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From <Jan.Mate...@rzf.fin-nrw.de>
Subject AW: How to get each file's name in a fileset?
Date Mon, 12 Dec 2005 06:48:17 GMT
try nested mappers

01: <project>
02:     <property name="schema.dir" location="schema"/>
03:     <pathconvert property="ports" pathsep="${line.separator}">
04:         <fileset dir="${schema.dir}" includes="**/*.wsdl"/>
05:         <map from="${schema.dir}${file.separator}" to=""/>
06:         <mapper>
07:             <chainedmapper>
08:                 <globmapper from="*.wsdl" to="*"/>
09:                 <flattenmapper/>
10:             </chainedmapper>
11:         </mapper>
12:     </pathconvert>
13:     <echo>PORTS:</echo>
14:     <echo>${ports}</echo>
15: </project>


02: Define the schema dir as location, so I can reuse the absolute path in the <map>
element.
03: Use ${line.separator} for better output. Use like you want.
05: Deletes the leading absolute path. Because 02 gets only the path we also have to delete
the
    dir separator there.
07: Apply two mappers in chain. Otherwise you┬┤ll double your output....
08: Delete the suffix
09: Delete directory hierarchy.


C:\TEMP\ant-mapper>tree /A /F
C:.
|   build.xml
|
\---schema
    |   123.wsdl
    |   4711.wsdl
    |   XXX.wsdl
    |
    \---dev
            9998.wsdl
            9999.wsdl


C:\TEMP\ant-mapper>ant
Buildfile: build.xml
     [echo] PORTS:
     [echo] 123
     [echo] 4711
     [echo] XXX
     [echo] 9998
     [echo] 9999

BUILD SUCCESSFUL
Total time: 0 seconds
C:\TEMP\ant-mapper>


Jan



>-----Urspr├╝ngliche Nachricht-----
>Von: Ding Shukai [mailto:dingshukai@gmail.com] 
>Gesendet: Montag, 12. Dezember 2005 05:54
>An: user@ant.apache.org
>Betreff: How to get each file's name in a fileset?
>
>hi,
>In my project, the ant buidfile needs to know each file's name 
>in a fileset.
>For example, there exists some files like XXX.wsdl in the 
>schema dir; if I could get the string XXX, which is a porttype 
>name, then I neednot to specify the porttype names in 
>build.properties. Could anyone Help me, please? Thanks~
>

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