设置WebRequest的正文数据

Wil*_*eja 115 c# httpwebrequest

我在ASP.NET中创建一个Web请求,我需要向主体添加一堆数据.我怎么做?

var request = HttpWebRequest.Create(targetURL);
request.Method = "PUT";
response = (HttpWebResponse)request.GetResponse();
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Tor*_*son 99

HttpWebRequest.GetRequestStream

代码示例来自http://msdn.microsoft.com/en-us/library/d4cek6cc.aspx

string postData = "firstone=" + inputData;
ASCIIEncoding encoding = new ASCIIEncoding ();
byte[] byte1 = encoding.GetBytes (postData);

// Set the content type of the data being posted.
myHttpWebRequest.ContentType = "application/x-www-form-urlencoded";

// Set the content length of the string being posted.
myHttpWebRequest.ContentLength = byte1.Length;

Stream newStream = myHttpWebRequest.GetRequestStream ();

newStream.Write (byte1, 0, byte1.Length);
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从我自己的代码之一:

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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Eva*_*ski 47

这应该有所帮助:

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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  • 为了回答@Yogurtu的完整性问题,`newStream`指向的`Stream`对象直接写入请求的主体.可以通过调用`HttpWReq.GetRequestStream()`来访问它.无需在请求中设置任何其他内容. (3认同)

rea*_*sil 9

这个话题的答案都很棒。但是我想提出另一个。很可能您已经获得了一个 api 并希望将其添加到您的 c# 项目中。使用 Postman,您可以在那里设置和测试 api 调用,一旦它正常运行,您只需单击“代码”,您一直在处理的请求就会写入 ac# 代码段。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 包 RestSharp,您可以轻松安装它。