使用ant，在不知道完整路徑的情況下重命名目錄？

Question

給定一個帶有未知目錄的zip文件，如何重命名或將該目錄移動到規範化路徑？

<!-- 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> 

我不是一個螞蟻用戶，任何洞察力都會有所幫助。

謝謝了，-Matt

Answer 1

這隻會只要dirset只返回1項工作。

<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

如果在包3d28djh3目錄中沒有子目錄（或任何一旦你提取它的叫法），可以使用

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

否則，使用正則表達式映射器，用於移動任務擺脫包xxx目錄：

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