SQL:两个没有重复的聚合函数

Question

如何使用两个聚合函数(例如 double string_agg 或只是 sum )，但确保结果不包含由另一个聚合函数引起的重复(由第二次连接引起)？我使用 PostgreSQL。

示例

我有三个表:

create table boxes
(
    id   bigserial primary key,
    name varchar(255)
);

create table animals
(
    id     bigserial primary key,
    name   varchar(255),
    age    numeric,
    box_id bigint constraint animals_boxes_id references boxes
);

create table vegetables
(
    id     bigserial primary key,
    name   varchar(255),
    weight numeric,
    box_id bigint constraint vegatables_box_id references boxes
);

一些输入数据:

insert into boxes (name) values ('First box');
insert into animals (box_id, name, age) values (1, 'Cat', 2);
insert into animals (box_id, name, age) values (1, 'Cat', 3);
insert into animals (box_id, name, age) values (1, 'Dog', 5);
insert into vegetables (box_id, name, weight) values (1, 'Tomato', 20);
insert into vegetables (box_id, name, weight) values (1, 'Cucumber', 30);
insert into vegetables (box_id, name, weight) values (1, 'Potato', 50);

我想在框中获取动物名称:

select b.name                                 as box_name,
       string_agg(a.name, ', ' order by a.id) as animal_names
from boxes as b
         left join animals a on b.id = a.box_id
group by b.name;

它有效:

<表类=“s-表”> <标题> box_name animal_names <正文> 第一个框 猫，猫，狗

但我也想得到蔬菜名称。但它不起作用:

select b.name                                 as box_name,
       string_agg(a.name, ', ' order by a.id) as animal_names,
       string_agg(v.name, ', ' order by v.id) as vegatable_names
from boxes as b
         left join animals a on b.id = a.box_id
         left join vegetables v on b.id = v.box_id
group by b.name;

它会产生重复的动物名称和蔬菜名称:

<表类=“s-表”> <标题> box_name animal_names vegatable_names <正文> 第一个框 猫，猫，猫，猫，猫，猫，狗，狗，狗 番茄，番茄，番茄， cucumber ， cucumber ， cucumber ，土 bean ，土 bean ，土 bean

结果应该是:

<表类=“s-表”> <标题> box_name animal_names vegatable_names <正文> 第一个框 猫，猫，狗 番茄、 cucumber 、土 bean

我不能只添加 distinct删除重复项，因为:

• 表中的名称可以重复(两只动物的名称为 Cat )。如果我使用 distinct它产生Cat, Dog而不是Cat, Cat, Dog .
• 我使用order by在string_agg (添加 ERROR: in an aggregate with DISTINCT, ORDER BY expressions must appear in argument list 时会导致 distinct )。即使我删除 order by ( string_agg(distinct a.name, ', ') ) 我无法使用它，因为第一点。

其他信息

它适用于所有聚合函数:string_agg , array_agg , json_object_agg甚至sum .

动物的总年龄:

select sum(a.age)
from boxes as b
         left join animals a on b.id = a.box_id
         -- left join vegetables v on b.id = v.box_id
group by b.name;

如果没有第二次连接，它会正确计算( 10 )，但计算错误( 30 ) - 由于重复。

Answer 1

基本问题解释如下:

• Two SQL LEFT JOINS produce incorrect result

对于较小的选择，每行聚合通常会更快。
使用LATERAL子查询(更通用):

SELECT b.name AS box_name, a.*, v.*
FROM   boxes b
LEFT  JOIN LATERAL (
   SELECT string_agg(a.name, ', ' ORDER BY a.id) AS animal_names
   FROM   animals a
   WHERE  a.box_id = b.id
   ) a ON true
LEFT  JOIN LATERAL (
   SELECT string_agg(v.name, ', ' ORDER BY v.id) AS vegetable_names
   FROM   vegetables v
   WHERE  v.box_id = b.id
   ) v ON true;

或者使用相关子查询(更简单，通常更快一点):

SELECT b.name AS box_name
    , (SELECT string_agg(a.name, ', ' ORDER BY a.id)
       FROM   animals a
       WHERE  a.box_id = b.id)  AS animal_names
    , (SELECT string_agg(v.name, ', ' ORDER BY v.id)
       FROM   vegetables v
       WHERE  v.box_id = b.id) AS vegetable_names
FROM   boxes b;

参见:

• What is the difference between a LATERAL JOIN and a subquery in PostgreSQL?
• Multiple array_agg() calls in a single query

聚合整个表时，速度更快:

SELECT b.name AS box_name, a.animal_names, v.vegetable_names
FROM   boxes b
LEFT   JOIN (
   SELECT box_id, string_agg(a.name, ', ') AS animal_names   
   FROM  (
      SELECT box_id, id, name
      FROM   animals a
      ORDER  BY box_id, id
      ) a
   GROUP  BY 1
   ) a ON a.box_id = b.id
LEFT   JOIN (
   SELECT box_id, string_agg(v.name, ', ') AS vegetable_names 
   FROM  (
      SELECT box_id, id, name
      FROM   vegetables v
      ORDER  BY box_id, id
      ) v
   GROUP  BY 1   
   ) v ON v.box_id = b.id;

fiddle

请注意我如何在子查询中排序，这通常比按聚合排序更快。可选的优化。

旁白:关于测试设置中的 varchar(255):

• Should I add an arbitrary length limit to VARCHAR columns?