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905

answers:

3

Hi,

I'm trying to open a file in android like this :

  try
   {
      FileInputStream fIn = context.openFileInput(FILE);
      DataInputStream in = new DataInputStream(fIn);
      BufferedReader br = new BufferedReader(new InputStreamReader(in));
      if(in!=null)
          in.close();
   }
   catch(Exception e)
   {  }

, but in case the file does not exists a file not found exception is thrown . I'd like to know how could I test if the file exists before attempting to open it.

+4  A: 

The documentation says Context.openFileInput either returns an inputStream (file found) or throws a FileNotFoundException (not found)

http://developer.android.com/reference/android/content/Context.html#openFileInput(java.lang.String)

So it looks like the exception is your "test".

You could also try using standard

java.io.File file = new java.io.File(PATHTOYOURCONTEXT , FILE);
if (file.exists()) {
    FileInputStream fIn = new FileInputStream(file);
}

But that is not recommended. Context.openFileInput() and Context.openFileOutput() make sure you stay in your applications storage context on the device, and that all of your files get deleted when your app gets uninstalled.

deadsven
A: 

why dont you just catch the FileNotFound exception and take that as the file not being present.

Prashast
That's the way I actually do , but I was wondering if there is another way
rantravee
+1  A: 

I think the best way to know if a file exists, without actually trying to open it, is as follows:

File file = getContext().getFileStreamPath(FILE_NAME);
if(file.exists()) ...

Hope that helps, bye!

lencinhaus