使用Ant查找具有最新版本号的文件

Question

I want to use ant to find the file with the latest version number. For example, I have a file directory named tomcat with the following files:

• apache-tomcat-6.0.37.zip
• apache-tomcat-6.0.38.zip
• apache-tomcat-6.0.39.zip

I want ant to determine that apache-tomcat-6.0.39.zip is the latest file. Is there a way to do this?

Thanks!

Answer 1

Use resource collections, see last and sort . fe :

<project>

 <path id="foo">
  <last>
    <sort>
      <fileset dir="C:/some/path" includes="**/apache-tomcat-*.zip"/>
    </sort>
  </last>
 </path>

  <echo>$${foo} => ${toString:foo}</echo>

</project>

output :

[echo] ${foo} => C:\some\path\apache-tomcat-6.0.39.zip

Answer 2

• use <pathconvert> to get file names from the relevant <fileset> .
• use <sortlist> to sort the filenames in natural String order .
• pick the last filename from the list.

try this:

<fileset dir="${your.base.dir}" id="one">
   <include name="**/apache-tomcat-.*.zip"/>
</fileset>
<pathconvert property="orig.list" refid="one" pathsep=","/>
<sortlist property="sorted.list" value="${orig.list}" delimiter="," />
<propertyregex property="result" input="${sorted.list}" regexp=",?([^,]+?)$" select="\1"/>

inputList: <property name="one" value="ab-2,ab-5,ab-1,ab-3,ab-4"/>
outputList: [echo] ab-5