我在文件夹中创建了一个文本文件，然后将该文件夹压缩并保存在相同的位置进行测试。我希望在创建后直接将这个zip文件下载到用户的机器上。我正在使用dotnetzip库，并已完成以下操作:
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"));
}
请问有人能建议如何让用户在其端下载压缩文件吗？
你可以使用控制器的 File 方法来返回一个文件，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");
}
}
如果不需要将zip文件保存到服务器上，则无需将其写入文件中：
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");
}
}
.ToArray()，否则会得到一个空的zip文件。 - Richard Vanbergen首先，考虑一种不在服务器磁盘上创建任何文件的方法。这是不好的做法。我建议在内存中创建一个文件并压缩它。希望你会发现下面的示例有用。
/// <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");
}
上述代码注意事项：
MemoryStream实例需要检查它是否打开、有效等。我省略了这些步骤。为了使代码更加健壮，我宁愿传递文件内容的字节数组而不是MemoryStream实例，但这对于本示例来说过于繁琐。对 Klaus 的解决方案进行修正：（由于我无法添加评论，只能添加另一个答案！）
这个解决方案很好，但是我得到的是损坏的 zip 文件，我意识到这是因为在完成 zip 对象之前返回了，所以它没有关闭 zip 并导致 zip 文件损坏。
所以我们只需要将 return 行移动到使用 zip 块之后即可使其工作。
最终结果如下：
/// <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");
}
对于那些只想从App_Data文件夹返回现有Zip文件的人（只需将zip文件放在其中），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 是一个扩展方法：
public static byte[] GetFile(string name)
{
string path = AppDomain.CurrentDomain.GetData("DataDirectory").ToString();
string filenanme = path + "/" + name;
byte[] bytes = File.ReadAllBytes(filenanme);
return bytes;
}
主页控制器的索引视图如下：
@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();
}
GET请求的控制器操作，返回一个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);
}