rq js
2010-08-20 333333333333333333333333330000000000111111111请问有没有可以统计出JS里有多少个3,多少个0,多少个1。0是从第几位开始的。1又是从第几位开始的。谢谢
2010-08-20 333333333333333333333333330000000000111111111请问有没有可以统计出JS里有多少个3,多少个0,多少个1。0是从第几位开始的。1又是从第几位开始的。谢谢
declare @s varchar(200)
set @s = '333333333333333333333333330000000000111111111'
select len(@s)-len(replace(@s,'3','')) as '3的个数',len(@s)-len(replace(@s,'0','')) as '0的个数',len(@s)-len(replace(@s,'1','')) as '1的个数',charindex('0',@s,1) as '0开始的位置',charindex('1',@s,1) as '1开始的位置'自己换成表中的字段就行了。
select len(js)-len(replace(js,'3','')) as '3的个数',len(js)-len(replace(js,'0','')) as '0的个数',len(js)-len(replace(js,'1','')) as '1的个数',charindex('0',js,1) as '0开始的位置',charindex('1',js,1) as '1开始的位置' from tb
insert into @t select '2010-8-20','3333333333333330000000000111111111'select convert(char(10),rq,120) as rq,
len_3 =len(js)-len(replace(js,'3','')),
len_0 =len(js)-len(replace(js,'0','')),
begin_0 = charindex('0',js),
len_1=len(js)-len(replace(js,'1','')),
begin_1= charindex('1',js)
from @t
select rq,
len(left(js,charindex('0',js)-1)) as [3的个数],
len(replace(replace(js,'3',''),'1','')) as [0的个数],
charindex('0',js) as [0的开始位置],
len(replace(replace(js,'3',''),'0','')) as [1的个数],
charindex('1',js) as [1的开始位置]
from tb
select rq,
charindex('0',js) as [0的开始位置],
charindex('1',js) as [1的开始位置],
len(left(js,charindex('0',js)-1)) as [3的个数],
len(replace(replace(js,'3',''),'1','')) as [0的个数],
len(replace(replace(js,'3',''),'0','')) as [1的个数]
from tb