.NET:使用数据和读取响应发送POST的最简单方法

时间:2010-11-03 15:21:46

标签: c# .net http http-post httpresponse

令我惊讶的是,在.NET BCL中我无法做到这么简单,我可以告诉他们:

byte[] response = Http.Post
(
    url: "http://dork.com/service",
    contentType: "application/x-www-form-urlencoded",
    contentLength: 32,
    content: "home=Cosby&favorite+flavor=flies"
);

上面的假设代码使用数据生成HTTP POST,并在静态类Post上返回Http方法的响应。

既然我们没有这么简单,那么下一个最佳解决方案是什么?

如何发送带有数据的HTTP POST并获取响应的内容?

8 个答案:

答案 0 :(得分:285)

   using (WebClient client = new WebClient())
   {

       byte[] response =
       client.UploadValues("http://dork.com/service", new NameValueCollection()
       {
           { "home", "Cosby" },
           { "favorite+flavor", "flies" }
       });

       string result = System.Text.Encoding.UTF8.GetString(response);
   }

您将需要以下内容:

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

如果您坚持使用静态方法/类:

public static class Http
{
    public static byte[] Post(string uri, NameValueCollection pairs)
    {
        byte[] response = null;
        using (WebClient client = new WebClient())
        {
            response = client.UploadValues(uri, pairs);
        }
        return response;
    }
}

然后简单地说:

var response = Http.Post("http://dork.com/service", new NameValueCollection() {
    { "home", "Cosby" },
    { "favorite+flavor", "flies" }
});

答案 1 :(得分:73)

使用HttpClient:就Windows 8应用程序开发问题而言,我遇到了这个问题。

var client = new HttpClient();

var pairs = new List<KeyValuePair<string, string>>
    {
        new KeyValuePair<string, string>("pqpUserName", "admin"),
        new KeyValuePair<string, string>("password", "test@123")
    };

var content = new FormUrlEncodedContent(pairs);

var response = client.PostAsync("youruri", content).Result;

if (response.IsSuccessStatusCode)
{


}

答案 2 :(得分:47)

使用WebRequest。来自Scott Hanselman

public static string HttpPost(string URI, string Parameters) 
{
   System.Net.WebRequest req = System.Net.WebRequest.Create(URI);
   req.Proxy = new System.Net.WebProxy(ProxyString, true);
   //Add these, as we're doing a POST
   req.ContentType = "application/x-www-form-urlencoded";
   req.Method = "POST";
   //We need to count how many bytes we're sending. 
   //Post'ed Faked Forms should be name=value&
   byte [] bytes = System.Text.Encoding.ASCII.GetBytes(Parameters);
   req.ContentLength = bytes.Length;
   System.IO.Stream os = req.GetRequestStream ();
   os.Write (bytes, 0, bytes.Length); //Push it out there
   os.Close ();
   System.Net.WebResponse resp = req.GetResponse();
   if (resp== null) return null;
   System.IO.StreamReader sr = 
         new System.IO.StreamReader(resp.GetResponseStream());
   return sr.ReadToEnd().Trim();
}

答案 3 :(得分:33)

private void PostForm()
{
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create("http://dork.com/service");
    request.Method = "POST";
    request.ContentType = "application/x-www-form-urlencoded";
    string postData ="home=Cosby&favorite+flavor=flies";
    byte[] bytes = Encoding.UTF8.GetBytes(postData);
    request.ContentLength = bytes.Length;

    Stream requestStream = request.GetRequestStream();
    requestStream.Write(bytes, 0, bytes.Length);

    WebResponse response = request.GetResponse();
    Stream stream = response.GetResponseStream();
    StreamReader reader = new StreamReader(stream);

    var result = reader.ReadToEnd();
    stream.Dispose();
    reader.Dispose();
}

答案 4 :(得分:12)

就个人而言,我认为执行http帖子并获得响应的最简单方法是使用WebClient类。这个类很好地抽象了细节。在MSDN文档中甚至还有一个完整的代码示例。

http://msdn.microsoft.com/en-us/library/system.net.webclient(VS.80).aspx

在您的情况下,您需要UploadData()方法。 (同样,代码示例包含在文档中)

http://msdn.microsoft.com/en-us/library/tdbbwh0a(VS.80).aspx

UploadString()可能也会起作用,它会将它抽象出一个级别。

http://msdn.microsoft.com/en-us/library/system.net.webclient.uploadstring(VS.80).aspx

答案 5 :(得分:7)

我知道这是一个老线程,但希望它能帮到某个人。

public static void SetRequest(string mXml)
{
    HttpWebRequest webRequest = (HttpWebRequest)HttpWebRequest.CreateHttp("http://dork.com/service");
    webRequest.Method = "POST";
    webRequest.Headers["SOURCE"] = "WinApp";

    // Decide your encoding here

    //webRequest.ContentType = "application/x-www-form-urlencoded";
    webRequest.ContentType = "text/xml; charset=utf-8";

    // You should setContentLength
    byte[] content = System.Text.Encoding.UTF8.GetBytes(mXml);
    webRequest.ContentLength = content.Length;

    var reqStream = await webRequest.GetRequestStreamAsync();
    reqStream.Write(content, 0, content.Length);

    var res = await httpRequest(webRequest);
}

答案 6 :(得分:5)

你可以使用类似这样的伪代码:

request = System.Net.HttpWebRequest.Create(your url)
request.Method = WebRequestMethods.Http.Post

writer = New System.IO.StreamWriter(request.GetRequestStream())
writer.Write("your data")
writer.Close()

response = request.GetResponse()
reader = New System.IO.StreamReader(response.GetResponseStream())
responseText = reader.ReadToEnd

答案 7 :(得分:3)

鉴于其他答案已有数年历史,目前我的想法可能会有所帮助:

最简单的方法

private async Task<string> PostAsync(Uri uri, HttpContent dataOut)
{
    var client = new HttpClient();
    var response = await client.PostAsync(uri, dataOut);
    return await response.Content.ReadAsStringAsync();
    // For non strings you can use other Content.ReadAs...() method variations
}

更实际的示例

通常,我们正在处理已知类型和JSON,因此您可以通过许多实现来进一步扩展该思想,例如:

public async Task<T> PostJsonAsync<T>(Uri uri, object dtoOut)
{
    var content = new StringContent(JsonConvert.SerializeObject(dtoOut));
    content.Headers.ContentType = MediaTypeHeaderValue.Parse("application/json");

    var results = await PostAsync(uri, content); // from previous block of code

    return JsonConvert.DeserializeObject<T>(results); // using Newtonsoft.Json
}

如何称呼它的例子:

var dataToSendOutToApi = new MyDtoOut();
var uri = new Uri("https://example.com");
var dataFromApi = await PostJsonAsync<MyDtoIn>(uri, dataToSendOutToApi);