Question

我正在尝试在postgresql上编写以下查询：

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？

由于

Answer 1

我不确定我是否理解你的意图，但也许以下内容会接近你想要的内容：

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。

分享并享受。

Answer 2

我只是在这里回答我需要的最终sql的格式化版本，基于我上面评论中发布的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在您不使用LATERAL时执行一个好的计划，在模拟之下，两种情况下查询数据结果相同，但成本差异很大

表格结构

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);

在没有JOIN的子查询中使用GROUP BY执行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)

使用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版本为PostgreSQL 10.3 (Debian 10.3-1.pgdg90+1)

Answer 4

我知道这已经过时了，但是自Postgresql 9.3以来，您可以选择使用关键字＆＃34; LATERAL＆＃34;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

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

4 个答案: