Sql 按日期统计前5个值

我有一个缺陷描述和生产日期字段表，如下所示：

+----------+--------+ |缺陷描述|生产日| +----------+--------+ |吹制垫片| 2020-01-01| |吹制垫片| 2020-01-01| |吹制垫片| 2020-01-01| |爆胎| 2020-01-01| |吹风灯| 2020-01-01| |吹风灯| 2020-01-01| |吹风灯| 2020-01-01| |吹风灯| 2020-01-01| |爆胎| 2020-02-01| |爆胎| 2020-02-01| |爆胎| 2020-02-01| +----------+--------+ 我如何计算每天前5个缺陷的数量：

+--------+----------+----+ |生产日|缺陷|描述|计数| +--------+----------+----+ |2020-01-01吹风灯4| |2020-01-01 |吹制垫片| 3| |2020-01-01 |爆胎| 1| |2020-02-01爆胎3| +--------+----------+----+

---

这将在SQL Server中工作

;WITH daysCTE as
(
    SELECT DISTINCT productionday
    FROM YourTable
)
SELECT daysCTE.productionday, t.*
FROM daysCTE 
CROSS APPLY 
(
    SELECT TOP 5
       defect_description, 
       Count(*) NumberOfDefects
    FROM   YourTable
    WHERE  productionday= daysCTE.productionday
    GROUP BY defect_description
    ORDER BY 2 DESC
) t

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下面的查询应该在Postgres SQL中工作

select * from(
select *,rank() over (partition by productionday order by c desc ) as rank
from (
select defect_description, productionday, count(*) as c
        from tab
        group by productionday,defect_description
) A
) B
where rank <=5

---

我想这正是你想要的：

select t.*
from (select defect_description, productionday, count(*) as c,
             row_number() over (partition by productionday order by count(*) desc) as seqnum
      from tab
      group by productionday, defect_description
     ) t
where seqnum <= 5

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