使用ant，在不知道完整路径的情况下重命名目录？

Question

给定一个包含未知目录的 zip 文件，我如何重命名该目录或将该目录移动到规范化路径？

<!-- Going to fetch some stuff -->
<target name="get.remote">

    <!-- Get the zipfile -->
    <get src="http://myhost.com/package.zip"
         dest="package.zip"/>

    <!-- Unzip the file -->
    <unzip src="package.zip"
           dest="./"/>

    <!-- Now there is a package-3d28djh3 directory.  The part after package- is
         a hash and cannot be known ahead of time -->

    <!-- Remove the zipfile -->
    <delete file="package.zip"/>

    <!-- Now we need to rename "package-3d28djh3" to "package".  My best attempt
         is below, but it just moves package-3d28djh3 into package instead of
         renaming the directory. -->

    <!-- Make a new home for the contents. -->
    <mkdir dir="package" />

    <!-- Move the contents -->
    <move todir="package/">
      <fileset dir=".">
        <include name="package-*/*"/>
      </fileset>
    </move>

</target>

我不是蚂蚁用户，任何见解都会有所帮助。

非常感谢，-马特

Answer 1

这只有在 dirset 只返回 1 项时才有效。

<project name="Test rename" basedir=".">

  <target name="rename">
    <path id="package_name">
      <dirset dir=".">
        <include name="package-*"/>
      </dirset>
    </path>
    <property name="pkg-name" refid="package_name" />
    <echo message="renaming ${pkg-name} to package" />
    <move file="${pkg-name}" tofile="package" />
  </target>

</project>

Answer 2

如果 package-3d28djh3 目录中没有子目录（或解压后调用的任何子目录），您可以使用

<move todir="package" flatten="true" />
  <fileset dir=".">
    <include name="package-*/*"/>
  </fileset>
</move>

否则，使用正则表达式映射器进行移动任务并摆脱 package-xxx 目录：

<move todir="package">
  <fileset dir=".">
    <include name="package-*/*"/>
  </fileset>
  <mapper type="regexp" from="^package-.*/(.*)" to="\1"/>
</move>