我有一个目录,D:\dist\lib ,下面有一些文件夹以及很多文件(不都是*.jar),我想把其中的*.jar文件输出到xml文件中
格式如下,其中只有path的内容是变化的。- <?xml version="1.0" encoding="UTF-8" ?>
- <classpath>
- <classpathentry kind="lib" path="dist/lib/jstl.jar"/>
- <classpathentry kind="lib" path="dist/lib/jta101.jar"/>
- <classpathentry kind="lib" path="dist/lib/jTDS2.jar"/>
- <classpathentry kind="lib" path="dist/lib/junit.jar"/>
- </classpath>
复制代码 我现在可以输出XML文件了,但是如何取得所有文件名,循环输出呢?- set classpath_name=.classpath
- echo ^<?xml version="1.0" encoding="UTF-8" ?^> >%classpath_name%
- echo ^<classpath^> >>%classpath_name%
- echo ^ ^<classpathentry kind="lib" path="dist/lib/jstl.jar"/^> >>%classpath_name%
- echo ^ ^<classpathentry kind="lib" path="dist/lib/jta101.jar"/^> >>%classpath_name%
- echo ^ ^<classpathentry kind="lib" path="dist/lib/jTDS2.jar"/^> >>%classpath_name%
- echo ^ ^<classpathentry kind="lib" path="dist/lib/junit.jar"/^> >>%classpath_name%
- echo ^</classpath^> >>%classpath_name%
复制代码 我看了一个用for /r 取得文件全路径的- set DstDir=D:\dist\lib
- for /r %DstDir% %%i in (*.jar) do (
- @echo %%i
- rem )
复制代码 但是会带上D:\...的路径,如何去掉呢?
[ 本帖最后由 michael_hy 于 2010-4-17 09:11 编辑 ] |