c# http请求方法名称"POST /sample.xml"
本文关键字:quot POST sample xml http 请求 方法 | 更新日期: 2023-09-27 18:10:07
Post样例xml方法异常消息:403
http请求名称编辑但不工作
HttpWebRequest request;
HttpWebResponse myHttpWebResponse;
WebResponse rsp = null;
Uri baseUri = new Uri("http://192.168.1.75:80");
string postData = "<?xml version='"1.0'" encoding='"utf-8'" ?>"...";
request = (HttpWebRequest)WebRequest.Create(baseUri.AbsoluteUri);
//WebHeaderCollection myWebHeaderCollection = request.Headers;
//myWebHeaderCollection.Add("Content-Type:text/xml charset=utf8");
//myWebHeaderCollection.Add("Content-Length:124");
request.Method = "POST /sample.xml";
UTF8Encoding encoding = new UTF8Encoding();
byte[] body = encoding.GetBytes(postData);
request.ContentType = "text/xml charset=utf8";
request.Accept = "text/html";
// request.Accept = "/setup.xml";
request.ContentLength = body.Length;
Stream newStream = request.GetRequestStream();
newStream.Write(body, 0, body.Length);
newStream.Close();
myHttpWebResponse = (HttpWebResponse)request.GetResponse();
myHttpWebResponse.Close();
右门柱:
POST/sample.xml HTTP/1.0的content - type: text/xml内容长度:124
…
请帮忙:
request.Method = "POST /sample.xml";
应为
request.Method = "POST";
和
Uri baseUri = new Uri("http://192.168.1.75:80");
应该是
Uri baseUri = new Uri("http://192.168.1.75:80/sample.xml");
此外,处理程序/sample.xml必须存在于您的服务器上,并且a=必须接受POST请求。