请问:如何在一个路径字符串中截取文件名

Question

例如
D:\GDMBackUP\2007-4-27 9.50.54 e52bb812-0cb8-41ab-ba5f-6b1efbcda896\GDM-V1.0-4-27-2007-9.50.54上午.BAK 
中取出 
GDM-V1.0-4-27-2007-9.50.54上午.BAK

Answer 1

declare @d varchar(1000)
set @d='D:\GDMBackUP\2007-4-27 9.50.54 e52bb812-0cb8-41ab-ba5f-6b1efbcda896\GDM-V1.0-4-27-2007-9.50.54上午.BAK'select reverse(left(reverse(@d),charindex('\',reverse(@d))-1))

Answer 2

declare @a varchar(150)
set @a='D:\GDMBackUP\2007-4-27 9.50.54 e52bb812-0cb8-41ab-ba5f-6b1efbcda896\GDM-V1.0-4-27-2007-9.50.54上午.BAK' 
select right(@a,charindex('\',reverse(@A))-1)