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如何从Web API 2 HttpGet发送zip文件

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我试图弄清楚如何从给定的文件夹路径创建新的zip文件，并将其发送回发件人 .

需要的是该文件将在请求它的发件人处下载 . 我已经看到了很多答案，但没有人帮我确切的答案 .

我的代码：

Guid folderGuid = Guid.NewGuid（）; string folderToZip = ConfigurationManager.AppSettings [“folderToZip”] folderGuid.ToString（）;

Directory.CreateDirectory（folderToZip）;

string directoryPath = ConfigurationManager.AppSettings [“DirectoryPath”]; string combinedPath = Path.Combine（directoryPath，id）;

DirectoryInfo di = new DirectoryInfo（combinedPath）; if（di.Exists）{//提供zip文件的路径和名称以创建字符串zipFile = folderToZip“\”folderGuid“.zip”;

//call the ZipFile.CreateFromDirectory() method
ZipFile.CreateFromDirectory(combinedPath, zipFile, CompressionLevel.Fastest, true);

var result = new HttpResponseMessage(HttpStatusCode.OK);
using (ZipArchive zip = ZipFile.Open(zipFile, ZipArchiveMode.Read))
{
    zip.CreateEntryFromFile(folderFiles, "file.zip");
}

var stream = new FileStream(zipFile, FileMode.Open);
result.Content = new StreamContent(stream);
result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
result.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
    FileName = "file.zip"
};
log.Debug("END ExportFiles()");
return ResponseMessage(result);
c# zipfile ziparchive

1 回答

  • 1

    在你的控制器中：

    using System.IO.Compression.FileSystem; // Reference System.IO.Compression.FileSystem.dll

[HttpGet]
[Route("api/myzipfile"]
public dynamic DownloadZip([FromUri]string dirPath)
{
if(!System.IO.Directory.Exists(dirPath))
   return this.NotFound();

    var tempFile = System.IO.Path.Combine(System.IO.Path.GetTempPath(), Guid.NewGuid()); // might want to clean this up if there are a lot of downloads
    ZipFile.CreateFromDirectory(dirPath, tempFile);
    HttpResponseMessage response = new HttpResponseMessage(HttpStatusCode.OK);
    response.Content = new StreamContent(new FileStream(tempFile, FileMode.Open, FileAccess.Read));
    response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
    response.Content.Headers.ContentDisposition.FileName = fileName;
    response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");

  return response;
}

    UPD：解决方案已存在的文件

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