如何在MultiScale图像中居中

时间:2010-09-08 06:38:15

标签: .net silverlight image xaml deepzoom

找到了将图像定位在多尺度图像中的功能......但我不确定如何获得实际图像宽度(单个或多个)并根据它更改_msi.ViewportOrigin x param。

有2条线影响图像位置......一条是

_msi.ViewportOrigin = new Point(0, 0);

另一个是:

//if (layout == ImageLayout.Vertical) //single column
//    X = ((_msi.ViewportWidth - subImages[i].Width) / 2);

我可以改变其中任何一个..但需要一些帮助。

上述代码段的代码:

 private void ArrangeImagesTile(ImageLayout layout)
        {
            if (_msi.ActualWidth <= 0 || _msi.ActualHeight <= 0)
                return;

            _lastMousePos = new Point(0, 0);
            _msi.ViewportOrigin = new Point(0, 0);
            _msi.ViewportWidth = 1;


            Storyboard moveStoryboard = initStoryboard();

            double containerAspectRatio = this._msi.ActualWidth / this._msi.ActualHeight;
            double spaceBetweenImages = 0.005;

            List<SubImage> subImages = new List<SubImage>();
            _imagesToShow.ForEach(subImage => subImages.Add(new SubImage(subImage)));

            // Capture the total width of all images
            double totalImagesWidth = 0.0;
            subImages.ForEach(subImage => totalImagesWidth += subImage.Width);

            // Calculate the total number of rows required to display all the images
            int numRows = 1; // layout - horizontal
            if (layout == ImageLayout.One)
                numRows = 1; //(int)Math.Sqrt((totalImagesWidth / containerAspectRatio) + 1);
            else if (layout == ImageLayout.Four) //.Vertical)
                numRows = 2; // subImages.Count;

            // Assign images to each row
            List<Row> rows = new List<Row>(numRows);
            for (int i = 0; i < numRows; i++)
                rows.Add(new Row(spaceBetweenImages));

            double widthPerRow = totalImagesWidth / numRows;
            double imagesWidth = 0;

            // Separate the images into rows. The total width of all images in a row should not exceed widthPerRow
            for (int i = 0, j = 0; i < numRows; i++, imagesWidth = 0)
            {
                while (imagesWidth < widthPerRow && j < subImages.Count)
                {
                    rows[i].AddImage(subImages[j]);
                    subImages[j].RowNum = i;
                    imagesWidth += subImages[j++].Width;
                }
            }

            // At this point in time the subimage height is 1 
            // If we assume that the total height is also 1 we need to scale the subimages to fit within a total height of 1
            // If the total height is 1, the total width is aspectRatio. Hence (aspectRatio)/(total width of all images in a row) is the scaling factor.
            // Added later: take into account spacing between images
            rows.ForEach(Row => Row.Scale(containerAspectRatio));

            // Calculate the total height, with space between images, of the images across all rows
            // Also adjust the colNum for each image
            double totalImagesHeight = (numRows - 1) * spaceBetweenImages;
            rows.ForEach(Row => totalImagesHeight += Row.Height);

            // The totalImagesHeight should not exceed 1. 
            // if it does, we need to scale all images by a factor of (1 / totalImagesHeight)
            if (totalImagesHeight > 1)
            {
                subImages.ForEach(subImage => subImage.Scale(1 / (totalImagesHeight + spaceBetweenImages)));
                totalImagesHeight = (numRows - 1) * spaceBetweenImages;
                rows.ForEach(Row => totalImagesHeight += Row.Height);
            }

            // Calculate the top and bottom margin
            double margin = (1 - totalImagesHeight) / 2;

            if (_imagesToHide != null)
            {
                // First hide all the images that should not be displayed
                _imagesToHide.ForEach(subImage =>
                {
                    //Do not use opacity for this as it slows down the animation after a few arranges
                    subImage.ViewportWidth = 0;
                });
            }

            // Then display the displayable images to scale
            for (int i = 0; i < _imagesToShow.Count; i++)
            {
                double X = rows[subImages[i].RowNum].CalcX(subImages[i].ColNum);
                //if (layout == ImageLayout.Vertical) //single column
                //    X = ((_msi.ViewportWidth - subImages[i].Width) / 2);

                double Y = margin;
                for (int j = 0; j < subImages[i].RowNum; j++)
                    Y += spaceBetweenImages + rows[j].Height;

                _imagesToShow[i].ViewportWidth = containerAspectRatio / subImages[i].Width;
                animateImage(moveStoryboard, _imagesToShow[i], new Point(-(X / subImages[i].Width), -(Y / subImages[i].Width)));    // for animation, use this statement instead of the next one                
                _imagesToShow[i].Opacity = 1.0;
            }

            if (ImagesRearranged != null)
            {
                ImagesRearranged(this, EventArgs.Empty);
            }

            // Play Storyboard
            moveStoryboard.Begin();
        }

以前的代码参考,在msi中打开图像时进入上述功能:

后端:

private void RootMultiScaleImage_Loaded(object sender, RoutedEventArgs e)
        {
            // Use the mid point of the image to zoom from    
            var xx = (MultiScaleImage) sender;
            xx.ZoomAboutLogicalPoint(1, 0.5, 0.5);
        }

前端:

 <ControlTemplate x:Key="DeepZoomerControlTemplate" TargetType="zoom:DeepZoomer">
            <Grid>
<MultiScaleImage x:Name="RootMultiScaleImage" Loaded="RootMultiScaleImage_Loaded" />

1 个答案:

答案 0 :(得分:2)

我同意这是相当令人困惑的,但是使用viewPortWidth和viewPortOrigin你应该能够做到。

  • 首先,您必须检查ViewPortWidth是否为&gt; 1(这意味着您的图像目前相对于父级是“更窄”。如果不是这样,您可以检查ViewPortHeight&gt; 1(图像是否更短,您必须垂直居中)。

    < / LI>
  • 假设您发现ViewPortWidth为&gt; 1,即右侧有空白区域并想要水平居中视口,您可以将一个负值设置为ViewPortOrigin以将视口移动到右侧。

示例:ViewPortWidth为3.这意味着您的图片正在填充可用宽度的1/3。你必须将它的宽度向右移动一次。 ViewportOrigin变为(-1,0)。

另一个例子:ViewPortWidth是4.您的图片填充了可用宽度的1/4。如果将ViewPortOrigin设置为-1.5,视口实际上向右移动了1.5倍宽度,实际上移到了中心。

通用公式* 应为ViewPortOrigin.x = - (ViewPortWidth - 1)/ 2

我建议您查看doc并在纸上画一些草图,直到找到答案。