How to get Bitmap from an Uri?

问题:

How to get a Bitmap object from an Uri (if I succeed to store it in /data/data/MYFOLDER/myimage.png or file///data/data/MYFOLDER/myimage.png) to use it in my application?

Does anyone have an idea on how to accomplish this?

回答1:

. . IMPORTANT: See answer from @Mark Ingram below and @pjv for at better solution. . .

You could try this:

public Bitmap loadBitmap(String url)
{
    Bitmap bm = null;
    InputStream is = null;
    BufferedInputStream bis = null;
    try 
    {
        URLConnection conn = new URL(url).openConnection();
        conn.connect();
        is = conn.getInputStream();
        bis = new BufferedInputStream(is, 8192);
        bm = BitmapFactory.decodeStream(bis);
    }
    catch (Exception e) 
    {
        e.printStackTrace();
    }
    finally {
        if (bis != null) 
        {
            try 
            {
                bis.close();
            }
            catch (IOException e) 
            {
                e.printStackTrace();
            }
        }
        if (is != null) 
        {
            try 
            {
                is.close();
            }
            catch (IOException e) 
            {
                e.printStackTrace();
            }
        }
    }
    return bm;
}

But remember, this method should only be called from within a thread (not GUI -thread). I a AsyncTask.

回答2:

Here\'s the correct way of doing it:

protected void onActivityResult(int requestCode, int resultCode, Intent data)
{
    super.onActivityResult(requestCode, resultCode, data);
    if (resultCode == RESULT_OK)
    {
        Uri imageUri = data.getData();
        Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), imageUri);
    }
}

If you need to load very large images, the following code will load it in in tiles (avoiding large memory allocations):

BitmapRegionDecoder decoder = BitmapRegionDecoder.newInstance(myStream, false);  
Bitmap region = decoder.decodeRegion(new Rect(10, 10, 50, 50), null);

See the answer here

回答3:

Here\'s the correct way of doing it, keeping tabs on memory usage as well:

protected void onActivityResult(int requestCode, int resultCode, Intent data)
{
  super.onActivityResult(requestCode, resultCode, data);
  if (resultCode == RESULT_OK)
  {
    Uri imageUri = data.getData();
    Bitmap bitmap = getThumbnail(imageUri);
  }
}

public static Bitmap getThumbnail(Uri uri) throws FileNotFoundException, IOException{
  InputStream input = this.getContentResolver().openInputStream(uri);

  BitmapFactory.Options onlyBoundsOptions = new BitmapFactory.Options();
  onlyBoundsOptions.inJustDecodeBounds = true;
  onlyBoundsOptions.inDither=true;//optional
  onlyBoundsOptions.inPreferredConfig=Bitmap.Config.ARGB_8888;//optional
  BitmapFactory.decodeStream(input, null, onlyBoundsOptions);
  input.close();

  if ((onlyBoundsOptions.outWidth == -1) || (onlyBoundsOptions.outHeight == -1)) {
    return null;
  }

  int originalSize = (onlyBoundsOptions.outHeight > onlyBoundsOptions.outWidth) ? onlyBoundsOptions.outHeight : onlyBoundsOptions.outWidth;

  double ratio = (originalSize > THUMBNAIL_SIZE) ? (originalSize / THUMBNAIL_SIZE) : 1.0;

  BitmapFactory.Options bitmapOptions = new BitmapFactory.Options();
  bitmapOptions.inSampleSize = getPowerOfTwoForSampleRatio(ratio);
  bitmapOptions.inDither = true; //optional
  bitmapOptions.inPreferredConfig=Bitmap.Config.ARGB_8888;//
  input = this.getContentResolver().openInputStream(uri);
  Bitmap bitmap = BitmapFactory.decodeStream(input, null, bitmapOptions);
  input.close();
  return bitmap;
}

private static int getPowerOfTwoForSampleRatio(double ratio){
  int k = Integer.highestOneBit((int)Math.floor(ratio));
  if(k==0) return 1;
  else return k;
}

The getBitmap() call from Mark Ingram\'s post also calls the decodeStream(), so you don\'t lose any functionality.

References:

• Android: Get thumbnail of image on SD card, given Uri of original image

• Handling large Bitmaps

回答4:

try
{
    Bitmap bitmap = MediaStore.Images.Media.getBitmap(c.getContentResolver() , Uri.parse(paths));
}
catch (Exception e) 
{
    //handle exception
}

and yes path must be in a format of like this

file:///mnt/sdcard/filename.jpg

回答5:

 private void uriToBitmap(Uri selectedFileUri) {
        try {
            ParcelFileDescriptor parcelFileDescriptor =
                    getContentResolver().openFileDescriptor(selectedFileUri, \"r\");
            FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
            Bitmap image = BitmapFactory.decodeFileDescriptor(fileDescriptor);


            parcelFileDescriptor.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

回答6:

You can retrieve bitmap from uri like this

Bitmap bitmap = null;
try {
    bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), imageUri);
} catch (IOException e) {
    e.printStackTrace();
}

回答7:

The solution is too easy:

Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), uri);

回答8:

Uri imgUri = data.getData();
Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), imgUri);

回答9:

Use startActivityForResult metod like below

        startActivityForResult(new Intent(Intent.ACTION_PICK).setType(\"image/*\"), PICK_IMAGE);

And you can get result like this:

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (resultCode != RESULT_OK) {
        return;
    }
    switch (requestCode) {
        case PICK_IMAGE:
            Uri imageUri = data.getData();
            try {
                Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), imageUri);
            } catch (IOException e) {
                e.printStackTrace();
            }
         break;
    }
}

回答10:

you can do this structure:

protected void onActivityResult(int requestCode, int resultCode, Intent imageReturnedIntent) {
    super.onActivityResult(requestCode, resultCode, imageReturnedIntent);
    switch(requestCode) {
        case 0:
            if(resultCode == RESULT_OK){
                    Uri selectedImage = imageReturnedIntent.getData();
                    Bundle extras = imageReturnedIntent.getExtras();
                    bitmap = extras.getParcelable(\"data\");
            }
            break;
   }

by this you can easily convert a uri to bitmap. hope help u.

回答11:

Full method to get image uri from mobile gallery.

protected void onActivityResult(int requestCode, int resultCode, Intent data) {
        super.onActivityResult(requestCode, resultCode, data);

  if (requestCode == PICK_IMAGE_REQUEST && resultCode == RESULT_OK && data != null && data.getData() != null) {
                Uri filePath = data.getData();

     try { //Getting the Bitmap from Gallery
           Bitmap bitmap = MediaStore.Images.Media.getBitmap(getContentResolver(), filePath);
           rbitmap = getResizedBitmap(bitmap, 250);//Setting the Bitmap to ImageView
           serImage = getStringImage(rbitmap);
           imageViewUserImage.setImageBitmap(rbitmap);
      } catch (IOException e) {
           e.printStackTrace();
      }


   }
}