SQL练习题之联接
联接算法是指sqlserver用于处理联接的物理策略,它支持三种联接算法,嵌套循环(nested loops)合并(merge)以及哈希算法.
练习题:
create table dbo.arrays(
arrid varchar(10) not null primary key,
array varchar(8000) not null
)
go
insert into arrays(arrid,array)
values
('a','20,223,2544,25567,14'),
('B','30,-23433,28'),
('C','12,10,8099,12,1200,13,12,14,10,9'),
('D','-4,--6,-45678,-2')
arrid array
A 20,223,2544,25567,14
B 30,-23433,28
C 12,10,8099,12,1200,13,12,14,10,9
D -4,--6,-45678,-2
想要查询出的结果是:
arrid pos element
A 1 20
A 2 223
.......
本文摘自inside sqlserver 2008 tsql query
SELECT arrid,
array,
substring(array,n,CHARINDEX(',',array+',',n)-n) as element
,n
,ROW_NUMBER() over(partition by arrid order by n) as rowno
from dbo.arrays
join dbo.num
on n<=DATALENGTH(array)+1
and SUBSTRING(','+array,n,1)=','
with split
as
(
select arrid, 1 as pos ,1 as startpos,
CHARINDEX(',',array+',')-1 as endpos
from dbo.arrays
where DATALENGTH(array)>0
union all
select a.arrid,
a.pos+1,
a.endpos+2,
CHARINDEX(',',b.array+',',a.endpos+2)-1
from split a join
dbo.arrays b
on a.arrid = b.arrid
and CHARINDEX(',',b.array+',',a.endpos+2)>0
)
select
a.arrid,
a.pos,
a.startpos,
a.endpos,
SUBSTRING(b.array,a.startpos,a.endpos-a.startpos+1) as ele,
b.array
from split a join dbo.arrays b
on a.arrid=b.arrid
order by a.arrid
使用了两种不同的方法,第一种方法使用了数字辅助表来生成副本.第二种方法使用CTE.
第二种方法的定位点是查出字符串中的第一个元素,递归部分是查出下一个元素的值,至到找不到,.
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