But all this file creation code is in pure java where I have not used openFileOutput but I am using FileOutputStream constructor and all classes in java are in package com.temp.test and to this constructor I want to simply give only file name and not any specific path so that it will create file private to android application's package which is using apis from my jar file. So I don't know android application's package name in advance , it may vary.
Please give me its solution.
-- On Sat, Apr 2, 2011 at 12:20 PM, TreKing <trekingapp@gmail.com> wrote:
For a given Android App ... no it can't ...
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TreKing - Chicago transit tracking app for Android-powered devices
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