How to write a file using Intent.ACTION_CREATE_DOCUMENT

Question

In simple words, I want to convert a viewgroup to a jpg image file. As Environment.getExternalStorageDirectory is deprecated, I use this intent Intent.ACTION_CREATE_DOCUMENT

private void createFile(String mimeType, String fileName) {
        Intent intent = new Intent(Intent.ACTION_CREATE_DOCUMENT);
        intent.addCategory(Intent.CATEGORY_OPENABLE);
        intent.setType(mimeType);
        intent.putExtra(Intent.EXTRA_TITLE, fileName);
        startActivityForResult(intent, WRITE_REQUEST_CODE);
    }

In the onActivityResult(); I get the Uri returned by the result. My problem is that with getExternalStorage() I'd use

Bitmap bitmap = Bitmap.createBitmap(
                    containerLayout.getWidth(),
                    containerLayout.getHeight(), Bitmap.Config.ARGB_8888);
            Canvas canvas = new Canvas(bitmap);
            containerLayout.draw(canvas);
            FileOutputStream fileOutupStream = null;


            try {
                fileOutupStream = new FileOutputStream(fileName);
                bitmap.compress(Bitmap.CompressFormat.JPEG, 100, fileOutupStream);
                fileOutupStream.flush();
                fileOutupStream.close();
                Toast.makeText(this, "saved " + fileName, Toast.LENGTH_LONG).show();
            } catch (Exception e) {
                Toast.makeText(this, "something went wrong" + e.getMessage(), Toast.LENGTH_SHORT).show();
                e.printStackTrace();
            }

Now I get the Uri returned by the result but, I don't know how to write the desired bitmap into this Uri

@Override
        protected void onActivityResult(int requestCode, int resultCode, @Nullable Intent data) {
            if (resultCode == RESULT_OK && requestCode == WRITE_REQUEST_CODE) {
                Uri resultUri = data.getData();
//need help

}
}

Emlen

Answer 1

You need to use getContentResolver().openOutputStream

        @Override
        protected void onActivityResult(int requestCode, int resultCode, @Nullable Intent data) {
            if (resultCode == RESULT_OK && requestCode == WRITE_REQUEST_CODE) {
                FileOutputStream fileOutupStream = getContentResolver().openOutputStream(data.getData());
            try {
                bitmap.compress(Bitmap.CompressFormat.JPEG, 100, fileOutupStream);
                fileOutupStream.flush();
                fileOutupStream.close();
                Toast.makeText(this, "saved " + fileName, Toast.LENGTH_LONG).show();
            } catch (Exception e) {
                Toast.makeText(this, "something went wrong" + e.getMessage(), Toast.LENGTH_SHORT).show();
                e.printStackTrace();
            }

}
}

Carouse

Why are you recreating fileOutupStream on line 6, after assigning it on line 4? – Gaol

I believe it's a copy paste mistake, i've just kept it from the code in the question – Carouse

getContentResolver().openOutputStream(data.getData()) this creates an OutputStream, not a FileOutputStream – Tell

Why do you suggest to use ContentResolver? Is not a plain FileInputStream(sourceFile) enough? – Carty

Answer 2

This is a minimal example for copying a file from one location to another using Intent.ACTION_CREATE_DOCUMENT on older APIs (before API 21). You should add request code and result code handling if necessary.

    ...
    // Intent creation
    int REQUEST_CODE_ARBITRARY = 1;
    Intent intent = new Intent(Intent.ACTION_CREATE_DOCUMENT);
    // will trigger exception if no  appropriate category passed
    intent.addCategory(Intent.CATEGORY_OPENABLE);
    // or whatever mimeType you want
    intent.setType("*text/plain");
    intent.putExtra(Intent.EXTRA_TITLE, "file_name_to_save_as");
    startActivityForResult(intent, REQUEST_CODE_ARBITRARY);

    // Handling result
    @Override
    protected void onActivityResult(int requestCode, int resultCode, Intent data) {
        super.onActivityResult(requestCode, resultCode, data);

        // Example file to copy
        File sourceFile = getDatabasePath("MyDatabaseName");

        InputStream is = null;
        OutputStream os = null;

        // Note: you may use try-with resources if your API is 19+
        try {
            // InputStream constructor takes File, String (path), or FileDescriptor
            is = new FileInputStream(sourceFile);
            // data.getData() holds the URI of the path selected by the picker
            os = getContentResolver().openOutputStream(data.getData());

            byte[] buffer = new byte[1024];
            int length;
            while ((length = is.read(buffer)) > 0) {
                os.write(buffer, 0, length);
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
                os.close();
            } catch (IOException e) {
                //
            }
        }
    }

Shorter try-catch on API 19+:

            try (InputStream is = new FileInputStream(sourceFile); OutputStream os = getContentResolver().openOutputStream(data.getData())) {
                byte[] buffer = new byte[1024];
                int length;
                while ((length = is.read(buffer)) > 0) {
                    os.write(buffer, 0, length);
                }
            } catch (IOException e) {
                e.printStackTrace();
            }

Tell

In my case, though i have given "file_name_to_save_as", It is prompting me to enter the file name with save button. Any idea, what could have went wrong ? – Monogram