Android: save a file from an existing URI

Question

How to save a media file (say .mp3) from an existing URI, which I am getting from an Implicit Intent?

Answer 1

You can do it using

new File(uri.getPath());

Answer 2

Use this method, it works

void savefile(URI sourceuri)
{
    String sourceFilename= sourceuri.getPath();
    String destinationFilename = android.os.Environment.getExternalStorageDirectory().getPath()+File.separatorChar+"abc.mp3";

    BufferedInputStream bis = null;
    BufferedOutputStream bos = null;

    try {
      bis = new BufferedInputStream(new FileInputStream(sourceFilename));
      bos = new BufferedOutputStream(new FileOutputStream(destinationFilename, false));
      byte[] buf = new byte[1024];
      bis.read(buf);
      do {
        bos.write(buf);
      } while(bis.read(buf) != -1);
    } catch (IOException e) {
      e.printStackTrace();
    } finally {
      try {
        if (bis != null) bis.close();
        if (bos != null) bos.close();
      } catch (IOException e) {
            e.printStackTrace();
      }
    }
}

Answer 3

When receiving a android.net.Uri from an external source, the best way to save the file is from the stream:

try (InputStream ins = activity.getContentResolver().openInputStream(source_uri)) {
    File dest = new File(destination_path);
    createFileFromStream(ins, dest);
} catch (Exception ex) {
    Log.e("Save File", ex.getMessage());
    ex.printStackTrace();
}

createFileFromStream method:

public static void createFileFromStream(InputStream ins, File destination) {
    try (OutputStream os = new FileOutputStream(destination)) {
        byte[] buffer = new byte[4096];
        int length;
        while ((length = ins.read(buffer)) > 0) {
            os.write(buffer, 0, length);
        }
        os.flush();
    } catch (Exception ex) {
        Log.e("Save File", ex.getMessage());
        ex.printStackTrace();
    }
}

Answer 4

I have used following code to save a file from an existing Uri given back from an Intent to an Uri that my App hosts:

 private void copyFile(Uri pathFrom, Uri pathTo) throws IOException {
        try (InputStream in = getContentResolver().openInputStream(pathFrom)) {
            if(in == null) return;
            try (OutputStream out = getContentResolver().openOutputStream(pathTo)) {
                if(out == null) return;
                // Transfer bytes from in to out
                byte[] buf = new byte[1024];
                int len;
                while ((len = in.read(buf)) > 0) {
                    out.write(buf, 0, len);
                }
            }
        }
    }

Answer 5

1.Create a file from a URI path as:

File from = new File(uri.toString());

2.Create another File where you want the file to save as:

File to = new File("target file path");

3.Rename the file as:

from.renameTo(to);

With this the file from default path is automatically deleted and created at the new path.

Answer 6

private static String FILE_NAM  = "video";
String outputfile = getFilesDir() + File.separator+FILE_NAM+"_tmp.mp4";

InputStream in = getContentResolver().openInputStream(videoFileUri);
private static File createFileFromInputStream(InputStream inputStream, String fileName) {

   try{
      File f = new File(fileName);
      f.setWritable(true, false);
      OutputStream outputStream = new FileOutputStream(f);
      byte buffer[] = new byte[1024];
      int length = 0;

      while((length=inputStream.read(buffer)) > 0) {
        outputStream.write(buffer,0,length);
      }

      outputStream.close();
      inputStream.close();

      return f;
   }catch (IOException e) {
       System.out.println("error in creating a file");
       e.printStackTrace();
   }

return null;

   }