Re: How can I select different classpath for building different apps?

[Antoine Lévy-Lambert]

[EMAIL PROTECTED] wrote:

All,

In one single build.xml, how can I select different set of classpaths? for 
Tomcat and for other apps?

I try to do the followings:

<target name="tomcat" if="system.CATALINA_HOME">

       <antcall target="clean"/>
       <antcall target="init-tomcat"/>
       <antcall target="compile-newserver"/>
       <antcall target="dist-tomcat"/>
       <antcall target="deploy-tomcat"/>
</target>

In the init-tomcat target, I have the followings:

<target name="init-tomcat">

       <echo message="CATALINA_HOME is ${system.CATALINA_HOME}"/>

       <path id="compile.classpath">

       <pathelement location="${catalina.home}/common/classes"/>
       <fileset dir="${catalina.home}/common/endorsed">
               <include name="*.jar"/>
       </fileset>
       <fileset dir="${catalina.home}/common/lib">
               <include name="*.jar"/>
       </fileset>
       <pathelement location="${catalina.home}/shared/classes"/>
       <fileset dir="${catalina.home}/shared/lib">
               <include name="*.jar"/>
       </fileset>
       </path>

</target>

Unfortunately, it does not work setting classpath like this... Any advice? 

Thanks

Barry

 

ur target above init-tomcat creates a piece of data with id 
"compile.classpath". But then you need to reference your classpath, for 
instance in <javac>

<javac classpathref="compile.classpath" ...>
<!-- other nested elements -->
</javac>
will work.

Antoine

---------------------------------------------------------------------
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]