Question

如何直接从zipfile加载图像源。

我使用DotNetZip。

以下2个代码示例。

代码示例1无法加载并显示图像文件。

代码示例1 我将一个zipEntry加载到MemoryStream，并创建一个BitmapImage用作Imagesource但它失败了。??? 只是为了测试我尝试从内存流中保存图像，以确保加载Zip条目。它保存正确：

        MemoryStream memoryStream1 = new MemoryStream();

        ZipFile zip = ZipFile.Read(@"D:\myZipArcive.zip");

        ZipEntry myEntry = zip["myfile.jpg"];
        myEntry.Extract(memoryStream1);
        memoryStream1.Close();

        System.IO.File.WriteAllBytes(@"D:\saved.jpg", memoryStream1.ToArray()); 
        //save file just for testing

        BitmapImage src = new BitmapImage();
        src.BeginInit();
        src.CacheOption = BitmapCacheOption.OnLoad;
        src.StreamSource = memoryStream1;
        src.EndInit();

        myImage.Source = src as ImageSource; //no image is displayed

代码示例2 我将一个文件加载到MemoryStream，并创建一个BitmapImage，用作带有成功的Imagesource。

        FileStream fileStream = File.OpenRead(@"D:\myFile.jpg");

        MemoryStream memoryStream2 = new MemoryStream();
        memoryStream2.SetLength(fileStream.Length);
        fileStream.Read(memoryStream2.GetBuffer(), 0, (int)fileStream.Length);

        BitmapImage src = new BitmapImage();
        src.BeginInit();
        src.CacheOption = BitmapCacheOption.OnLoad;
        src.StreamSource = memoryStream2;       
        src.EndInit();

        myImage.Source = src as ImageSource; //The image is displayed

示例1中的MemoryStream有什么问题当我执行示例1时，我收到一条调试消息： “在mscorlib.dll中发生'System.NotSupportedException'类型的第一次机会异常”

Answer 1

Stream似乎被Extract方法关闭了。但是，您可以从第一个缓冲区创建第二个MemoryStream：

var zipFile = ZipFile.Read(@"D:\myZipArcive.zip");
var zipEntry = zipFile["myfile.jpg"];
var zipStream = new MemoryStream();
zipEntry.Extract(zipStream);

var bitmap = new BitmapImage();

using (var sourceStream = new MemoryStream(zipStream.ToArray()))
{
    bitmap.BeginInit();
    bitmap.CacheOption = BitmapCacheOption.OnLoad;
    bitmap.StreamSource = sourceStream;
    bitmap.EndInit();
}

myImage.Source = bitmap;

更新：因为您事先知道图像缓冲区的未压缩大小，所以可以通过手动提供两个MemoryStream操作的缓冲区来略微提高性能：

var buffer = new byte[zipEntry.UncompressedSize];
var zipStream = new MemoryStream(buffer); // here
zipEntry.Extract(zipStream);
...
using (var sourceStream = new MemoryStream(buffer)) // and here
...

Answer 2

这也有效：

var zipFile = ZipFile.Read(@"D:\myZipArcive.zip");
var zipEntry = zipFile["myfile.jpg"];
var bitmap = new BitmapImage();
bitmap.BeginInit();
bitmap.CacheOption = BitmapCacheOption.OnLoad;
bitmap.StreamSource = zipEntry.OpenReader();
bitmap.EndInit();
myImage.Source = bitmap;

Wpf：从zip加载图像源而不保存文件

2 个答案: