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
(芬兰湾的科特林) 所以,截至2020年4月7日,上述选项都没用,但以下是对我有效的:
如果你想在val中存储位图,并设置一个imageView,使用这个: val bitmap = BitmapFactory.decodeFile(currentPhotoPath).also {bitmap -> imageView.setImageBitmap(bitmap)} 如果你只是想设置位图为和imageView,使用这个: BitmapFactory.decodeFile(currentPhotoPath).also {bitmap -> imageView.setImageBitmap(bitmap)}
其他回答
InputStream imageStream = null;
try {
imageStream = getContext().getContentResolver().openInputStream(uri);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
final Bitmap selectedImage = BitmapFactory.decodeStream(imageStream);
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
下面是正确的方法,同时也要注意内存的使用情况:
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;
}
Mark Ingram的文章中的getBitmap()调用也调用decodeStream(),因此不会丢失任何功能。
引用:
Android:在SD卡上获取图像的缩略图,给定原始图像的Uri 处理大位图
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