在一个查询中加入 2 个 sql avg 表答案

【问题标题】：join 2 sql avg table in one query在一个查询中加入 2 个 sql avg 表
【发布时间】：2023-01-13 04:47:55
【问题描述】：

所以我有这个 SQL 代码，它显示了第二个表中也存在的所有蛋白质的平均长度，称为 mrna_pieces。

select AVG(protein_length)
FROM protein
WHERE exists
(select protein_id from mrna_pieces where mrna_brokstukken.protein_id = protein.protein_id)

问题是我也想显示完全相同但对于第二个表中不存在的所有蛋白质。

select AVG(protein_length)
FROM protein
WHERE exists
(select protein_id from mrna_pieces where mrna_brokstukken.protein_id != protein.protein_id)

但我希望这两个部分像这样放在一张桌子上 table example

我试过这个

select AVG(eiwit_lengte) AS avglengthwith, AVG(eiwit_lengte) AS avglengthwithout
FROM eiwit
WHERE exists
(select eiwit_id from mrna_brokstukken where mrna_brokstukken.eiwit_id != eiwit.eiwit_id)
WHERE exists
(select eiwit_id from mrna_brokstukken where mrna_brokstukken.eiwit_id = eiwit.eiwit_id)

但这给了我以下错误：错误：pq：“WHERE”处或附近的语法错误

【问题讨论】：

标签： sql postgresql

【解决方案1】：

只需同时选择两个：

select (
  select AVG(protein_length)
  FROM protein
  WHERE exists
  (select protein_id from mrna_pieces where mrna_brokstukken.protein_id = protein.protein_id)
) as avgwith,
(
  select AVG(protein_length)
  FROM protein
  WHERE exists
  (select protein_id from mrna_pieces where mrna_brokstukken.protein_id != protein.protein_id)
)
as avgwithout

【讨论】：

【解决方案2】：

因为这两个查询都会返回一个标量值，您可以简单地使用 bot 作为子查询

SELECT

    (select AVG(protein_length)
    FROM protein
    WHERE exists
    (select protein_id from mrna_pieces where mrna_brokstukken.protein_id = protein.protein_id) ) AVG_1,
    (select AVG(protein_length)
    FROM protein
    WHERE exists
    (select protein_id from mrna_pieces where mrna_brokstukken.protein_id != protein.protein_id)) AVG_2

【讨论】：