求SQL语句

[look_007]

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

[hao1hao2hao3]

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。

[hao1hao2hao3]

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 

[playwarcraft]

3，0，1 都是连续的吗？

declare @t table(rq datetime,js varchar(100))
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

[cjzm83]

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

[lcqtgb]

上面的应该那个位置不会对。。。。replace都搞光字符了。。能正确的找出字符所出现的位置呢。。借用上楼的兄弟代码一用

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

[蓝色风影]

学习了...