某表tb,一列a
a
str1
str2
str3
str4
str5
str6
str7
str8
....
想把每两行的字符串连接起来,如
a
str1str2
str3str4
str5str6
......
再求高手
最佳答案1:
create table tb(a
varchar(10))
insert
into tb values('str1')
insert
into tb values('str2')
insert
into tb values('str3')
insert
into tb values('str4')
insert
into tb values('str5')
insert
into tb values('str6')
insert
into tb values('str7')
insert
into tb values('str8')
go
select a
= replace(stuff((select
','
+ a
from
(
select t.* , px
= (row_number()
over(order
by a)
- 1)/2
from tb t
) m where m.px
= n.px
for xml path('')) ,
1 ,
1 ,
''),',','')
from
(
select t.* , px
= (row_number()
over(order
by a)
- 1)/2
from tb t
) n
group
by px
drop
table tb
/*
a
------------
str1str2
str3str4
str5str6
str7str8
(4 行受影响)
*/
最佳答案2:
create table tb(a
varchar(10))
insert
into tb values('str1')
insert
into tb values('str2')
insert
into tb values('str3')
insert
into tb values('str4')
insert
into tb values('str5')
insert
into tb values('str6')
insert
into tb values('str7')
insert
into tb values('str8')
go
--创建函数解决:
create
function dbo.f_str(@px
int)
returns varchar(1000)
as
begin
declare
@str varchar(1000)
select
@str =
isnull(@str ,
'')
+ cast(a
as varchar)
from (select t.* , px
= (((select
count(1)
from tb
where a < t.a)
+ 1)
- 1)/2
from tb t) m
where px
= @px
return
@str
end
go
--调用函数
select dbo.f_str(px) a
from
(
select t.* , px
= (((select
count(1)
from tb
where a < t.a)
+ 1)
- 1)/2
from tb t
) m
group
by px
drop
function dbo.f_str
drop
table tb
/*
a
----------------------
str1str2
str3str4
str5str6
str7str8
(所影响的行数为 4 行)
*/