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我想从手机图库中挑选一张图片作为用户的个人资料图片上传到应用程序中。我想获取它的 URI，以便将其存储在用户数据库中。

activityResultLauncher = registerForActivityResult(
            new ActivityResultContracts.StartActivityForResult(), result -> {
                if (result.getResultCode() == RESULT_OK && result.getData()!= null) {
                    Bundle data = result.getData().getExtras();
                    Uri myUri = (Uri) data.get("data");
                    profilePic.setImageURI(myUri);
                }
        });

    uploadPicture.setOnClickListener(view -> {
        Intent imagePickerIntent = new Intent(Intent.ACTION_PICK, 
MediaStore.Images.Media.EXTERNAL_CONTENT_URI);
        imagePickerIntent.setType("image/*");
        activityResultLauncher.launch(imagePickerIntent);
    });

现在我可以在这里输入代码并打开图库并浏览图片，但是当我选择一个并尝试从图库返回我的应用程序时应用程序崩溃。谁能告诉我如何修复我的代码？谢谢

4

1 回答 1

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启动Activity的方法。

 private void selectImage(){
    Intent intent = new Intent(Intent.ACTION_PICK, MediaStore.Images.Media.EXTERNAL_CONTENT_URI);
    activityResultLauncher.launch(intent);
}

然后，按如下方式覆盖 onRequestPermissionsResult()：

 @Override
public void onRequestPermissionsResult(int requestCode, @NonNull String[] permissions, @NonNull int[] grantResults) {
    super.onRequestPermissionsResult(requestCode, permissions, grantResults);
    if (requestCode == REQUEST_CODE_STORAGE_PERMISSION && grantResults.length > 0){
        if (grantResults[0] == PackageManager.PERMISSION_GRANTED){
            selectImage();
        }else {
            Toast.makeText(this, "Permission Denied!", Toast.LENGTH_SHORT).show();
        }
    }
}

另一种，方法如下：

 private void displayResult(){
    activityResultLauncher = registerForActivityResult(new ActivityResultContracts.StartActivityForResult(),
            result -> {
                if (result.getResultCode() == Activity.RESULT_OK){
                    Intent data = result.getData();
                    if (data != null){
                        Uri selectedImageUri = data.getData();
                        if (selectedImageUri != null){
                            try {
                                InputStream inputStream = getContentResolver().openInputStream(selectedImageUri);
                                Bitmap bitmap = BitmapFactory.decodeStream(inputStream);
                                noteBinding.imageNote.setImageBitmap(bitmap);
                                selectedImagePath = getPathFormatUri(selectedImageUri);
                            }catch (Exception e){
                                Toast.makeText(CreateNoteActivity.this, e.getMessage(), Toast.LENGTH_SHORT).show();
                            }
                        }
                    }
                }
            });
}

返回图像的 url/路径的另一种方法。

 private String getPathFormatUri(Uri contentUri){
    String filePath;
    Cursor cursor = getContentResolver()
            .query(contentUri,null,null,null,null);
    if (cursor == null){
        filePath = contentUri.getPath();
    }else {
        cursor.moveToFirst();
        int index = cursor.getColumnIndex("_data");
        filePath = cursor.getString(index);
        cursor.close();
    }

    return filePath;
}
于 2021-11-29T03:59:08.350 回答