Question

我在文件夹中创建了一个文本文件并压缩了该文件夹并保存了@same位置以供测试。我想在创建后直接在用户计算机上下载该zip文件。我正在使用dotnetzip库并完成了以下操作：

medium

有人可以建议如何在用户结束时下载zip文件。

Answer 1

您可以使用控制器的File方法返回文件，例如：

public ActionResult Download()
{
    using (ZipFile zip = new ZipFile())
    {
        zip.AddDirectory(Server.MapPath("~/Directories/hello"));
        zip.Save(Server.MapPath("~/Directories/hello/sample.zip"));
        return File(Server.MapPath("~/Directories/hello/sample.zip"), 
                                   "application/zip", "sample.zip");
    }
}

如果不需要存储zip文件，则无需将其写入服务器上的文件中：

public ActionResult Download()
{
    using (ZipFile zip = new ZipFile())
    {
        zip.AddDirectory(Server.MapPath("~/Directories/hello"));

        MemoryStream output = new MemoryStream();
        zip.Save(output);
        return File(output.ToArray(), "application/zip", "sample.zip");
    }  
}

Answer 2

首先，考虑一种不在服务器磁盘上创建任何文件的方法。不好的做法。我建议创建一个文件并将其压缩到内存中。希望，你会发现下面的例子很有用。

/// <summary>
///     Zip a file stream
/// </summary>
/// <param name="originalFileStream"> MemoryStream with original file </param>
/// <param name="fileName"> Name of the file in the ZIP container </param>
/// <returns> Return byte array of zipped file </returns>
private byte[] GetZippedFiles(MemoryStream originalFileStream, string fileName)
{
    using (MemoryStream zipStream = new MemoryStream())
    {
        using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
        {
            var zipEntry = zip.CreateEntry(fileName);
            using (var writer = new StreamWriter(zipEntry.Open()))
            {
                originalFileStream.WriteTo(writer.BaseStream);
            }
            return zipStream.ToArray();
        }
    }
}

/// <summary>
///     Download zipped file
/// </summary>
[HttpGet]
public FileContentResult Download()
{
    var zippedFile = GetZippedFiles(/* your stream of original file */, "hello");
    return File(zippedFile, // We could use just Stream, but the compiler gets a warning: "ObjectDisposedException: Cannot access a closed Stream" then.
                "application/zip",
                "sample.zip");
}

上述代码注释：

传递MemoryStream实例需要检查它是否已打开，有效等等。我省略了它们。我宁愿传递文件内容的字节数组而不是MemoryStream实例，以使代码更健壮，但对于这个例子来说太过分了。
它没有显示如何在内存中创建所需的上下文（您的文件）。我会参考MemoryStream课程来获取指示。

Answer 3

只是对Klaus解决方案的修复:(因为我无法添加评论，我必须添加另一个答案！）

解决方案很棒，但对我来说它提供了损坏的zip文件，我意识到这是因为返回是在最终确定zip对象之前所以它没有关闭zip并导致拉链损坏。

所以为了解决这个问题，我们需要在使用zip块之后移动返回线，这样才能正常工作。最终结果是：

/// <summary>
///     Zip a file stream
/// </summary>
/// <param name="originalFileStream"> MemoryStream with original file </param>
/// <param name="fileName"> Name of the file in the ZIP container </param>
/// <returns> Return byte array of zipped file </returns>
private byte[] GetZippedFiles(MemoryStream originalFileStream, string fileName)
{
    using (MemoryStream zipStream = new MemoryStream())
    {
        using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
        {
            var zipEntry = zip.CreateEntry(fileName);
            using (var writer = new StreamWriter(zipEntry.Open()))
            {
                originalFileStream.WriteTo(writer.BaseStream);
            }
        }
        return zipStream.ToArray();
    }
}

/// <summary>
///     Download zipped file
/// </summary>
[HttpGet]
public FileContentResult Download()
{
    var zippedFile = GetZippedFiles(/* your stream of original file */, "hello");
    return File(zippedFile, // We could use just Stream, but the compiler gets a warning: "ObjectDisposedException: Cannot access a closed Stream" then.
                "application/zip",
                "sample.zip");
}

Answer 4

对于那些只想从App_Data文件夹返回现有Zip文件的人（只在那里转储zip文件），在Home控制器中创建此操作方法：

    public FileResult DownLoad(string filename)
    {
        var content = XFile.GetFile(filename);
        return File(content, System.Net.Mime.MediaTypeNames.Application.Zip, filename);

    }

获取文件是一种扩展方法：

   public static byte[] GetFile(string name)
    {
        string path = AppDomain.CurrentDomain.GetData("DataDirectory").ToString();
        string filenanme = path + "/" + name;
        byte[] bytes = File.ReadAllBytes(filenanme);
        return bytes;
    }

家庭控制器索引视图如下所示：

@model  List<FileInfo>

<table class="table">
    <tr>
        <th>
            @Html.DisplayName("File Name")
        </th>
        <th>
            @Html.DisplayName("Last Write Time")
        </th>
        <th>
            @Html.DisplayName("Length (mb)")
        </th>
        <th></th>
    </tr>

    @foreach (var item in Model)
    {
        <tr>
            <td>
                @Html.ActionLink("DownLoad","DownLoad",new {filename=item.Name})
            </td>
            <td>
                @Html.DisplayFor(modelItem => item.Name)
            </td>
            <td>
                @Html.DisplayFor(modelItem => item.LastWriteTime)
            </td>
            <td>
                @Html.DisplayFor(modelItem => item.Length)
            </td>
        </tr>
    }
</table>

主索引文件操作方法：

    public ActionResult Index()
    {
        var names = XFile.GetFileInformation();
        return View(names);
    }

GetFileInformation是一种扩展方法：

    public static List<FileInfo> GetFileInformation()
    {
        string path = AppDomain.CurrentDomain.GetData("DataDirectory").ToString();
        var dirInfo = new DirectoryInfo(path);
        return dirInfo.EnumerateFiles().ToList();
    }

Answer 5

创建一个GET - 仅控制器操作，返回FileResult，如下所示：

[HttpGet]
public FileResult Download()
{   
    // Create file on disk
    using (ZipFile zip = new ZipFile())
    {
        zip.AddDirectory(Server.MapPath("~/Directories/hello"));
        //zip.Save(Response.OutputStream);
        zip.Save(Server.MapPath("~/Directories/hello/sample.zip"));
    }

    // Read bytes from disk
    byte[] fileBytes = System.IO.File.ReadAllBytes(
        Server.MapPath("~/Directories/hello/sample.zip"));
    string fileName = "sample.zip";

    // Return bytes as stream for download
    return File(fileBytes, "application/zip", fileName);
}

使用DotNetZip通过ASP.NET MVC下载zip文件

5 个答案: