Source Code : Define path with file set

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Define path with file set

 
<?xml version="1.0" encoding="UTF-8"?>
<project name="Classpath Sample" default="compile" basedir=".">
  <property environment="env"/>
  <property name="tomcatHome" value="${env.TOMCAT_HOME}"/>
  <property name="dir.src" value="src"/>
  <property name="dir.build" value="build"/>
  <property name="dir.lib" value="lib"/>
  
  <!-- Define a classpath for use throughout the buildfile -->
  <path id="project.classpath">
    <pathelement location="${dir.src}"/>
    <!-- include Tomcat libraries -->
    <fileset dir="${tomcatHome}/common/lib">
      <include name="*.jar"/>
    </fileset>    
    <!-- include our own libraries -->
    <fileset dir="${dir.lib}">
      <include name="*.jar"/>
    </fileset>
  </path>
  
  <target name="clean">
    <delete dir="${dir.build}"/>
  </target>
  
  <target name="prepare">
    <mkdir dir="${dir.build}"/>
    
    <!-- just create a dummy directory for this fake buildfile -->
    <mkdir dir="${dir.lib}"/>
  </target>
  
  <target name="compile" depends="prepare">
    <!-- use <pathconvert> to convert the path into a property -->
    <pathconvert targetos="windows" property="windowsPath" refid="project.classpath"/>
    <!-- now echo the path to the console -->
    <echo>Windows path = ${windowsPath}</echo>
        
    <!-- Here is how to use the classpath for compiling -->
    <javac destdir="${dir.build}">
      <src path="${dir.src}"/>
      <classpath refid="project.classpath"/>
    </javac>
  </target>

</project>

           
         
  

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