Android: save a file from an existing URI

Solution 1:

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();
      }
    }
}

Solution 2:

If Uri is received from Google Drive, it can be a Virtual File Uri too. Check this article from CommonsWare for more information. So you have to consider that condition too while saving file from Uri.

To find if file Uri is virtual or not you can use

private static boolean isVirtualFile(Context context, Uri uri) {
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT) {
        if (!DocumentsContract.isDocumentUri(context, uri)) {
            return false;
        }
        Cursor cursor = context.getContentResolver().query(
                uri,
                new String[]{DocumentsContract.Document.COLUMN_FLAGS},
                null, null, null);
        int flags = 0;
        if (cursor.moveToFirst()) {
            flags = cursor.getInt(0);
        }
        cursor.close();
        return (flags & DocumentsContract.Document.FLAG_VIRTUAL_DOCUMENT) != 0;
    } else {
        return false;
    }
}

You can get the stream data from this virtual file like this:

private static InputStream getInputStreamForVirtualFile(Context context, Uri uri, String mimeTypeFilter)
        throws IOException {

    ContentResolver resolver = context.getContentResolver();
    String[] openableMimeTypes = resolver.getStreamTypes(uri, mimeTypeFilter);
    if (openableMimeTypes == null || openableMimeTypes.length < 1) {
        throw new FileNotFoundException();
    }
    return resolver
            .openTypedAssetFileDescriptor(uri, openableMimeTypes[0], null)
            .createInputStream();
}

For finding MIME type try

private static String getMimeType(String url) {
    String type = null;
    String extension = MimeTypeMap.getFileExtensionFromUrl(url);
    if (extension != null) {
        type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
    }
    return type;
}

Overall, you can use

public static boolean saveFile(Context context, String name, Uri sourceuri, String destinationDir, String destFileName) {

    BufferedInputStream bis = null;
    BufferedOutputStream bos = null;
    InputStream input = null;
    boolean hasError = false;

    try {
        if (isVirtualFile(context, sourceuri)) {
            input = getInputStreamForVirtualFile(context, sourceuri, getMimeType(name));
        } else {
            input = context.getContentResolver().openInputStream(sourceuri);
        }

        boolean directorySetupResult;
        File destDir = new File(destinationDir);
        if (!destDir.exists()) {
            directorySetupResult = destDir.mkdirs();
        } else if (!destDir.isDirectory()) {
            directorySetupResult = replaceFileWithDir(destinationDir);
        } else {
            directorySetupResult = true;
        }

        if (!directorySetupResult) {
            hasError = true;
        } else {
            String destination = destinationDir + File.separator + destFileName;
            int originalsize = input.available();

            bis = new BufferedInputStream(input);
            bos = new BufferedOutputStream(new FileOutputStream(destination));
            byte[] buf = new byte[originalsize];
            bis.read(buf);
            do {
                bos.write(buf);
            } while (bis.read(buf) != -1);
        }
    } catch (Exception e) {
        e.printStackTrace();
        hasError = true;
    } finally {
        try {
            if (bos != null) {
                bos.flush();
                bos.close();
            }
        } catch (Exception ignored) {
        }
    }

    return !hasError;
}

private static boolean replaceFileWithDir(String path) {
    File file = new File(path);
    if (!file.exists()) {
        if (file.mkdirs()) {
            return true;
        }
    } else if (file.delete()) {
        File folder = new File(path);
        if (folder.mkdirs()) {
            return true;
        }
    }
    return false;
}

Call this method from an AsycTask. Let me know if this helps.

Solution 3:

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;
}