Question

我从字节数组中在控制器中创建一个zip文件，然后将zip文件作为fileresult返回。当我下载zip文件并解压缩文件时，它已损坏。我是这样做的：

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

你可能不幸遇到 DotNetZip 中的一个漏洞。例如有问题取决于文件大小（https://dotnetzip.codeplex.com/workitem/14087）。

不幸的是， DotNetZip 存在一些关键问题，似乎不再积极维护项目。更好的替代方案是使用SharpZipLib（如果您遵守基于GPL的许可证），或.NET ports of zlib之一。

如果您使用的是.NET 4.5，则可以使用System.IO.Compression命名空间中的内置类。以下示例可以在ZipArchive类的文档中找到：

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;
    }
}

使用dotnetzip压缩文件，打开时它已损坏

2 个答案: