Generate manifest class-path from <classpath> in Ant - ant

Question

In the build file below, the jar target refers to the jar.class.path property for the manifest class-path. The compile target refers to project.class.path
There is redundancy here, because jar.class.path and project.class.path are very similar. They must be both updated when libraries are added, which can be a pain if the list of libraries gets very long. Is there a better way? Any solution must be cross-platform and always use relative paths.
Edit:
It should generate the JAR classpath from a fileset and not the other way around, so I can use wildcards to e.g. include all JAR files in a directory.
<?xml version="1.0"?>
<project name="Higgins" default="jar" basedir=".">
<property name="jar.class.path" value="lib/forms-1.2.0.jar lib/BrowserLauncher.jar"/>
<path id="project.class.path">
<pathelement location="build"/>
<fileset dir="lib">
<include name="forms-1.2.0.jar"/>
<include name="BrowserLauncher.jar"/>
</fileset>
</path>
<target name="prepare">
<mkdir dir="build"/>
</target>
<target name="compile" depends="prepare" description="Compile core sources">
<javac srcdir="src"
includes="**"
destdir="build"
debug="true"
source="1.5">
<classpath refid="project.class.path"/>
</javac>
</target>
<target name="jar" depends="compile" description="Generates executable jar file">
<jar jarfile="higgins.jar">
<manifest>
<attribute name="Main-Class" value="nl.helixsoft.higgins.Main"/>
<attribute name="Class-Path" value="${jar.class.path}"/>
</manifest>
<fileset dir="build" includes="**/*.class"/>
<fileset dir="src" includes="**/*.properties"/>
</jar>
</target>
</project>

Answer 1

<path id="build.classpath">
<fileset dir="${basedir}">
<include name="lib/*.jar"/>
</fileset>
</path>
<pathconvert property="manifest.classpath" pathsep=" ">
<path refid="build.classpath"/>
<mapper>
<chainedmapper>
<flattenmapper/>
<globmapper from="*.jar" to="lib/*.jar"/>
</chainedmapper>
</mapper>
</pathconvert>
<target depends="compile" name="buildjar">
<jar jarfile="${basedir}/${test.jar}">
<fileset dir="${build}" />
<manifest>
<attribute name="Main-Class" value="com.mycompany.TestMain"/>
<attribute name="Class-Path" value="${manifest.classpath}"/>
</manifest>
</jar>
</target>
For further information check out this article.

Answer 2

Assuming Ant 1.7 or above, you can use the manifestclasspath task.
<path id="dep.runtime">
<fileset dir="./lib">
<include name="**/*.jar" />
</fileset>
</path>
<property name="dep_cp" value="${toString:dep.runtime}" />
<target name="default">
<manifestclasspath property="manifest_cp" jarfile="myjar.jar">
<classpath refid="dep.runtime" />
</manifestclasspath>
<echo message="Build Classpath: ${dep_cp}" />
<echo message="Manifest Classpath: ${manifest_cp}" />
</target>

Answer 3

If you just want a common subpath shared between two (or more) paths, that is easy to do:
<path id="lib.path>
<fileset dir="lib">
<include name="forms-1.2.0.jar"/>
<include name="BrowserLauncher.jar"/>
</fileset>
</path>
<path id="project.class.path">
<pathelement location="build"/>
<path refid="lib.path"/>
</path>
<property name="jar.class.path" refid="lib.path"/>
EDIT Sorry, I misunderstood the question. Try this:
<property name="jar.class.path" value="lib/forms-1.2.0.jar lib/BrowserLauncher.jar"/>
<path id="project.class.path">
<pathelement location="build"/>
<fileset dir="." includes="${jar.class.path}"/>
</path>

Answer 4

You can use <pathconvert> to convert a path (which can contain a fileset) into a plain string. You'll likely need to <echo> that string to a file, use either <replace> or <replaceregexp> to chop the leading path bits, then finally use <loadfile> to load the manipulated string into the final property.
Implementation left as an exercise to the reader.