压缩文件，dotnetzip，并打开时，它已损坏(Compress file with dotnet

创建在从字节数组控制器zip文件和我返回zip文件作为fileresult。 当我下载ZIP文件并解压文件时，它已损坏。 我做这样说：

byte[] fileBytes =array
MemoryStream fileStream = new MemoryStream(fileBytes);
MemoryStream outputStream = new MemoryStream();
fileStream.Seek(0, SeekOrigin.Begin);

using (ZipFile zipFile = new ZipFile())
{
    zipFile.AddEntry(returnFileName, fileStream);
    zipFile.Save(outputStream);
}

outputStream.Position = 0;

FileStreamResult fileResult = new FileStreamResult(outputStream, System.Net.Mime.MediaTypeNames.Application.Zip);
fileResult.FileDownloadName = returnFileName + ".zip";

return fileResult;

Answer 1:

你可能会不走运击中DotNetZip的开放的错误之一。 还有例如取决于文件大小（一期https://dotnetzip.codeplex.com/workitem/14087 ）。

不幸的是，DotNetZip有一些关键问题和项目似乎不再积极进行维护。 更好的办法是使用SharpZipLib（如果你符合他们的基于GPL的许可证），或之一的zlib .NET端口 。

如果你在.NET 4.5，你可以在使用内置类System.IO.Compression命名空间。 下面的示例可以在的文档中找到ZipArchive类：

using System;
using System.IO;
using System.IO.Compression;

namespace ConsoleApplication
{
    class Program
    {
        static void Main(string[] args)
        {
            using (var zipToOpen = 
                new FileStream(@"c:\tmp\release.zip", FileMode.Open))
            {
                using (var archive = 
                     new ZipArchive(zipToOpen, ZipArchiveMode.Update))
                {
                    var readmeEntry = archive.CreateEntry("Readme.txt");
                    using (var writer = new StreamWriter(readmeEntry.Open()))
                    {
                            writer.WriteLine("Information about this package.");
                            writer.WriteLine("========================");
                    }
                }
            }
        }
    }
}

Answer 2:

public class HomeController : Controller
{
    public FileResult Index()
    {
        FileStreamResult fileResult = new FileStreamResult(GetZippedStream(), System.Net.Mime.MediaTypeNames.Application.Zip);
        fileResult.FileDownloadName = "result" + ".zip";
        return fileResult;
    }

    private static Stream GetZippedStream()
    {
        byte[] fileBytes = Encoding.ASCII.GetBytes("abc");
        string returnFileName = "something";

        MemoryStream fileStream = new MemoryStream(fileBytes);
        MemoryStream resultStream = new MemoryStream();

        using (ZipFile zipFile = new ZipFile())
        {
            zipFile.AddEntry(returnFileName, fileStream);
            zipFile.Save(resultStream);
        }

        resultStream.Position = 0;
        return resultStream;
    }
}

文章来源: Compress file with dotnetzip, and when open it is corrupted