我如何能使一个HTTP POST请求和发送数据的主体?


当前回答

如果你喜欢一个流畅的API,你可以使用Tiny.RestClient。在NuGet上可以买到。

var client = new TinyRestClient(new HttpClient(), "http://MyAPI.com/api");
// POST
var city = new City() { Name = "Paris", Country = "France" };
// With content
var response = await client.PostRequest("City", city)
                           .ExecuteAsync<bool>();

其他回答

为什么这不是完全无关紧要的?执行请求并不是处理结果。而且似乎还涉及到一些。net Bug——参见HttpClient中的Bug。GetAsync应该抛出WebException,而不是TaskCanceledException

我最终得到了这样的代码:

static async Task<(bool Success, WebExceptionStatus WebExceptionStatus, HttpStatusCode? HttpStatusCode, string ResponseAsString)> HttpRequestAsync(HttpClient httpClient, string url, string postBuffer = null, CancellationTokenSource cts = null) {
    try {
        HttpResponseMessage resp = null;

        if (postBuffer is null) {
            resp = cts is null ? await httpClient.GetAsync(url) : await httpClient.GetAsync(url, cts.Token);

        } else {
            using (var httpContent = new StringContent(postBuffer)) {
                resp = cts is null ? await httpClient.PostAsync(url, httpContent) : await httpClient.PostAsync(url, httpContent, cts.Token);
            }
        }

        var respString = await resp.Content.ReadAsStringAsync();
        return (resp.IsSuccessStatusCode, WebExceptionStatus.Success, resp.StatusCode, respString);

    } catch (WebException ex) {
        WebExceptionStatus status = ex.Status;
        if (status == WebExceptionStatus.ProtocolError) {
            // Get HttpWebResponse so that you can check the HTTP status code.
            using (HttpWebResponse httpResponse = (HttpWebResponse)ex.Response) {
                return (false, status, httpResponse.StatusCode, httpResponse.StatusDescription);
            }
        } else {
            return (false, status, null, ex.ToString());
        }

    // https://devblogs.microsoft.com/dotnet/net-5-new-networking-improvements/
    } catch (TaskCanceledException ex) when (ex.InnerException is TimeoutException) {
        return (false, ex.ToString(), null, WebExceptionStatus.Timeout);

    } catch (TaskCanceledException ex) {
        return (false, ex.ToString(), null, WebExceptionStatus.RequestCanceled);

    } catch (Exception ex) {
        return (false, WebExceptionStatus.UnknownError, null, ex.ToString());
    }
}

这将根据postBuffer是否为空来执行GET或POST操作。

如果Success为true,响应将在ResponseAsString中。

如果Success为false,你可以检查WebExceptionStatus, HttpStatusCode和ResponseAsString,看看哪里出了问题。

当使用Windows.Web.Http命名空间时,对于POST而不是FormUrlEncodedContent,我们编写HttpFormUrlEncodedContent。同样,响应类型为HttpResponseMessage。其余的就像Evan Mulawski写的那样。

简单的GET请求

using System.Net;

...

using (var wb = new WebClient())
{
    var response = wb.DownloadString(url);
}

简单的POST请求

using System.Net;
using System.Collections.Specialized;

...

using (var wb = new WebClient())
{
    var data = new NameValueCollection();
    data["username"] = "myUser";
    data["password"] = "myPassword";

    var response = wb.UploadValues(url, "POST", data);
    string responseInString = Encoding.UTF8.GetString(response);
}

如果你喜欢一个流畅的API,你可以使用Tiny.RestClient。在NuGet上可以买到。

var client = new TinyRestClient(new HttpClient(), "http://MyAPI.com/api");
// POST
var city = new City() { Name = "Paris", Country = "France" };
// With content
var response = await client.PostRequest("City", city)
                           .ExecuteAsync<bool>();

MSDN有一个样本。

using System;
using System.IO;
using System.Net;
using System.Text;

namespace Examples.System.Net
{
    public class WebRequestPostExample
    {
        public static void Main()
        {
            // Create a request using a URL that can receive a post. 
            WebRequest request = WebRequest.Create("http://www.contoso.com/PostAccepter.aspx");
            // Set the Method property of the request to POST.
            request.Method = "POST";
            // Create POST data and convert it to a byte array.
            string postData = "This is a test that posts this string to a Web server.";
            byte[] byteArray = Encoding.UTF8.GetBytes(postData);
            // Set the ContentType property of the WebRequest.
            request.ContentType = "application/x-www-form-urlencoded";
            // Set the ContentLength property of the WebRequest.
            request.ContentLength = byteArray.Length;
            // Get the request stream.
            Stream dataStream = request.GetRequestStream();
            // Write the data to the request stream.
            dataStream.Write(byteArray, 0, byteArray.Length);
            // Close the Stream object.
            dataStream.Close();
            // Get the response.
            WebResponse response = request.GetResponse();
            // Display the status.
            Console.WriteLine(((HttpWebResponse)response).StatusDescription);
            // Get the stream containing content returned by the server.
            dataStream = response.GetResponseStream();
            // Open the stream using a StreamReader for easy access.
            StreamReader reader = new StreamReader(dataStream);
            // Read the content.
            string responseFromServer = reader.ReadToEnd();
            // Display the content.
            Console.WriteLine(responseFromServer);
            // Clean up the streams.
            reader.Close();
            dataStream.Close();
            response.Close();
        }
    }
}