[英]Downloading of zip file through ASP.NET MVC using DotNetZip
I have created a text file in a folder and zipped that folder and saved @same location for test purpose. 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。 I wanted to download that zip file directly on user machine after it is created.
我想在创建后直接在用户计算机上下载该zip文件。 I am using dotnetzip library and have done following:
我正在使用dotnetzip库并完成以下操作:
Response.Clear();
Response.ContentType = "application/zip";
Response.AddHeader("content-disposition", "filename=" + "sample.zip");
using (ZipFile zip = new ZipFile())
{
zip.AddDirectory(Server.MapPath("~/Directories/hello"));
zip.Save(Server.MapPath("~/Directories/hello/sample.zip"));
}
Can someone please suggest how the zip file can be downloaded at user's end.? 有人可以建议如何在用户端下载zip文件。
You may use the controller's File
method to return a file, like: 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");
}
}
If the zip file is not required otherwise to be stored, it is unnecessary to write it into a file on the server: 如果不需要存储zip文件,则无需将其写入服务器上的文件:
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");
}
}
First of all, consider a way without creating any files on the server's disk. 首先,考虑一种不在服务器磁盘上创建任何文件的方法。 Bad practise.
不好的做法。 I'd recommend creating a file and zipping it in memory instead.
我建议创建一个文件并将其压缩到内存中。 Hope, you'll find my example below useful.
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");
}
Notes to the code above: 上面代码的注释:
MemoryStream
instance requires checks that it's open, valid and etc. I omitted them. MemoryStream
实例需要检查它是否打开,有效等等。我省略了它们。 I'd rather passed a byte array of the file content instead of a MemoryStream
instance to make the code more robust, but it'd be too much for this example. MemoryStream
实例来使代码更健壮,但对于这个例子来说太过分了。 just a fix to Klaus solution: (as I can not add comment I have to add another answer!) 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!)
The solution is great but for me it gave corrupted zip file and I realized that it is because of return is before finalizing zip object so it did not close zip and result in a corrupted zip. 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并导致损坏的zip。
so to fix we need to just move return line after using zip block so it works. 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,以便它可以工作。 the final result is :
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");
}
For those just wanting to return an existing Zip file from the App_Data folder (just dump in your zip files there), in the Home controller create this action method: 对于那些只想从App_Data文件夹中返回现有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);
}
Get File is an extention method: 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;
}
Home controller Index view looks like this: 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>
The main index file action method: 主索引文件操作方法:
public ActionResult Index()
{
var names = XFile.GetFileInformation();
return View(names);
}
Where GetFileInformation is an extension method: GetFileInformation是一个扩展方法:
public static List<FileInfo> GetFileInformation()
{
string path = AppDomain.CurrentDomain.GetData("DataDirectory").ToString();
var dirInfo = new DirectoryInfo(path);
return dirInfo.EnumerateFiles().ToList();
}
Create a GET
-only controller action that returns a FileResult
, like this: 创建一个返回
FileResult
的GET
FileResult
控制器操作,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);
}
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