Question

我正在尝试从图库中读取图像，然后将该图像用于输入图像，下面是我的代码

@Override
public void onCreate(Bundle savedInstanceState) 
    {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        mGetImageButton = (Button) findViewById(R.id.button_getImage);
        mGetImageButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                // SET action AND miniType
                Intent intent = new Intent();
                intent.setAction(Intent.ACTION_PICK);
                intent.setType("image/*");
                // REQUEST Uri of image
                startActivityForResult(intent, REQUEST_IMAGE);
            }
        });
        mImageViewForGallery = (ImageView) findViewById(R.id.imageView2);
    }

        @Override
        protected void onActivityResult(int requestCode, int resultCode, Intent data) {
            super.onActivityResult(requestCode, resultCode, data);
            if (resultCode != Activity.RESULT_OK) {return;}

            if (requestCode == REQUEST_IMAGE) {
                Uri uri = data.getData();
                // SET image
                mImageViewForGallery.setImageURI(uri);
                Drawable drawable = mImageViewForGallery.getDrawable();
                InputStream is;
                is = this.getResources().openRawResource(R.drawable.img1);
                Bitmap bmInImg = BitmapFactory.decodeStream(is);

            }
        }

在上面的代码is = this.getResources().openRawResource(R.drawable.img1);正在从可绘制的文件夹名称img1中读取图像，但现在我的图像是我从图库中选择的图像，我如何将该图像作为输入流，就像我尝试的那样

InputStream is;
is=uri;

但它显示错误，对于来自c ++的java来说是新手。

编辑，在@Shawn回答后，我将此代码放在行onActivityResult之后的Drawable drawable = mImageViewForGallery.getDrawable();函数中：

        InputStream is = this.getContentResolver().openInputStream(uri);
        Bitmap bmInImg = BitmapFactory.decodeStream(is);
        InputStream Vign = this.getResources().openRawResource(R.drawable.p);
        Bitmap bmInImg2 = BitmapFactory.decodeStream(Vign);
        mPhotoIntArray = new int[bmInImg.getWidth() * bmInImg.getHeight()];
        nPhotoIntArray = new int[bmInImg.getWidth() * bmInImg.getHeight()];
        vPhotoIntArray = new int[bmInImg2.getWidth() * bmInImg2.getHeight()];

但它在this.getContentResolver().openInputStream(uri);

上显示错误

错误：

Unhandled exception type FileNotFoundException

在代码中使用InputStream Vign = this.getResources().openRawResource(R.drawable.p);时出现错误。

Answer 1

如果你有图像的Uri，你可以使用内容解析器解决它：

InputStream is = context.getContentResolver().openInputStream(uri);

别忘了关闭小溪。（并检查null）。上下文有时候是你的活动或＆＃34;这＆＃34;。

如何将图库图像作为输入流

1 个答案: