Question

Tbl 1

ID Name Email
1  A    a@b.c
2  B    b@c.d

Tbl2

ID Related_ID Value Name
1  1          Z     Name1
2  1          Y     Name2
3  2          X     Name1
4  2          W     Name2
5  2          G     Name3

我可以写一个查询来显示

ID Name Email Value1
1  A    a@b.c Z
1  A    a@b.c Y

我怎么写它才会变得像

ID Name Email Value1 Value2
1  A    a@b.c Z      Y
2  B    b@c.d X      W

Answer 1

可以使用PIVOT声明来完成以下是您可以查看的几个链接 http://blog.sqlauthority.com/2008/06/07/sql-server-pivot-and-unpivot-table-examples/

Answer 2

以下是使用动态sql获取所需内容的示例。

请注意，CTE测试和临时表#tmp2应映射到原始表。

create table #tmp (ID int, Related_ID int, Value nvarchar(50), Name nvarchar(50), rn int)
delete #tmp

create table #tmp2 (ID int, Name nvarchar(50), Email nvarchar(50))
delete #tmp2

;with test(ID, Related_ID, Value, Name)
as
(
select 1,  1,          'Z',     'Name1'
union all
select 2,  1,          'Y',     'Name2'
union all
select 3,  2,          'X',     'Name1'
union all
select 4,  2,          'W',     'Name2'
union all
select 5,  2,          'G',     'Name3'
)

insert into #tmp
select *, row_number() OVER (partition by Related_ID order by ID) as rn
from test

insert into #tmp2
select 1,  'A',    'a@b.c'
union all
select 2,  'B',    'b@c.d'


declare @d nvarchar(MAX)
        ,@e nvarchar(MAX)

SET     @d = 'select a.ID, a.Name, a.Email '
SET     @e =   ',min(case b.rn when >rn< then Value else null end) as [Value>rn<]'

select  @d = @d + (
select t from(
select  distinct replace(@e,'>rn<', b.rn) as [text()]
from    #tmp b
for xml path('')) c(t)
)

set     @d = @d + '
from    #tmp2 a
join    #tmp b
    on  a.ID = b.Related_ID
  group by a.ID, a.Name, a.Email'

exec sp_executesql @d

drop table #tmp
drop table #tmp2

Answer 3

declare @t1 table(id int, name varchar(20), email varchar(32)) 
declare @t2 table(id int, related_id int, value varchar(10), name varchar(10))

insert @t1 values(1,'A', 'a@b.c')
insert @t1 values(2,'B', 'b@c.d')

insert @t2 values(1, 1, 'Z', 'Name1')
insert @t2 values(2, 1, 'Y', 'Name2')
insert @t2 values(3, 2, 'X', 'Name1')
insert @t2 values(4, 2, 'W', 'Name2')
insert @t2 values(5, 2, 'G', 'Name3')

;with a as
(
select value, related_id, ROW_NUMBER() over(PARTITION BY related_id order by value desc) rn
from @t2
), b as
(
select value, related_id, rn from a where rn in (1,2)
)
select t5.ID, t5.Name, t5.Email, t3.value Value1,  t4.value Value2 
from @t1 t5
join b t3
on t3.related_id = t5.id and t3.rn = 1
join b t4
on t4.related_id = t5.id and t4.rn = 2

结果：

ID     Name       Email  Value1   Value2
------ ------ ---------- -------- ----------
1      A      a@b.c      Z        Y
2      B      b@c.d      X        W

需要在SQL Server中进行调整

3 个答案: