c# - 如何从 Web API 2 HttpGet 发送 zip 文件

Question

我正在尝试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。

需要的是文件将被下载到请求它的发件人处。 我看过很多答案，但没有一个能帮助我找到确切的答案。

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); 如果(di.Exists) { //提供要创建的 zip 文件的路径和名称 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);

Answer 1

在你的 Controller 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:文件已存在的解决方法