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Question

我正在尝试使用 SevenZipSharp 压缩和解压缩内存流。压缩工作正常但解压缩不是。我认为 SevenZipSharp 无法从流中确定存档类型。

SevenZipCompressor compress = new SevenZip.SevenZipCompressor();
compress.CompressionLevel = CompressionLevel.Normal;
compress.CompressionMethod = CompressionMethod.Lzma

using (MemoryStream memStream = new MemoryStream())
{
    compress.CompressFiles(memStream, @"d:\Temp1\MyFile.bmp");

    using (FileStream file = new FileStream(@"d:\arch.7z", FileMode.Create, System.IO.FileAccess.Write))
    {
        memStream.CopyTo(file);
    }
}

//works till here, file is created
Console.Read();

using (FileStream file = new FileStream(@"d:\arch.7z", FileMode.Open, System.IO.FileAccess.Read))
{
    using (MemoryStream memStream = new MemoryStream())
    {
        file.CopyTo(memStream);

        //throws exception here on this line
        using (var extractor = new SevenZipExtractor(memStream))
        {
            extractor.ExtractFiles(@"d:\x", 0);
        }
    }
}

Answer 1

尝试查看您的输出文件是否可以使用 7Zip 客户端加载。我猜它会失败。

问题出在写入内存流上。比如说，你向流写入 100 个字节，它将位于位置 100。当你使用 CopyTo 时，流将从 current 位置复制，而不是流的开头。

因此您必须在 reading/writing 之后将位置重置为 0，以允许下一个 reader 读取所有数据。例如在创建 7Zip 文件时：

using (MemoryStream memStream = new MemoryStream())
{
    // Position starts at 0
    compress.CompressFiles(memStream, @"d:\Temp1\MyFile.bmp");
    // Position is now N

    memStream.Position = 0; // <-- Reset the position to 0.

    using (FileStream file = new FileStream(@"d:\arch.7z", FileMode.Create, System.IO.FileAccess.Write))
    {
        // Will copy all data in the stream from current position till the end of the stream.
        memStream.CopyTo(file);
    }
}

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c#

memorystream

sevenzipsharp