使用System.IO.Compression创建的zip文件无效 [英] Invalid zip file after creating it with System.IO.Compression

问题描述

我正在尝试创建一个包含一个或多个文件的zip文件.
我正在使用.NET Framework 4.5，更具体地说是使用System.IO.Compression命名空间.
目的是允许用户通过ASP.NET MVC应用程序下载zip文件.
该zip文件正在生成并发送到客户端，但是当我尝试通过双击打开它时，出现以下错误:
Windows无法打开该文件夹.压缩(压缩)文件夹...无效.
这是我的代码:

I'm trying to create a zip file that contains one or more files.
I'm using the .NET framework 4.5 and more specifically System.IO.Compression namespace.
The objective is to allow a user to download a zip file through a ASP.NET MVC application.
The zip file is being generated and sent to the client but when I try to open it by doing double click on it I get the following error:
Windows cannot open the folder. The compressed (zipped) folder ... is invalid.
Here's my code:

[HttpGet]
public FileResult Download()
{
    var fileOne = CreateFile(VegieType.POTATO);
    var fileTwo = CreateFile(VegieType.ONION);
    var fileThree = CreateFile(VegieType.CARROT);

    IEnumerable<FileContentResult> files = new List<FileContentResult>() { fileOne, fileTwo, fileThree };
    var zip = CreateZip(files);

    return zip;
}

private FileContentResult CreateFile(VegieType vType)
{
    string fileName = string.Empty;
    string fileContent = string.Empty;

    switch (vType)
    {
        case VegieType.BATATA:
            fileName = "batata.csv";
            fileContent = "THIS,IS,A,POTATO";
            break;
        case VegieType.CEBOLA:
            fileName = "cebola.csv";
            fileContent = "THIS,IS,AN,ONION";
            break;
        case VegieType.CENOURA:
            fileName = "cenoura.csv";
            fileContent = "THIS,IS,A,CARROT";
            break;
        default:
            break;
    }

    var fileBytes = Encoding.GetEncoding(1252).GetBytes(fileContent);
    return File(fileBytes, MediaTypeNames.Application.Octet, fileName);
}

private FileResult CreateZip(IEnumerable<FileContentResult> files)
{
    byte[] retVal = null;

    if (files.Any())
    {
        using (MemoryStream zipStream = new MemoryStream())
        {
            using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, false))
            {
                foreach (var f in files)
                {
                    var entry = archive.CreateEntry(f.FileDownloadName, CompressionLevel.Fastest);
                    using (var entryStream = entry.Open())
                    {
                        entryStream.Write(f.FileContents, 0, f.FileContents.Length);
                        entryStream.Close();
                    }
                }

                zipStream.Position = 0;
                retVal = zipStream.ToArray();
            }
        }
    }

    return File(retVal, MediaTypeNames.Application.Zip, "horta.zip");
}

任何人都可以阐明为什么Windows双击双击后我的zip文件无效.
最后，我可以使用7-Zip打开它.

Can anyone please shed some light on why is windows saying that my zip file is invalid when I double click on it.
A final consideration, I can open it using 7-Zip.

推荐答案

在处置ZipArchive对象之后，需要通过ToArray获取MemoryStream缓冲区.否则，您最终将导致存档损坏.

You need to get the MemoryStream buffer via ToArray after the ZipArchive object gets disposed. Otherwise you end up with corrupted archive.

并且请注意，我已更改ZipArchive构造函数的参数，以在添加条目时使其保持打开状态.

And please note that I have changed the parameters of ZipArchive constructor to keep it open when adding entries.

在处理ZipArchive时会有一些校验和，因此如果您之前阅读过MemoryStream，它仍然不完整.

There is some checksumming going on when the ZipArchive is beeing disposed so if you read the MemoryStream before, it is still incomplete.

    private FileResult CreateZip(IEnumerable<FileContentResult> files)
    {
        byte[] retVal = null;

        if (files.Any())
        {
            using (MemoryStream zipStream = new MemoryStream())
            {
                using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
                {
                    foreach (var f in files)
                    {
                        var entry = archive.CreateEntry(f.FileDownloadName, CompressionLevel.Fastest);
                        using (BinaryWriter writer = new BinaryWriter(entry.Open()))
                        {                                   
                            writer.Write(f.FileContents, 0, f.FileContents.Length);
                            writer.Close();
                        }
                    }

                    zipStream.Position = 0;
                }
                retVal = zipStream.ToArray();
            }
        }

        return File(retVal, MediaTypeNames.Application.Zip, "horta.zip");
    }

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