我有一张桌子
EmployeeID IndividualPay FamilyPay IsActive
1 200 300 true
1 100 150 false
Run Code Online (Sandbox Code Playgroud)
但我希望输出如下(我想使用此输出与其他表的内连接)
EmployeeID IndPay_IsActive IndPay_IsNotActive FamilyPay_IsActive FamilyPay_IsNotActive
1 200 100 300 150
Run Code Online (Sandbox Code Playgroud)
我已经调查过PIVOT,但不知道如何在我的情况下使用它.
Tar*_*ryn 26
这种类型的转换称为枢轴.您没有指定正在使用的数据库,但您可以CASE在任何系统中使用带有表达式的聚合函数:
select employeeid,
max(case when IsActive = 'true' then IndividualPay end) IndPay_IsActive,
max(case when IsActive = 'false' then IndividualPay end) IndPay_IsNotActive,
max(case when IsActive = 'true' then FamilyPay end) FamilyPay_IsActive,
max(case when IsActive = 'false' then FamilyPay end) FamilyPay_IsNotActive
from yourtable
group by employeeid
Run Code Online (Sandbox Code Playgroud)
根据您的数据库,如果您同时访问PIVOT和UNPIVOT函数,则可以使用它们来获取结果.该UNPIVOT函数将IndividualPay和FamilyPay列转换为行.完成后,您可以使用以下PIVOT函数创建四个新列:
select *
from
(
select employeeid,
case when isactive = 'true'
then col+'_IsActive'
else col+'_IsNotActive' end col,
value
from yourtable
unpivot
(
value
for col in (IndividualPay, FamilyPay)
) unpiv
) src
pivot
(
max(value)
for col in (IndividualPay_IsActive, IndividualPay_IsNotActive,
FamilyPay_IsActive, FamilyPay_IsNotActive)
) piv
Run Code Online (Sandbox Code Playgroud)
两者都给出相同的结果:
| EMPLOYEEID | INDIVIDUALPAY_ISACTIVE | INDIVIDUALPAY_ISNOTACTIVE | FAMILYPAY_ISACTIVE | FAMILYPAY_ISNOTACTIVE |
----------------------------------------------------------------------------------------------------------------
| 1 | 200 | 100 | 300 | 150 |
Run Code Online (Sandbox Code Playgroud)