使用 PostgreSQL 13 上的另一列使用 string_agg 订购 DISTINCT？

Question

我有一个emails表：

CREATE TABLE public.emails (
  id                bigint NOT NULL PRIMARY KEY GENERATED BY DEFAULT AS IDENTITY
    (MAXVALUE 9223372036854775807),
  name  text not null
);

我有一个contacts表：

CREATE TABLE public.contacts (
  id                bigint NOT NULL PRIMARY KEY GENERATED BY DEFAULT AS IDENTITY
    (MAXVALUE 9223372036854775807),
  email_id            bigint NOT NULL,
  full_name            text NOT NULL,
  ordering  int  not null
);

并记录如下：

insert into emails (name) VALUES ('dennis1');
insert into emails (name) VALUES ('dennis2');

insert into contacts (id, email_id, full_name, ordering) VALUES (5, 1, 'dennis1', 9);
insert into contacts (id, email_id, full_name, ordering) VALUES (6, 2, 'dennis1', 5);
insert into contacts (id, email_id, full_name, ordering) VALUES (7, 2, 'dennis5', 1);
insert into contacts (id, email_id, full_name, ordering) VALUES (8, 1, 'john', 2);

我的查询获取数据如下：

SELECT
  "emails"."name",
  STRING_AGG(DISTINCT CAST("contacts"."id" AS TEXT), ','
  ORDER BY CAST("contacts"."id" AS TEXT)) AS "contact_ids"
FROM "emails"
  INNER JOIN "contacts"
    ON ("contacts"."email_id" = "emails"."id")
WHERE "emails"."id" > 0
GROUP BY "emails"."name"
ORDER BY "emails"."name" DESC LIMIT 50
  

实际结果

name       contact_ids
dennis2    6,7
dennis1    5,8

预期结果

name       contact_ids
dennis2    7,6
dennis1    8,5

我想根据ordering列作为DESC订购contact_ids ，但我不想获取ordering列。 只需使用它来订购联系人的 ID。

如何根据ordering列对contact_ids的每个id进行排序？

演示： https://dbfiddle.uk/?rdbms=postgres_12&fiddle=4d6851ec67b579608427bb399eae5891

Answer 1

如果您在加入之前聚合，则不需要 DISTINCT，您可以随意订购任何您想要的东西：

select em.name,
       c.contact_ids 
from emails em
  join (       
    select email_id, string_agg(id::text, ',' order by ordering desc) as contact_ids
    from contacts
    group by email_id
  ) c on c.email_id = em.id
order by em.name desc 
limit 50;

在线示例

Answer 2

演示：db<>小提琴

我想，省略排序列是因为问题，你不能在聚合中使用它DISTINCT 。

因此，您可能可以在聚合之前执行DISTINCT ：

SELECT
    "name",
    STRING_AGG(CAST("id" AS TEXT), ',' 
        ORDER BY "ordering") AS "contact_ids"
FROM (
  SELECT DISTINCT ON ("contacts"."id")
    "emails"."name",
    "contacts"."id",
    "contacts"."ordering"
  FROM "emails"
    INNER JOIN "contacts"
      ON ("contacts"."email_id" = "emails"."id")
  WHERE "emails"."id" > 0
  ORDER BY "contacts"."id"
) s
GROUP BY "name"
ORDER BY "name" DESC LIMIT 50