如何从Uri中获得位图对象(如果我成功地将它存储在 /data/data/MYFOLDER/myimage.png或文件///data/data/MYFOLDER/myimage.png)在我的应用程序中使用它?

有人知道怎么做到吗?


当前回答

下面是正确的做法:

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);
    }
}

如果你需要加载非常大的图像,下面的代码将以tile的形式加载它(避免大的内存分配):

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

点击这里查看答案

其他回答

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();
    }
}

使用startActivityForResult方法,如下所示

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

你可以得到这样的结果:

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

yes路径必须是这样的格式

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

在2022年使用线圈库。

https://coil-kt.github.io/coil/compose/

对于喷气背包组合

            AsyncImage(
                model = uriOfImage,
                contentDescription = null,
            )
* For getting bitmap from uri. Work for me perfectly.
    
    
    public static Bitmap decodeUriToBitmap(Context mContext, Uri sendUri) {
          Bitmap getBitmap = null;
          try {
            InputStream image_stream;
            try {
              image_stream = mContext.getContentResolver().openInputStream(sendUri);
              getBitmap = BitmapFactory.decodeStream(image_stream);
            } catch (FileNotFoundException e) {
              e.printStackTrace();
            }
          } catch (Exception e) {
            e.printStackTrace();
          }
          return getBitmap;
        }