如何在select子句中使用像SQL Server这样的from子句加入Postgresql子查询？

Question

我试图在postgresql上写下面的查询:

select name, author_id, count(1), 
    (select count(1)
    from names as n2
    where n2.id = n1.id
        and t2.author_id = t1.author_id
    )               
from names as n1
group by name, author_id

这肯定适用于Microsoft SQL Server,但它在postegresql上完全不起作用.我稍微阅读了它的文档,似乎我可以将其重写为:

select name, author_id, count(1), total                     
from names as n1, (select count(1) as total
    from names as n2
    where n2.id = n1.id
        and n2.author_id = t1.author_id
    ) as total
group by name, author_id

但是这会在postegresql上返回以下错误:"FROM中的子查询不能引用相同查询级别的其他关系".所以我被卡住了.有谁知道我怎么能做到这一点？

谢谢

Answer 1

我不确定我是否理解你的意图,但也许以下内容可能与你想要的一致:

select n1.name, n1.author_id, count_1, total_count
  from (select id, name, author_id, count(1) as count_1
          from names
          group by id, name, author_id) n1
inner join (select id, author_id, count(1) as total_count
              from names
              group by id, author_id) n2
  on (n2.id = n1.id and n2.author_id = n1.author_id)

不幸的是,这增加了按id以及name和author_id对第一个子查询进行分组的要求,我认为不需要.我不知道如何解决这个问题,因为你需要有id可以加入第二个子查询.也许其他人会提出更好的解决方案.

分享和享受.

Answer 2

我只是在这里回答我需要的最终sql的格式化版本,基于我在上面的评论中发布的Bob Jarvis答案:

select n1.name, n1.author_id, cast(count_1 as numeric)/total_count
  from (select id, name, author_id, count(1) as count_1
          from names
          group by id, name, author_id) n1
inner join (select author_id, count(1) as total_count
              from names
              group by author_id) n2
  on (n2.author_id = n1.author_id)

Answer 3

补充@Bob Jarvis和@dmikam回答,Postgres在你不使用LATERAL时没有执行一个好的计划,在模拟之下,在这两种情况下查询数据结果是相同的,但成本是非常不同的

表结构

CREATE TABLE ITEMS (
    N INTEGER NOT NULL,
    S TEXT NOT NULL
);

INSERT INTO ITEMS
  SELECT
    (random()*1000000)::integer AS n,
    md5(random()::text) AS s
  FROM
    generate_series(1,1000000);

CREATE INDEX N_INDEX ON ITEMS(N);

执行JOIN与GROUP BY子查询无LATERAL

EXPLAIN 
SELECT 
    I.*
FROM ITEMS I
INNER JOIN (
    SELECT 
        COUNT(1), n
    FROM ITEMS
    GROUP BY N
) I2 ON I2.N = I.N
WHERE I.N IN (243477, 997947);

结果

Merge Join  (cost=0.87..637500.40 rows=23 width=37)
  Merge Cond: (i.n = items.n)
  ->  Index Scan using n_index on items i  (cost=0.43..101.28 rows=23 width=37)
        Index Cond: (n = ANY ('{243477,997947}'::integer[]))
  ->  GroupAggregate  (cost=0.43..626631.11 rows=861418 width=12)
        Group Key: items.n
        ->  Index Only Scan using n_index on items  (cost=0.43..593016.93 rows=10000000 width=4)

运用 LATERAL

EXPLAIN 
SELECT 
    I.*
FROM ITEMS I
INNER JOIN LATERAL (
    SELECT 
        COUNT(1), n
    FROM ITEMS
    WHERE N = I.N
    GROUP BY N
) I2 ON 1=1 --I2.N = I.N
WHERE I.N IN (243477, 997947);

结果

Nested Loop  (cost=9.49..1319.97 rows=276 width=37)
  ->  Bitmap Heap Scan on items i  (cost=9.06..100.20 rows=23 width=37)
        Recheck Cond: (n = ANY ('{243477,997947}'::integer[]))
        ->  Bitmap Index Scan on n_index  (cost=0.00..9.05 rows=23 width=0)
              Index Cond: (n = ANY ('{243477,997947}'::integer[]))
  ->  GroupAggregate  (cost=0.43..52.79 rows=12 width=12)
        Group Key: items.n
        ->  Index Only Scan using n_index on items  (cost=0.43..52.64 rows=12 width=4)
              Index Cond: (n = i.n)

我的Postgres版本是 PostgreSQL 10.3 (Debian 10.3-1.pgdg90+1)

Answer 4

我知道这是旧的,但是自Postgresql 9.3以来,有一个选项可以使用关键字"LATERAL"在JOINS中使用RELATED子查询,因此问题中的查询看起来像:

SELECT 
    name, author_id, count(*), t.total
FROM
    names as n1
    INNER JOIN LATERAL (
        SELECT 
            count(*) as total
        FROM 
            names as n2
        WHERE 
            n2.id = n1.id
            AND n2.author_id = n1.author_id
    ) as t ON 1=1
GROUP BY 
    n1.name, n1.author_id