Question

我有一个查询导致一个不同长度的字段（6,8,10，...）和各自的计数如下所示：

"region","repeatLength","count"
"promoter","6","272387"
"promoter","8","86929"
"promoter","10","28337"
"promoter","12","8873"
"promoter","14","3080"
"promoter","16","1098"
"promoter","18","475"
"promoter","20","206"
"promoter","22","133"
"promoter","24","75"
"promoter","26","42"
"promoter","28","32"
"promoter","30","16"
"promoter","32","6"
"promoter","34","9"

此表由此次调用生成：

select region, repeatLength, count(*) as count 
from alignedRepeats
group by region, repeatLength;

我希望能够压缩这些结果，以便重复长度＆lt; 18是完整的，但是通过对计数字段求和，将重复长度> = 18聚合成一行。这可以在单个SQL查询中进行，而无需创建临时表并将它们联合起来吗？

很抱歉，如果这是一个简单的问题，我是一个SQL新手。

Answer 1

select region, 
    case when repeatLength >= 18 then ">=18" else repeatLength end as repeatLength, 
    count(*) as count  
from alignedRepeats 
group by region, 
    case when repeatLength >= 18 then ">=18" else repeatLength end;

Answer 2

您不必创建临时表。但是你可以使用联合：

select region, repeatLength, 1 as Colname
from alignedRepeats
where repeatLength < 18
union
select region, repeatLength, count(*)
from alignedRepeats
where repeatLength >= 18
group by region, repeatLength;

不确定您想要包含的最后一个字段，因此'1为Colname'

Answer 3

正确答案（已测试）

select
    region,
    repeatLength,
    count(*) as count
from
    alignedRepeats
where
    repeatLength < 18
group by
    region,
    repeatLength
union
select
    region,
    999,
    count(*)
from
    alignedRepeats
where
    repeatLength >= 18
group by
    region

注意：使用“999”表示聚合行。

Answer 4

引入另一个字段highGroup，如果repeatlength＆lt; = 18则为0，否则为1，然后在外部查询中的highGroupfield上进行分组。

想要总结一些字段并让其他人单独使用SQL

4 个答案: