如何像 SQL Server 一样在 select 子句中使用 join in from 子句执行 Postgresql 子查询?

发布于 2024-09-04 11:14:22 字数 701 浏览 5 评论 0原文

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

这当然可以在 Microsoft SQL Server 上运行,但在 postegresql 上根本不起作用。我读了一下它的文档,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

但这会在 postegresql 上返回以下错误:“FROM 中的子查询无法引用相同查询级别的其他关系”。所以我被困住了。有谁知道我怎样才能实现这一目标?

谢谢

I am trying to write the following query on postgresql:

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

This would certainly work on Microsoft SQL Server but it does not at all on postegresql. I read its documentation a bit and it seems I could rewrite it as:

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

But that returns the following error on postegresql: "subquery in FROM cannot refer to other relations of same query level". So I'm stuck. Does anyone know how I can achieve that?

Thanks

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评论(5

筱武穆 2024-09-11 11:14:22

我不确定我是否完全理解您的意图,但也许以下内容会接近您想要的:

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,我认为这是不需要的。不过,我不确定如何解决这个问题,因为您需要有 id 才能加入第二个子查询。也许其他人会想出更好的解决方案。

I'm not sure I understand your intent perfectly, but perhaps the following would be close to what you want:

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)

Unfortunately this adds the requirement of grouping the first subquery by id as well as name and author_id, which I don't think was wanted. I'm not sure how to work around that, though, as you need to have id available to join in the second subquery. Perhaps someone else will come up with a better solution.

网名女生简单气质 2024-09-11 11:14:22

补充 @Bob Jarvis@dmikam 答案,当您不使用 LATERAL 时,Postgres 不会执行良好的计划,在模拟下面,在这两种情况下查询数据结果相同,但成本却截然不同

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);

在不使用 LATERAL 的子查询中使用 GROUP BY 执行 JOIN

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< 的 /code>

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)

Complementing @Bob Jarvis and @dmikam answer, Postgres don't perform a good plan when you don't use LATERAL, below a simulation, in both cases the query data results are the same, but the cost are very different

Table structure

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);

Performing JOIN with GROUP BY in subquery without 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);

The results

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)

Using 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);

Results

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)

My Postgres version is PostgreSQL 10.3 (Debian 10.3-1.pgdg90+1)

给我一枪 2024-09-11 11:14:22

我知道这已经很旧了,但是自从 Postgresql 9.3< /a> 有一个选项可以使用关键字“LATERAL”在 JOINS 内部使用 RELATED 子查询,因此问题中的查询如下所示:

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

I know this is old, but since Postgresql 9.3 there is an option to use a keyword "LATERAL" to use RELATED subqueries inside of JOINS, so the query from the question would look like:

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
云之铃。 2024-09-11 11:14:22

我只是根据上面评论中发布的 Bob Jarvis 答案,用我需要的最终 sql 的格式化版本来回答:

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)

I am just answering here with the formatted version of the final sql I needed based on Bob Jarvis answer as posted in my comment above:

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)
审判长 2024-09-11 11:14:22
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 distinct(author_id), count(1) as total_count
              from names) n2
  on (n2.author_id = n1.author_id)
Where true

如果更多的内连接使用distinct,因为更多的连接组性能很慢

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 distinct(author_id), count(1) as total_count
              from names) n2
  on (n2.author_id = n1.author_id)
Where true

used distinct if more inner join, because more join group performance is slow

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