是.
@s你的字符串在哪里......
select
SUBSTRING (@s,
CHARINDEX(',',@s,CHARINDEX(',',@s)+1)+1,
CHARINDEX(',',@s,CHARINDEX(',',@s,CHARINDEX(',',@s)+1)+1)
-CHARINDEX(',',@s,CHARINDEX(',',@s)+1)-1)
Run Code Online (Sandbox Code Playgroud)
或者更一般地......
;with cte as
(
select 1 as Item, 1 as Start, CHARINDEX(',',@s, 1) as Split
union all
select cte.Item+1, cte.Split+1, nullif(CHARINDEX(',',@s, cte.Split+1),0) as Split
from cte
where cte.Split<>0
)
select SUBSTRING(@s, start,isnull(split,len(@s)+1)-start)
from cte
where Item = 3
Run Code Online (Sandbox Code Playgroud)
现在正确存储您的数据:)
| 归档时间: |
|
| 查看次数: |
2545 次 |
| 最近记录: |