Question

我有一个Emptbl我有EmpType列。

在EmpType中，我有以下数据：

现在我想只获得那些具有相同EmpType的emp数据，例如以下数据：

并希望将此数据显示为0123和1245

那么如何获得以上数据？我使用UNION但它没有获得有效的数据。

由于

Answer 1

试试这个：

select substring(emptype, 2, len(emptype))
from emptbl
group by substring(emptype, 2, len(emptype))
having count(*) > 1

硬编码2基于您的样本数据。相反，如果您在数字部分之前有任意数量的字母，例如'ABCDEFG0123'，您可以使用patindex来获取子字符串的起始索引，如此;

select substring(emptype, patindex('%[0-9]%',emptype), len(emptype)

Answer 2

;with CTE as (Select Name,SUBSTRING(Name,2,5) as Number, ROW_NUMBER() 
        OVER (PARTITION By SUBSTRING(Name,2,5) ORDER BY Name) AS Row
from #Temp)

Select Temp.Name
From CTE C
     Cross Apply (Select Name FRom CTE T Where T.Number=C.Number) as Temp
Where C.Row>1

以下是fiddle sample

Answer 3

Select A.*
    From EmpTbl A
        Inner Join
    EmpTbl B
        On  SubString(A.EmpType, 2, 4) = SubString(B.EmpType, 2, 4) And
        SubString(A.EmpType, 1, 1) <> SubString(B.EmpType , 1, 1)

Answer 4

select top 4 id from
(
select id,rn=row_number()over(partition by right(id,3) order by right(id,3)) from #t
)x

SEE IT LIVE

Answer 5

这是最小的查询：

select substr(a.emptype, 2) num
join emptbl a
join emptbl b on substr(a.emptype, 2) = substr(b.emptype, 2)
and a.emptype != b.emptype

使用group by从表中获取数据

5 个答案: