How can I transform a Bitmap into a Uri?

Question

I'm trying to share images with Facebook, twitter, etc using SHARE INTENT from Android.

I found code to send a image to the share intent, but this code needs the URI of the bitmap: fullSizeImageUri

This is the full code:

private void startShareMediaActivity(Bitmap image) { boolean isVideo=false; String mimeType="bmp"; Uri fullSizeImageUri=null; Intent intent = new Intent(); intent.setAction(Intent.ACTION_SEND); intent.setType(mimeType); intent.putExtra(Intent.EXTRA_STREAM, fullSizeImageUri); try { startActivity(Intent.createChooser(intent, (isVideo ? "video" : "image"))); } catch (android.content.ActivityNotFoundException ex) { } }

How to transform a Bitmap into a Uri?

DragonFire · Accepted Answer · 2020-06-24 03:43:22Z

96

Here is the Colin's Blog who suggest the simple method to convert bitmap to Uri Click here

public Uri getImageUri(Context inContext, Bitmap inImage) { ByteArrayOutputStream bytes = new ByteArrayOutputStream(); inImage.compress(Bitmap.CompressFormat.JPEG, 100, bytes); String path = MediaStore.Images.Media.insertImage(inContext.getContentResolver(), inImage, "Title", null); return Uri.parse(path); }

edited Jun 24, 2020 at 3:43

DragonFire

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answered Apr 23, 2013 at 11:22

Ajay

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9 Comments

ocross Over a year ago

Note this creates an image that will stick around in the users gallery.

Roni Castro Over a year ago

path can be null, if so, Uri.parse(path) will make app crash

Oğuzhan Gedik Over a year ago

Dont forget to take WRITE_EXTERNAL_STORAGE permission. Otherwise insertImage method may return null. See : stackoverflow.com/a/35283353/3341089

Ashutosh Srivastava Over a year ago

Not workking in android oreo, String path = Images.Media.insertImage(inContext.getContentResolver(), inImage, "Title", null); path will get null.

Abraham Mathew Over a year ago

This method is deprecated

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KK_07k11A0585 · Accepted Answer · 2011-11-29 08:35:35Z

23

String FILENAME = "image.png"; String PATH = "/mnt/sdcard/"+ FILENAME; File f = new File(PATH); Uri yourUri = Uri.fromFile(f);

edited Nov 29, 2011 at 8:35

answered Nov 28, 2011 at 13:58

KK_07k11A0585

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5 Comments

Aurelian Cotuna Over a year ago

Parse always take as argument a String not a file. Next time read details about functions or maybe test them.

Prasad Over a year ago

can't we get Uri without saving the bitmap

reza rahmad Over a year ago

work in android studio 2022 bumblebee

KK_07k11A0585 Over a year ago

@rezarahmad Dear sheldon, This was answered 10 years back,

reza rahmad Over a year ago

@KK_07k11A0585 yes this still working until 2023 and more future years

Uzzam Altaf · Accepted Answer · 2022-08-30 18:57:13Z

The latest solution I found is this. Its in kotlin you can covert it in Java.

 // Get uri of images from camera function private fun getImageUri(inContext: Context?, inImage: Bitmap): Uri { val tempFile = File.createTempFile("temprentpk", ".png") val bytes = ByteArrayOutputStream() inImage.compress(Bitmap.CompressFormat.PNG, 100, bytes) val bitmapData = bytes.toByteArray() val fileOutPut = FileOutputStream(tempFile) fileOutPut.write(bitmapData) fileOutPut.flush() fileOutPut.close() return Uri.fromFile(tempFile) }

ocross · Accepted Answer · 2016-03-11 17:52:42Z

The above solution uses media store and stores the image in the users main image folder making it viewable through the gallery/photo viewer. This solution will store it as a temporary file in your apps data. In this example inImage is a Bitmap and title is a string for the name of the image file.

 File tempDir= Environment.getExternalStorageDirectory(); tempDir=new File(tempDir.getAbsolutePath()+"/.temp/"); tempDir.mkdir(); File tempFile = File.createTempFile(title, ".jpg", tempDir); ByteArrayOutputStream bytes = new ByteArrayOutputStream(); inImage.compress(Bitmap.CompressFormat.JPEG, 100, bytes); byte[] bitmapData = bytes.toByteArray(); //write the bytes in file FileOutputStream fos = new FileOutputStream(tempFile); fos.write(bitmapData); fos.flush(); fos.close(); return Uri.fromFile(tempFile);

How temporary is temporary? When will the file be deleted? Can I use the same file name for several images in quick succession?
instead of tempDir=new File(tempDir.getAbsolutePath()+"/.temp/"); you can always use a directory exclusively used by your app.. and just leave the images there
@Joel Also if you are using in a loop name the file name as "file"+i and on success of your process like an api response you can clear this folder -> that will have to be done manually

Suraj Rao · Accepted Answer · 2022-08-29 04:59:15Z

val context = LocalContext.current val uri = context.saveImage(bitmap) // here you will receive the Uri private fun Context.saveImage(bitmap: Bitmap): Uri? { var uri: Uri? = null try { val fileName = System.nanoTime().toString() + ".png" val values = ContentValues().apply { put(MediaStore.Images.Media.DISPLAY_NAME, fileName) put(MediaStore.Images.Media.MIME_TYPE, "image/png") if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.Q) { put(MediaStore.MediaColumns.RELATIVE_PATH, "DCIM/") put(MediaStore.MediaColumns.IS_PENDING, 1) } else { val directory = Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DCIM) val file = File(directory, fileName) put(MediaStore.MediaColumns.DATA, file.absolutePath) } } uri = contentResolver.insert(MediaStore.Images.Media.EXTERNAL_CONTENT_URI, values) uri?.let { contentResolver.openOutputStream(it).use { output -> bitmap.compress(Bitmap.CompressFormat.PNG, 100, output) } if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.Q) { values.apply { clear() put(MediaStore.Audio.Media.IS_PENDING, 0) } contentResolver.update(uri, values, null, null) } } return uri }catch (e: java.lang.Exception) { if (uri != null) { // Don't leave an orphan entry in the MediaStore contentResolver.delete(uri, null, null) } throw e } }

Pankaj Talaviya · Accepted Answer · 2017-02-17 08:50:16Z

1

pass bitmap and compressFormat like (PNG, JPG, etc...) and image quality in percentage

public Uri getImageUri(Bitmap src, Bitmap.CompressFormat format, int quality) { ByteArrayOutputStream os = new ByteArrayOutputStream(); src.compress(format, quality, os); String path = MediaStore.Images.Media.insertImage(getContentResolver(), src, "title", null); return Uri.parse(path); }

answered Feb 17, 2017 at 8:50

Pankaj Talaviya

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5 Comments

Roni Castro Over a year ago

path can be null, so your code would crash on Uri.parse(path)

Pankaj Talaviya Over a year ago

you check path is null or not. and after you pass in Uri

Pankaj Mundra Over a year ago

@PankajTalaviya This method is saving the image with .jpg format even we provide PNG format to compress.

Pankaj Talaviya Over a year ago

@PankajMundra Increase decrease quality percentage to compress image

Pankaj Mundra Over a year ago

So according to you what should be the appropriate percentage to save image as .png...because if it saves as jpeg the background of image is black.

Aurelian Cotuna · Accepted Answer · 2011-11-28 12:36:21Z

0

Well you can't transforma a bitmap file into a uri. Read more about URI here

URI is an Uniform Resource Identifier. But you can place the bitmap in an absolute or relative URI like this

Absolute: http://android.com/yourImage.bmp Relative: yourImage.bmp

answered Nov 28, 2011 at 12:36

Aurelian Cotuna

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2 Comments

NullPointerException Over a year ago

then how can achieve my needs? it's impossible?

Aurelian Cotuna Over a year ago

I guess you have to rethink your way to see this... check Facebook and twitter APIS to see how you can achieve that image sharing

mehrdad khosravi · Accepted Answer · 2016-04-13 09:01:43Z

String picName = "pic.jpg"; String PATH = Environment.getExternalStorageDirectory().getPath()+ picName; File f = new File(PATH); Uri yourUri = Uri.fromFile(f);

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