得到特定字符后面的所有字符

Question

105.13-158.4
类似这种格式取-后面所有字符
前后字符数量都不固定

Answer 1

select substring(col,charindex('-',col) + 1 , len(col)) from tb

Answer 2

SELECT RIGHT('105.13-158.4',LEN('105.13-158.4')-CHARINDEX('-','105.13-158.4'))
             
------------ 
158.4（所影响的行数为 1 行）

Answer 3

create table tb(col varchar(50))
insert into tb values('105.13-158.4') 
insert into tb values('105.13-158.53') 
insert into tb values('105.13-158.666')
goselect col = substring(col,charindex('-',col) + 1 , len(col)) from tbdrop table tb
/*col                                                
-------------------------------------------------- 
158.4
158.53
158.666（所影响的行数为 3 行）*/

Answer 4

declare @s varchar(29)
set @s='105.13-158.4'
select STUFF(@s,1,CHARINDEX('-',@s),'')
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
158.4

Answer 5

create table tb(col varchar(50))
insert into tb values('105.13-158.4') 
insert into tb values('105.13-158.53') 
insert into tb values('105.13-158.666')
go--方法一
select col = substring(col,charindex('-',col) + 1 , len(col)) from tb where charindex('-',col) > 0--方法二
select col = reverse(left(reverse(col),charindex('-',reverse(col))-1)) from tb where charindex('-',col) > 0drop table tb
/*col                                                
-------------------------------------------------- 
158.4
158.53
158.666（所影响的行数为 3 行）*/

Answer 6

substring(字段名称,charindex('-',字段名称) + 1 , len(字段名称)) 

Answer 7

好像问过了吧stuff + charindex

Answer 8

declare @s varchar(20)
set @s='105.13-158.4'
select substring(@s,charindex('-',@s) + 1 , len(@s)) 
/*--------------------
158.4(1 行受影响)*/

Answer 9

declare @col varchar(15)
set @col='105.13-158.4'
select substring(@col,CHARINDEX('-',@col)+1,LEN(@col)-CHARINDEX('-',@col)) as col
/*
col
---------------
158.4
*/

Answer 10

3q...
对SQL里的方法很不熟悉...