Question

我需要帮助，我有这种方法从Byte []

获取BitmapImage

public BitmapSource ByteToBitmapSource(byte[] image)
{
    BitmapImage imageSource = new BitmapImage();

    using (MemoryStream stream = new MemoryStream(image))
    {
        stream.Seek(0, SeekOrigin.Begin);
        imageSource.BeginInit();
        imageSource.StreamSource = stream;
        imageSource.CacheOption = BitmapCacheOption.OnLoad;
        imageSource.EndInit();
    }

    return imageSource;
}

imageSource.EndInit();抛出错误“我们发现没有适合完成此操作的成像组件。”

Answer 1

将Image.Source设置为XAML中的字节数组属性。

<Image x:Name="MyImage" Source="{Binding Path=MyByteArrayProperty}" />

如果真的希望你可以在这样的代码中执行此操作

  public void DecodePhoto(byte[] byteVal)
        {


                MemoryStream strmImg = new MemoryStream(byteVal);
                BitmapImage myBitmapImage = new BitmapImage();
                myBitmapImage.BeginInit();
                myBitmapImage.StreamSource = strmImg;
                myBitmapImage.DecodePixelWidth = 200;
                myBitmapImage.EndInit();
                MyImage.Source = myBitmapImage;

        }

Answer 2

您应该向我们提供有关您图片的更多信息我可以假设它不受系统支持，我建议你使用外部工具，例如imageMagik或任何其他特定于图像的转换器。

Answer 3

我做了类似的东西，但它不是BitmapImage，希望它可以帮助......

首先，我从路径中获取图像，因此我得到一个BMP并转换为一个字节[]：

private byte[] LoadImage(string szPathname)
  {
     try
     {
        Bitmap image = new Bitmap(szPathname, true);

        MemoryStream ms = new MemoryStream();
        image.Save(ms, ImageFormat.Bmp);
        return ms.ToArray();
     }catch (Exception){...}

     return null;
  }

我在我的代码中这样称呼它：

byte[] bitmapData1 = LoadImage(@"C:\Users\toto\Desktop\test1.bmp");

当我想将byte []转换为System.Windows.Controls.Image时，我这样做：

protected virtual void OnByteArrayChanged(DependencyPropertyChangedEventArgs e)
  {
     try
     {
        // PropertyChanged method
        BitmapImage bmpi = new BitmapImage();
        bmpi.BeginInit();
        bmpi.StreamSource = new MemoryStream(ByteArray);
        bmpi.EndInit();

        System.Windows.Controls.Image image1 = (get my image in my xaml);
        image1.Source = bmpi;
     }catch (Exception){...}
  }

Answer 4

这也可能有所帮助：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Data;
using System.Windows.Documents;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Media.Imaging;
using System.Windows.Navigation;
using System.Windows.Shapes;

using System.Drawing;
using System.Runtime.InteropServices;
using System.IO;
using System.ComponentModel;


public class MakeBitmapSource
{
    [DllImport("kernel32.dll", EntryPoint = "RtlMoveMemory")]
    public static extern void CopyMemory(IntPtr Destination, IntPtr Source, uint Length);



    public static BitmapSource FromNativePointer(IntPtr pData, int w, int h, int ch)
    {
        PixelFormat format = PixelFormats.Default;

        if (ch == 1) format = PixelFormats.Gray8; //grey scale image 0-255
        if (ch == 3) format = PixelFormats.Bgr24; //RGB
        if (ch == 4) format = PixelFormats.Bgr32; //RGB + alpha


        WriteableBitmap wbm = new WriteableBitmap(w, h, 96, 96, format, null);
        CopyMemory(wbm.BackBuffer, pData, (uint)(w * h * ch));

        wbm.Lock();
        wbm.AddDirtyRect(new Int32Rect(0, 0, wbm.PixelWidth, wbm.PixelHeight));
        wbm.Unlock();

        return wbm;
    }

    public static BitmapSource FromArray(byte[] data, int w, int h, int ch)
    {
        PixelFormat format = PixelFormats.Default;

        if (ch == 1) format = PixelFormats.Gray8; //grey scale image 0-255
        if (ch == 3) format = PixelFormats.Bgr24; //RGB
        if (ch == 4) format = PixelFormats.Bgr32; //RGB + alpha


        WriteableBitmap wbm = new WriteableBitmap(w, h, 96, 96, format, null);
        wbm.WritePixels(new Int32Rect(0, 0, w, h), data, ch * w, 0);

        return wbm;
    }
}

如何将Byte []转换为BitmapImage

4 个答案: