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>