C# 设置WebRequest'；人体数据

我正在ASP.NET中创建一个web请求，需要向正文中添加一组数据。我该怎么做

var request = HttpWebRequest.Create(targetURL);
request.Method = "PUT";
response = (HttpWebResponse)request.GetResponse();

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更新

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原创

var request = (HttpWebRequest)WebRequest.Create("https://example.com/endpoint");

string stringData = ""; // place body here
var data = Encoding.Default.GetBytes(stringData); // note: choose appropriate encoding

request.Method = "PUT";
request.ContentType = ""; // place MIME type here
request.ContentLength = data.Length;

var newStream = request.GetRequestStream(); // get a ref to the request body so it can be modified
newStream.Write(data, 0, data.Length);
newStream.Close();

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使用

HttpWebRequest.GetRequestStream

源代码示例

来自我自己的代码之一：

var request = (HttpWebRequest)WebRequest.Create(uri);
request.Credentials = this.credentials;
request.Method = method;
request.ContentType = "application/atom+xml;type=entry";
using (Stream requestStream = request.GetRequestStream())
using (var xmlWriter = XmlWriter.Create(requestStream, new XmlWriterSettings() { Indent = true, NewLineHandling = NewLineHandling.Entitize, }))
{
    cmisAtomEntry.WriteXml(xmlWriter);
}

try 
{    
    return (HttpWebResponse)request.GetResponse();  
}
catch (WebException wex)
{
    var httpResponse = wex.Response as HttpWebResponse;
    if (httpResponse != null)
    {
        throw new ApplicationException(string.Format(
            "Remote server call {0} {1} resulted in a http error {2} {3}.",
            method,
            uri,
            httpResponse.StatusCode,
            httpResponse.StatusDescription), wex);
    }
    else
    {
        throw new ApplicationException(string.Format(
            "Remote server call {0} {1} resulted in an error.",
            method,
            uri), wex);
    }
}
catch (Exception)
{
    throw;
}

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这个题目的答案都很好。不过我想提出另一个建议。最有可能的情况是，您已经获得了一个api，并希望将其应用到您的c#项目中。 使用Postman，您可以在那里设置和测试api调用，一旦api调用正常运行，您只需单击“代码”，就可以将您正在处理的请求写入c#snippet。像这样：

var client = new RestClient("https://api.XXXXX.nl/oauth/token");
client.Timeout = -1;
var request = new RestRequest(Method.POST);
request.AddHeader("Authorization", "Basic   N2I1YTM4************************************jI0YzJhNDg=");
request.AddHeader("Content-Type", "application/x-www-form-urlencoded");
request.AddHeader("Content-Type", "application/x-www-form-urlencoded");
request.AddParameter("grant_type", "password");
request.AddParameter("username", "development+XXXXXXXX-admin@XXXXXXX.XXXX");
request.AddParameter("password", "XXXXXXXXXXXXX");
IRestResponse response = client.Execute(request);
Console.WriteLine(response.Content);

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上面的代码依赖于nuget软件包RestSharp，您可以轻松安装该软件包。

Hi Torbjorn，我正在使用请求，以便获得“request.GetResponse（）；”，在上面的示例中，它是如何工作的？当您调用GetRequestStream（）时，它会调用服务器。因此，您必须将其添加到上述示例的末尾。是否有一种方法可以查看请求对象内部的全文以进行调试？我尝试序列化它，并尝试使用StreamReader，但无论我做什么，我都看不到我刚刚写入请求的数据。Fan-freaking-tastic@詹姆斯，你应该可以用fiddler或wireshark来查看完整的请求，你是否遗漏了什么？像httpWReq.Content=newStream；您没有在webRequest中使用newStream对象。要回答@Yogurtu的完整性问题，

Stream

对象指向直接写入请求正文的

newStream

。通过调用

HttpWReq.GetRequestStream（）

可以访问它。不需要在请求上设置任何其他内容。另外，在Postman窗口的右上方可以找到“Code”按钮，您可以在其中设置和测试api调用

var client = new RestClient("https://api.XXXXX.nl/oauth/token");
client.Timeout = -1;
var request = new RestRequest(Method.POST);
request.AddHeader("Authorization", "Basic   N2I1YTM4************************************jI0YzJhNDg=");
request.AddHeader("Content-Type", "application/x-www-form-urlencoded");
request.AddHeader("Content-Type", "application/x-www-form-urlencoded");
request.AddParameter("grant_type", "password");
request.AddParameter("username", "development+XXXXXXXX-admin@XXXXXXX.XXXX");
request.AddParameter("password", "XXXXXXXXXXXXX");
IRestResponse response = client.Execute(request);
Console.WriteLine(response.Content);