我必须在一行中显示单个人的多种收入,收入类型和雇主姓名值。因此,如果'A'有来自三个不同来源的三种不同收入,
id | Name | Employer | IncomeType | Amount
123 | XYZ | ABC.Inc | EarningsformJob | $200.00
123 | XYZ | Self | Self Employment | $300.00
123 | XYZ. | ChildSupport| Support | $500.00
我需要将它们显示为
id | Name | Employer1 | Incometype1| Amount1 | Employer2 | incometype2 | Amount2| Employer3 | Incometype3| Amount3.....
123 |XYZ | ABC.Inc |EarningsformJob | $200.00|Self | Self Employment | $300.00|ChildSupport| Support | $500.00.....
我需要'固定数量的列'(我们知道雇主,收入类型和金额将要重复多少次)逻辑和'列的动态显示'(这些列将要重复的未知次数)
感谢。
答案 0 :(得分:10)
由于您使用的是SQL Server,因此可以通过多种方法将数据行转换为列。
聚合函数/ CASE:您可以将带有CASE表达式的聚合函数与row_number()
一起使用。此版本要求您具有已知数量的值才能成为列:
select id,
name,
max(case when rn = 1 then employer end) employer1,
max(case when rn = 1 then IncomeType end) IncomeType1,
max(case when rn = 1 then Amount end) Amount1,
max(case when rn = 2 then employer end) employer2,
max(case when rn = 2 then IncomeType end) IncomeType2,
max(case when rn = 2 then Amount end) Amount2,
max(case when rn = 3 then employer end) employer3,
max(case when rn = 3 then IncomeType end) IncomeType3,
max(case when rn = 3 then Amount end) Amount3
from
(
select id, name, employer, incometype, amount,
row_number() over(partition by id order by employer) rn
from yourtable
) src
group by id, name;
PIVOT / UNPIVOT:您可以使用UNPIVOT和PIVOT函数来获取结果。在应用数据透视之前,UNPIVOT会将您的多列Employer
,IncomeType
和Amount
转换为多行。您没有具体说明SQL Server的版本,假设您具有已知数量的值,那么您可以在SQL Server 2005+中使用以下内容,它使用CROSS APPLY和UNION ALL进行取消:
select id, name,
employer1, incometype1, amount1,
employer2, incometype2, amount2,
employer3, incometype3, amount3
from
(
select id, name, col+cast(rn as varchar(10)) col, value
from
(
select id, name, employer, incometype, amount,
row_number() over(partition by id order by employer) rn
from yourtable
) t
cross apply
(
select 'employer', employer union all
select 'incometype', incometype union all
select 'amount', cast(amount as varchar(50))
) c (col, value)
) src
pivot
(
max(value)
for col in (employer1, incometype1, amount1,
employer2, incometype2, amount2,
employer3, incometype3, amount3)
) piv;
动态版本:最后,如果您有一个未知数量的值,那么您将需要使用动态SQL来生成结果。
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT ',' + QUOTENAME(col+cast(rn as varchar(10)))
from
(
select row_number() over(partition by id order by employer) rn
from yourtable
) d
cross apply
(
select 'employer', 1 union all
select 'incometype', 2 union all
select 'amount', 3
) c (col, so)
group by col, rn, so
order by rn, so
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT id, name,' + @cols + '
from
(
select id, name, col+cast(rn as varchar(10)) col, value
from
(
select id, name, employer, incometype, amount,
row_number() over(partition by id order by employer) rn
from yourtable
) t
cross apply
(
select ''employer'', employer union all
select ''incometype'', incometype union all
select ''amount'', cast(amount as varchar(50))
) c (col, value)
) x
pivot
(
max(value)
for col in (' + @cols + ')
) p '
execute(@query);
见SQL Fiddle with Demo。所有版本都给出了结果:
| ID | NAME | EMPLOYER1 | INCOMETYPE1 | AMOUNT1 | EMPLOYER2 | INCOMETYPE2 | AMOUNT2 | EMPLOYER3 | INCOMETYPE3 | AMOUNT3 |
-------------------------------------------------------------------------------------------------------------------------------------
| 123 | XYZ | ABC.Inc | EarningsformJob | 200 | ChildSupport | Support | 500 | Self | Self Employment | 300 |