Android （裁剪图像）将裁剪矩形从缩放图像映射到原始图像

我正试图在Android中实现图像裁剪，源代码来自AOSP gallery，应用程序代码实际上取自此。我已将ImageView的缩放类型设置为矩阵，并在图像矩阵上使用后平移和后缩放方法缩放和平移图像

在图像上绘制固定260x260像素的裁剪矩形，在ImageView中居中。此裁剪矩形的大小固定，无法移动。相反，允许移动和缩放图像放大/缩小，以便用户可以将图像的所需区域放置在裁剪矩形内

当用户通过dx、dy数量、posttranslatedx移动图像时，会在图像的矩阵上调用dy。然后通过调用matrix.maprect将裁剪矩形映射到当前矩阵。然后，通过将dx、dy转换为图像空间并相应移动裁剪矩形，计算裁剪矩形在原始图像上的新位置

下面是一些移动图像的源代码摘录：

// This matrix is recomputed when we go from the thumbnail image to
// the full size image.
protected Matrix baseMatrix = new Matrix();

// This is the supplementary transformation which reflects what
// the user has done in terms of zooming and panning.
//
// This matrix remains the same when we go from the thumbnail image
// to the full size image.
protected Matrix suppMatrix = new Matrix();

// This is the final matrix which is computed as the concatentation
// of the base matrix and the supplementary matrix.
private final Matrix displayMatrix = new Matrix();

//move image by dx dy
protected void panBy(float dx, float dy) {
    postTranslate(dx, dy);
    setImageMatrix(getImageViewMatrix());
}

 protected void postTranslate(float dx, float dy) {
    suppMatrix.postTranslate(dx, dy);
} 

// Combine the base matrix and the supp matrix to make the final matrix
protected Matrix getImageViewMatrix() {
    // The final matrix is computed as the concatentation of the base matrix
    // and the supplementary matrix
    displayMatrix.set(baseMatrix);
    displayMatrix.postConcat(suppMatrix);
    return displayMatrix;
}

这就是图像移动如何影响裁剪矩形视图

public void handlePan(float dx, float dy)
{
   RectF r = computeLayout();

   float xDelta = (dx * (cropRect.width() / r.width()));
   float yDelta = -(dy * (cropRect.height() / r.height()));
   cropRect.offset(xDelta, yDelta);

   // Put the cropping rectangle inside image rectangle
   cropRect.offset(
           Math.max(0, imageRect.left - cropRect.left),
           Math.max(0, imageRect.top  - cropRect.top));

   cropRect.offset(
           Math.min(0, imageRect.right  - cropRect.right),
           Math.min(0, imageRect.bottom - cropRect.bottom));
}

private RectF computeLayout() {
    RectF r = new RectF(cropRect.left, cropRect.top,
                        cropRect.right, cropRect.bottom);
    matrix.mapRect(r);
    return r;
}

裁剪视图通过以下getunrotatedmatrix函数获取图像的当前矩阵

这很好用。但我正在努力处理图像的缩放。我尝试将一个矩形映射到矩阵，然后将该矩形映射到原始图像，将裁剪矩形缩小到与图像放大相同的大小，反之亦然。但没有一个成功。在imageview中缩放图像后，如何在SD卡上获取裁剪矩形在原始图像上的位置？缩放操作如下所示：

protected void zoomTo(float scaleFactor, float centerX, float centerY) {
    float oldScale = getScale();
    float scale = oldScale*scaleFactor;

    if (scale > maxZoom) {
        scale = maxZoom;
        scaleFactor=maxZoom/oldScale;
    }
    else if(scale<0.5)
    {
        scale=0.5f;
        scaleFactor=0.5f/oldScale;
    }

    suppMatrix.postScale(scaleFactor, scaleFactor, centerX, centerY);
    setImageMatrix(getImageViewMatrix());
}

---

我知道在放大后，裁剪矩形必须变小，以指向原始图像的较小区域，反之亦然。但是如何精确地进行缩放并适当地定位裁剪矩形，考虑到缩放中心可以位于图像上的任何位置，甚至在裁剪矩形之外

使用反转矩阵将矩形映射回原来的位置，而不是放大后将其缩小到应该的位置。假设RectF clip是一个剪辑矩形，然后调用invertedMatrix.mapRectoutRect，clipok，我会尝试让您知道。不起作用。只需将矩形移出图像的边界即可。

protected void zoomTo(float scaleFactor, float centerX, float centerY) {
    float oldScale = getScale();
    float scale = oldScale*scaleFactor;

    if (scale > maxZoom) {
        scale = maxZoom;
        scaleFactor=maxZoom/oldScale;
    }
    else if(scale<0.5)
    {
        scale=0.5f;
        scaleFactor=0.5f/oldScale;
    }

    suppMatrix.postScale(scaleFactor, scaleFactor, centerX, centerY);
    setImageMatrix(getImageViewMatrix());
}