字符串A为
(01)(12)(1201)(1300)(15)表T中有个字段ID,Bianhao
数据
ID Bianhao
1 00
2 01
3 12
4 1300
……怎样验证A中包含的编号在T中是否存在
若不存在则从串A中删除该编号
如结果:
(01)(12)(1300)求SQL语句
(01)(12)(1201)(1300)(15)表T中有个字段ID,Bianhao
数据
ID Bianhao
1 00
2 01
3 12
4 1300
……怎样验证A中包含的编号在T中是否存在
若不存在则从串A中删除该编号
如结果:
(01)(12)(1300)求SQL语句
set @str='(01)(12)(1201)(1300)(15)'
select * into #temp from table where charindex(Bianhao,@str)>0select @str=stuff((select '('+Bianhao+')' from #temp for xml path('')),1,1,'')) from #temp
(
ID int,
Bianhao varchar(20)
)
insert into aa
select 1,'00'
union select 2,'01'
union select 3,'12'
union select 4,'1300'declare @str varchar(200)
set @str='(01)(12)(1201)(1300)(15)'
select * into #temp from aa where charindex('('+Bianhao+')',@str)>0select @str='('+stuff((select '('+Bianhao+')' from #temp for xml path('')),1,1,'') from #tempprint @str--------------
(4 行受影响)(3 行受影响)
(01)(12)(1300)
(
ID int,
Bianhao varchar(20)
)
insert into aa
select 1,'00'
union select 2,'01'
union select 3,'12'
union select 4,'1300'declare @temp table (bianhao varchar(32))
declare @str2 varchar(max)
set @str2=''
declare @str1 varchar(200)--set @str1='(021)(212)(12201)(13040)(153)(003)'
set @str1='(021)(212)(12201)(13040)(153)(003)(00)'
insert into @temp
select bianhao from aa where charindex('('+Bianhao+')',@str1)>0select @str2=@str2+ '(' +t.bianhao+ ')' from @temp t inner join aa b
on t.bianhao=b.bianhao
print @str2
set @str='(01)(12)(1201)(1300)(15)'
select * into #temp from T where charindex('('+Bianhao+')',@str)>0select @str='('+stuff((select '('+Bianhao+')' from #temp for xml path('')),1,1,'') from #tempselect @str