How can I print the filename which is being processes by xmltask?

Question

In the below ant script snippet, I am processing all the conn.xml files in a directory to find out if there is a DB entry other than MyDB. This code only sets the DB name of the last match which is still okay as there is only one such file normally. But I want to name the exact file (out of those all included files) which has this invalid entry.

<xmltask>
   <fileset dir="${srcdir}/../apps" includes="*/conn.xml"/>
   <copy path="//Reference[@name!='MyDB']/@name" attrvalue="true" property="bad_connection_name"/>
</xmltask>

What parameter I can use in "path" field of copy command which will print the current file name as well?

Answer 1

<target name="check_connection_violations">
    <xmltask source="${file}">
        <copy path="//Reference[@className='oracle.jdeveloper.db.adapter.DatabaseProvider' and @name!='MyDB']/@name" attrvalue="true" property="bad_connection_name"/>
    </xmltask>
    <if>
        <isset property="bad_connection_name"/>
    <then>
        <echo message="${file} has connection violation due to ${bad_connection_name} entry. ${line.separator}" file="${basedir}/conn_name_violation.txt" append="true"/>
    </then>
    </if>
</target>

The above one is newly added target. Here is the original snippet which calls it in a loop :

<foreach target="check_connection_violations" param="file">
    <fileset dir="${srcdir}/../apps" includes="*/conn.xml"/>
</foreach>