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Java 如何使用zipfileset正确创建ZIP文件?_Java_Ant - Fatal编程技术网

Java 如何使用zipfileset正确创建ZIP文件?

Java 如何使用zipfileset正确创建ZIP文件?,java,ant,Java,Ant,我想从Myproj目录及其子目录中收集所有.map文件,然后创建一个L.zip文件 这是我的密码 <target name="buildLFiles" > <zip destfile="../../bin/L.zip" update="true" > <zipfileset casesensitive="no" dir="../../../Myproj&q

我想从Myproj目录及其子目录中收集所有
.map
文件,然后创建一个
L.zip
文件

这是我的密码

<target name="buildLFiles" >    
  <zip destfile="../../bin/L.zip" update="true" >
     <zipfileset casesensitive="no" dir="../../../Myproj" includes= "****/*.MAP" />
  </zip>
</target>
电流输出:

L
|- a
   |- b 
      | - x1.MAP
所需输出:

L
|- x1.MAP

您可以在另一个目录中使用复制任务的
plant
属性展平结构,然后压缩该目录

<target name="buildLFiles">
     <property name="tmp.dir" value="../../bin/TMP" />
     <property name="zip.file" value="../../bin/L.zip" />
     <copy todir="${tmp.dir}" flatten="true">
         <fileset dir="../../../Myproj">
             <include name="**/*.MAP" />
         </fileset>
     </copy>
     <zip destfile="${zip.file}" update="true">
         <zipfileset casesensitive="no" dir="${tmp.dir}" includes="*.MAP" />
     </zip>
     <delete dir="${tmp.dir}" />
</target>

<target name="buildLFiles">
     <property name="tmp.dir" value="../../bin/TMP" />
     <property name="zip.file" value="../../bin/L.zip" />
     <copy todir="${tmp.dir}" flatten="true">
         <fileset dir="../../../Myproj">
             <include name="**/*.MAP" />
         </fileset>
     </copy>
     <zip destfile="${zip.file}" update="true">
         <zipfileset casesensitive="no" dir="${tmp.dir}" includes="*.MAP" />
     </zip>
     <delete dir="${tmp.dir}" />
</target>