单个查询中有多个array_agg（）调用 [英] Multiple array_agg() calls in a single query

问题描述

我正在尝试通过查询完成某些操作，但实际上并没有用。我的应用程序以前有一个mongo数据库，因此该应用程序用于在字段中获取数组，现在我们不得不更改为Postgres，并且我不想更改我的应用程序代码以保持v1的正常运行。

I'm trying to accomplish something with my query but it's not really working. My application used to have a mongo db so the application is used to get arrays in a field, now we had to change to Postgres and I don't want to change my applications code to keep v1 working.

为了在Postgres中的1个字段中获得数组，我使用了 array_agg（）函数。到目前为止，这个工作还不错。但是，我正要在另一个表中的字段中需要另一个数组。

In order to get arrays in 1 field within Postgres I used array_agg() function. And this worked fine so far. However, I'm at a point where I need another array in a field from another different table.

例如：

我有我的员工。员工有多个地址，并且有多个工作日。

I have my employees. employees have multiple address and have multiple workdays.

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例如：

Now this worked fine for me, this would result in for example:

| 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：

Now I want to join another table for working days so I do:

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

这将导致：

| 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}

但我需要得到结果：

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

我知道这与我的联接有关，因为有多个联接多个行，但我不知道该如何完成，有人可以给我正确的提示吗？

I understand it has to do with my joins, because of the multiple joins the rows multiple but I don't know how to accomplish this, can anyone give me the correct tip?

推荐答案

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，不必先增加行，然后就不必排序

DISTINCT is often applied to repair queries that are rotten from the inside, and that's often slow and / or incorrect. Don't multiply rows to begin with, then you don't have to sort out unwanted duplicates at the end.

一次加入多个n表（有很多）将结果集中的行相乘，就像 CROSS JOIN 或笛卡尔产品 通过代理人：

Joining to multiple n-tables ("has many") at once multiplies rows in the result set. That's like a CROSS JOIN or Cartesian product by proxy:

• 两个SQL LEFT JOINS产生错误的结果

• Two SQL LEFT JOINS produce incorrect result

有多种方法可以避免此错误。

There are various ways to avoid this mistake.

从技术上讲，只要您加入 one ，该查询就起作用表格中一次汇总多个行：

Technically, the query works as long as you join to one table with multiple rows at a time before you aggregate:

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;

最好也包含主键 id 和 GROUP BY 它，因为名称和 age 不一定独特。您可能会错误地合并两个雇员。

It's also best to include the primary key id and GROUP BY it, because name and age are not necessarily unique. You could merge two employees by mistake.

但是您可以在加入之前在子查询中进行汇总，除非您有选择地在雇员的条件：

But you can aggregate in a subquery before you join, that's superior unless you have selective WHERE conditions on employees:

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;

或将两者总计：

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;

最后一个通常更快，如果您检索全部或大部分

请注意，请使用 JOIN 而不是 LEFT JOIN 从结果中删除没有地址或没有工作日的员工。这可能是预期的，也可能不是预期的。切换到 LEFT JOIN 保留结果中的 all 名员工。

Note that using JOIN and not LEFT JOIN removes employees from the result who have no address or no workingdays. That may or may not be intended. Switch to LEFT JOIN to retain all employees in the result.

对于少量选择，我会考虑使用相关子查询：

For a small selection, I would consider correlated subqueries instead:

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，对于Postgres 9.3或更高版本，您可以使用横向连接为此：

Or, with Postgres 9.3 or later, you can use LATERAL joins for that:

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

• 在PostgreSQL中，LATERAL和子查询之间有什么区别？


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

这两个查询都会在结果中保留 all 个雇员。

Either query retains all employees in the result.

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