Opening a file or its location...

hi ....
we've created a software which converts word 2 xml....
here we browse the word file...select its destination...i.e where the user wants the xml file 2 be stored...after clicking the convert button the file gets converted....
NOW WE WANT THAT AFTER THE USER CLICKS THE CONVERT BUTTON THE FILE SHOULD BE CONVERTED AND A .....NEW WINDOW SHOULD OPEN WHICH SPECIFIES THE LOCATION OF THE FILE...
for eg....
if user selects the destination as....
C:\jdk1.3\bin
then that particular window should open which has the file in it....
CAN YOU GIVE ME THE CODE FOR THIS....

DO YOU MEAN YOU WANT... oh I'm sorry, I was shouting.
Do you mean you want to open a file browser window (like Explorer if you are on Windows) in the folder where your newly created xml file was just created, after converting?
If so, you will need to use one of the Runtime.exec methods (see documentation), and pass it a different command depending on your platform and file browser application of choice.
What have you got so far?

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