WCF WebInvoke POST 消息正文答案

【问题标题】：WCF WebInvoke POST message bodyWCF WebInvoke POST 消息正文
【发布时间】：2015-03-02 19:28:16
【问题描述】：

[OperationContract]
[WebInvoke(UriTemplate = "s={s}", Method = "POST")]
string EchoWithPost(string s);

我正在尝试使用WebRequest 使用此方法（WCF 服务）：

WebRequest request1 = WebRequest.Create("http://MyIP/Host");
request1.Method = "POST";
request1.ContentType = "application/x-www-form-urlencoded";
string postData1 = "s=TestString";

我不想在 url 中传递数据 (s=TestString)，我想做的是在消息正文中传递数据。

【问题讨论】：

标签： c# wcf post webinvoke

【解决方案1】：

首先你需要像这样改变你的服务合同：

[OperationContract]
[WebInvoke(UriTemplate = "EchoWithPost", Method = "POST")]
string EchoWithPost(string s);

请注意UriTemplate 如何不再期待 URL 中的任何变量值。

从客户端调用这样的操作：

// Set up request
string postData = @"""Hello World!""";
HttpWebRequest request = 
     (HttpWebRequest)WebRequest.Create("http://MyIP/Host/EchoWithPost");
request.Method = "POST";
request.ContentType = "text/json";
byte[] dataBytes = new ASCIIEncoding().GetBytes(postData);
request.ContentLength = dataBytes.Length;
using (Stream requestStream = request.GetRequestStream())
{
     requestStream.Write(dataBytes, 0, dataBytes.Length);
}

// Get and parse response
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
string responseString = string.Empty;
using (var responseStream = new StreamReader(response.GetResponseStream()))
{
     //responseData currently will be in XML format 
     //<string xmlns="http://schemas.microsoft.com/2003/10/Serialization/">Hello World!</string>
     var responseData = responseStream.ReadToEnd();
     responseString = System.Xml.Linq.XDocument.Parse(responseData).Root.Value;
}

// display response - Hello World!
Console.WriteLine(responseString);
Console.ReadKey();

【讨论】：