http://topic.csdn.net/u/20120204/10/8b902fd2-8909-4ed9-b534-4a1a72454eff.html#r_77443331
需求简介:
比如:YYYYN, 结果处理为:人1,人2,人3,人4
分析:
貌似就是一个字符串的替换,如果是'Y'替换成'人'+Y的所在位置的编号。
解决方案:
declare @T table (id int,col varchar(5)) insert into @T select 1,'YYYYN' union all select 2,'YNNNY' union all select 3,'NNNYN' union all select 4,'YYNYN' --1.第一种方法:case when --最直观的处理方式 select id,col=left(col,len(col)-1) from ( select id,col= case when left(col,1)='Y' then '人1,' else '' end+ case when substring(col,2,1)='Y' then '人2,' else '' end+ case when substring(col,3,1)='Y' then '人3,' else '' end+ case when substring(col,4,1)='Y' then '人4,' else '' end+ case when substring(col,5,1)='Y' then '人5,' else '' end from @T ) a /* id col ----------- -------------------- 1 人1,人2,人3,人4 2 人1,人5 3 人4 4 人1,人2,人4 */ --2.使用自定义函数 --创建函数 create function fn_GetPersonCount ( @str varchar(5) ) returns varchar(20) as begin declare @result varchar(20) set @result='' declare @i int set @i=0 while(charindex('Y',@str)>0) begin set @result=@result+'人'+ltrim(charindex('Y',@str)+@i)+',' set @str=right(@str,len(@str)-charindex('Y',@str)) set @i=@i+1 end set @result=left(@result,len(@result)-1) return @result end declare @T table (id int,col varchar(5)) insert into @T select 1,'YYYYN' union all select 2,'YNNNY' union all select 3,'NNNYN' union all select 4,'YYNYN' select id,col=dbo.fn_GetPersonCount(col) from @T /* id col ----------- -------------------- 1 人1,人2,人3,人4 2 人1,人5 3 人4 4 人1,人2,人4 */ --3.合并列值 declare @T table (id int,col varchar(5)) insert into @T select 1,'YYYYN' union all select 2,'YNNNY' union all select 3,'NNNYN' union all select 4,'YYNYN' ;with maco as ( select a.*,'人'+ltrim(b.number+1) as c1 from @T a,master..spt_values b where b.type='p' and b.number <5 and substring(col,0+right(number+1,1),1)='Y' ) select id, col= stuff((select ','+c1 from maco t where id=maco.id for xml path('')), 1, 1, '') from maco group by id /* id col ----------- -------------------- 1 人1,人2,人3,人4 2 人1,人5 3 人4 4 人1,人2,人4 */