Path to a file using openFileOutput

Question

I'm trying to copy a file to an android device using Wifi. For what i cant tell the creation of the file works, but i can't find it anywhere...

I've try to use the DDMS on Eclipse, but there's nothing on the Data folder... for what i read on other topics, the file should be located on the data folder of the App.. but there's no folder anywhere...

This is the code

package com.ejemplo.cliente;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.net.Socket;
import java.net.UnknownHostException;


import android.app.Activity;
import android.os.Bundle;
import android.widget.TextView;
import android.widget.Toast;

 public class ClienteSocket extends Activity {

 Socket socket;
 String IP = "192.168.1.101";
 Integer Puerto = 4444;
 TextView Mensaje;

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    Mensaje = (TextView)findViewById(R.id.Mensaje);

    try {
         socket = new Socket(IP,Puerto);
         Mensaje.setText("Conectado");
         InputStream input = socket.getInputStream();
         BufferedReader inReader = new BufferedReader(new                  InputStreamReader(socket.getInputStream()));
        BufferedWriter outReader = new BufferedWriter(new OutputStreamWriter(socket.getOutputStream()));
        String filename = inReader.readLine();

      if ( !filename.equals("") ){

          outReader.write("READY\n");
          outReader.flush();
          Mensaje.setText("Request Acepted");

                FileOutputStream wr = openFileOutput(filename,MODE_WORLD_READABLE);
                OutputStreamWriter osw = new OutputStreamWriter(wr); 


          byte[] buffer = new byte[socket.getReceiveBufferSize()];

          int bytesReceived = 0;

          while((bytesReceived = input.read(buffer))>0)
          {

             osw.write(bytesReceived);
             Toast.makeText(this, "Transfering", Toast.LENGTH_SHORT).show();
          }

          osw.flush();
          osw.close();
          Mensaje.setText("File Received");
      }

    } catch (UnknownHostException e) {
      Mensaje.setText("error 1" + e.getMessage());
    } catch (IOException e) {
      Mensaje.setText("error 2" + e.getMessage());
   }

   }
}

Falmarri · Accepted Answer · 2011-01-08 20:41:56Z

5

If it is indeed working, the file will be at /data/data/your.package.name/. You will only be able to access that folder on an emulator unless your phone is rooted.

answered Jan 8, 2011 at 20:41

Falmarri

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4 Comments

Diego Ulloa Over a year ago

and it's there anyway to copy the file to a more "public" folder?? so it can be reached by other applications... for example if i copy a picture, want to be display it by the Galery.. same with an Mp3..

Falmarri Over a year ago

For pictures there's a content provider you can put it into that's build into android. If you want to make it accessible to other applications, you either have to write your own content provider, or put it on the SD card

hackbod Over a year ago

If you want to put it in a world accessible (allowing anyone to read or write it), put it in external storage: developer.android.com/reference/android/os/…

Diego Ulloa Over a year ago

Does anyone can help me with the content provider for the IMAGES or give me link to an example to copy a file and make it available for other apps, using content providers??

lucidbrot · Accepted Answer · 2014-04-15 17:50:25Z

0

if you want to save it in the - for users accesible - external storage, use this File f = new File(Environment.getExternalStorageDirectory(),"Folder/name.ending"); to create the file to copy, also use that path. If you don't want the image to appear in the Gallery of the phone, add a file called ".nomedia" to the same directory as the image

answered Apr 15, 2014 at 17:50

lucidbrot

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