(street) AS streets
   FROM   address
   WHERE  employeeid = e.id
   GROUP  BY 1
   ) a ON true
LEFT   JOIN LATERAL (
   SELECT array_agg(day) AS days
   FROM   workingdays
   WHERE  employeeid = e.id
   GROUP  BY 1
   ) w ON true
WHERE  e.name = 'peter';  -- very selective

• What is the difference between LATERAL and a subquery in PostgreSQL?

任一查询都会在结果中保留 所有 员工。

Answer 2

每当您需要不重复的值时，请使用DISTINCT，如下所示：

SELECT name, age, array_agg(DISTINCT ad.street), array_agg(DISTINCT wd.day) FROM employees e 
JOIN address ad ON e.id = ad.employeeid 
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY name, age