Not to rain on the proverbial parade, but you don't need iteration or
third-party tasks to do this:
<pathconvert pathsep="${line.separator}">
<fileset dir="C:/test" includes="*.txt" />
<flattenmapper />
</pathconvert>
HTH,
Matt
P.S. You also didn't need to wrap your fileset in a path on your
iteration-based example.
--- On Wed, 6/24/09, nisse <[email protected]> wrote:
> From: nisse <[email protected]>
> Subject: RE: Using foreach
> To: [email protected]
> Date: Wednesday, June 24, 2009, 2:46 AM
>
> Thanks! Works great!
>
>
>
> Rebhan, Gilbert wrote:
> >
> >
> > -----Original Message-----
> > From: nisse [mailto:[email protected]]
>
> > Sent: Wednesday, June 24, 2009 9:21 AM
> > To: [email protected]
> > Subject: Using foreach
> >
> > /*
> >
> > [..]
> >
> >
> > This code prints for example "C:/Temp/MyFile.txt"
> > But I only want it to print "MyFile.txt", how do I do
> that in a simple
> > way??
> >
> > */
> >
> > use <basename>, example
> >
> > <for param="file">
> > <path>
> > <fileset dir="C:/test"
> includes="*.txt"/>
> > </path>
> > <sequential>
> > <var name="basename" unset="true"/>
> > <basename file="@{file}"
> property="basename"/>
> > <var name="filenames"
> value="${filenames},${basename}"/>
> > </sequential>
> > </for>
> >
> >
> > Regards, Gilbert
> >
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> >
> >
> >
>
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>
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