c# - 手动发布到 Asp.Net MVC Controller

Question

我有一个看起来像这样的 Controller ，用于将 xml 导入我的站点:

    [HttpPost]
    public ActionResult Import(string xml)
    {

我有一个独立的应用程序，它读取一个 xml 文件并将其发送到 url。它看起来像这样:

    static void Main(string[] args)
    {
        var reader = new StreamReader(@"myfile.xml");
        var request = WebRequest.Create("http://localhost:41379/mycontroller/import");
        request.Method = "POST";
        request.ContentType = "text/xml";
        StreamWriter sw = new StreamWriter(request.GetRequestStream());
        sw.Write(reader.ReadToEnd());
        sw.Close();
        var theResponse = (HttpWebResponse)request.GetResponse();
        StreamReader sr = new StreamReader(theResponse.GetResponseStream());
        var response = sr.ReadToEnd();
    }

Controller 被正确调用，但当我进入那里时，参数为空。我很确定我只是没有设置正确的内容类型或类似的东西。对 xml 进行编码以便框架能够获取它并将其正确地提供给 Controller 的正确方法是什么？

Answer 1

使用 WebClient.UploadFile 为自己省去很多痛苦.

我把你带错了路，我写了一个看起来工作正常的 Controller 和客户端:

Controller

public class HomeController : Controller
{
    public ActionResult Upload()
    {
        XDocument doc;
        using (var sr = new StreamReader(Request.InputStream))
        {
            doc = XDocument.Load(sr);
        }
        return Content(doc.ToString());
    }
}

客户端

static void Main(string[] args)
{

    var req = (HttpWebRequest)WebRequest.Create("http://host/Home/Upload");
    req.Method = "POST";
    req.ContentType = "text/xml";

    using (var stream = File.OpenRead("myfile.xml"))
    using (var requestStream = req.GetRequestStream()) {
        stream.CopyTo(requestStream);
    }


    using (var response = (HttpWebResponse) req.GetResponse())
    using (var responseStream = response.GetResponseStream())
    using (var sr = new StreamReader(responseStream))
    {
        XDocument doc = XDocument.Load(sr);
        Console.WriteLine(doc);
    }


    Console.ReadKey();
}