How do I get the URI of a file saved with FileOutputStream?

Question

How do I get the URI of a file saved with FileOutputStream?

Here's the code that saves the image file ... somewhere? How do I get the URI for the "webimage" file?

Bitmap bmp = ((BitmapDrawable)imageView.getDrawable()).getBitmap();
        String fileName = "webImage";//no .png or .jpg needed
        try {
            ByteArrayOutputStream bytes = new ByteArrayOutputStream();
            bmp.compress(Bitmap.CompressFormat.JPEG, 100, bytes);
            FileOutputStream fo = openFileOutput(fileName, Context.MODE_PRIVATE);
            fo.write(bytes.toByteArray());
            // remember close file output
            fo.close();
        } catch (Exception e) {
            e.printStackTrace();
        }

+3

java android uri android-file

Ethan allen 13 jul. 15 at 17:56

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3 answers

I thought, I think:

final File file = new File(getFilesDir(), "webImage");
Uri weburi = Uri.fromFile(file);

+2

Ethan allen 13 jul. 15 at 18:12

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You can use Uri.fromFile(imageFile)

where imageFile

is an instanceFile

0

Roberto Tellez Ibarra 13 jul. 15 at 18:05

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Bob snyder · Accepted Answer · 2015-07-13T18:14:51+0000

Use getFileStreamPath () like this:

String fileName = "webImage";
//...
Uri uri = Uri.fromFile(getFileStreamPath(fileName));

How do I get the URI of a file saved with FileOutputStream?

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