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?
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
Uri imgUri = data.getData();
Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), imgUri);
其他回答
从移动库中获取图像uri的完整方法。
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();
}
}
}
MediaStore.Images.Media.getBitmap在API 29中已弃用。推荐的方法是使用ImageDecoder。在API 28中添加的createSource。
下面是如何获得位图:
val bitmap = if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.P) {
ImageDecoder.decodeBitmap(ImageDecoder.createSource(requireContext().contentResolver, imageUri))
} else {
MediaStore.Images.Media.getBitmap(requireContext().contentResolver, imageUri)
}
重要:ImageDecoder.decodeBitmap读取EXIF方向,媒体。getBitmap不
Bitmap bitmap = null;
ContentResolver contentResolver = getContentResolver();
try {
if(Build.VERSION.SDK_INT < 28) {
bitmap = MediaStore.Images.Media.getBitmap(contentResolver, imageUri);
} else {
ImageDecoder.Source source = ImageDecoder.createSource(contentResolver, imageUri);
bitmap = ImageDecoder.decodeBitmap(source);
}
} catch (Exception e) {
e.printStackTrace();
}
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
InputStream imageStream = null;
try {
imageStream = getContext().getContentResolver().openInputStream(uri);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
final Bitmap selectedImage = BitmapFactory.decodeStream(imageStream);