how to get file name from URI

Question

Hello I am develop video player with android gallery. I receive URI from gallry. and I need to display video title, when play video. so if content has not title meta data. I

Answer 1

int actual_image_column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA); 
String filename = cursor.getString(actual_image_column_index);

This is example for Image . you can try same thing for your needs ( video).

Thanks & Best Luck :)

Answer 2

try
        {
            String uriString = "http://somesite.com/video.mp4";
            URI uri = new URI(uriString);

            URL videoUrl = uri.toURL();
            File tempFile = new File(videoUrl.getFile());
            String fileName = tempFile.getName();
        }
        catch (Exception e)
        {

        }

Answer 3

Here is the code for getting name of file from url

Uri u = Uri.parse("www.google.com/images/image.jpg");

File f = new File("" + u);

f.getName();

Answer 4

Since none of the answers really solve the questions, i put my solution here, hope it will be helpful.

     String fileName;
     if (uri.getScheme().equals("file")) {
            fileName = uri.getLastPathSegment();
        } else {
            Cursor cursor = null;
            try {
                cursor = getContentResolver().query(uri, new String[]{
                        MediaStore.Images.ImageColumns.DISPLAY_NAME
                }, null, null, null);

                if (cursor != null && cursor.moveToFirst()) {
                    fileName = cursor.getString(cursor.getColumnIndex(MediaStore.Images.ImageColumns.DISPLAY_NAME));
                    Log.d(TAG, "name is " + fileName);
                }
            } finally {

                if (cursor != null) {
                    cursor.close();
                }
            }
        }

Answer 5

Below code works for me in case of image from gallery, It should work for any file type.

    String fileName = "default_file_name";
    Cursor returnCursor =
            getContentResolver().query(YourFileUri, null, null, null, null);
    try {
        int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
        returnCursor.moveToFirst(); 
        fileName = returnCursor.getString(nameIndex);
        LOG.debug(TAG, "file name : " + fileName);
    }catch (Exception e){
        LOG.error(TAG, "error: ", e);
        //handle the failure cases here
    } finally {
        returnCursor.close();
    }

Here you can find detailed explanation and much more.

Answer 6

For kotlin just simply:

val fileName = File(uri.path).name