WCF WebInvoke方法POST

我有一个wcf服务，我想测试发布数据到它。但是我的函数的参数从来没有得到任何值。WCF WebInvoke方法POST

[OperationContract] 
[WebInvoke(UriTemplate = "TestPost", Method = "POST", 
     ResponseFormat = WebMessageFormat.Json, 
     RequestFormat = WebMessageFormat.Json, 
     BodyStyle = WebMessageBodyStyle.WrappedRequest)] 
int Test(string value); 

public int Test(string value) //Value stays null 
{ 
    return 0; 
}

的JSON请求我送，建立使用Fiddler2

http://localhost:49633/Service1.svc/TestPost 

User-Agent: Fiddler 
Host: localhost:49633 
Content-Length: 42 
Content-type: application/json 

{"value":{"name":"value","name1":"value"}}

我想要的参数包含一个JSON字符串，所以基本上我创建一个包含JSON对象JSON请求，因为我想稍后将JSON对象反序列化为我的自定义对象之一。任何想法为什么值参数保持为空？

感谢

来源

2012-03-15 Armand

的参数是一个字符串，所以它的值应该是一个字符串了。尝试传递'{“值”：“{\”name \“：\”value \“，\”name1 \“：\”value \“}”}'代替。虽然我很惊讶，但服务不会给你一个错误，而不是一个空的参数。 – shambulator 2012-03-15 09:22:04

您可以尝试将BodyStyle移除为WrappedRequest的默认值 – Rajesh 2012-03-15 10:06:01

我用下面的方法来发布JSON字符串上面定义的服务，它为我工作：

我的服务：

[WebInvoke(UriTemplate = "TestPost", ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json)] 
     public int Test(string value) 
     { 
      Console.Write(value); 
      return 1; 
     }

我的客户：

private string UseHttpWebApproach(string serviceUrl, string resourceUrl, string method, string requestBody) 
     { 
      string responseMessage = null;     
      var request = WebRequest.Create(string.Concat(serviceUrl, resourceUrl)) as HttpWebRequest; 
      if (request != null) 
      {      
       request.ContentType = "application/json"; 
       request.Method = method; 
      } 

      //var objContent = HttpContentExtensions.CreateDataContract(requestBody); 
      if(method == "POST" && requestBody != null) 
      {     
       byte[] requestBodyBytes = ToByteArrayUsingJsonContractSer(requestBody);     
       request.ContentLength = requestBodyBytes.Length; 
       using (Stream postStream = request.GetRequestStream()) 
        postStream.Write(requestBodyBytes, 0, requestBodyBytes.Length); 

      } 

      if (request != null) 
      { 
       var response = request.GetResponse() as HttpWebResponse; 
       if(response.StatusCode == HttpStatusCode.OK) 
       { 
        Stream responseStream = response.GetResponseStream(); 
        if (responseStream != null) 
        { 
         var reader = new StreamReader(responseStream); 

         responseMessage = reader.ReadToEnd(); 
        } 
       } 
       else 
       { 
        responseMessage = response.StatusDescription; 
       } 
      } 
      return responseMessage; 
     } 

private static byte[] ToByteArrayUsingJsonContractSer(string requestBody) 
     { 
      byte[] bytes = null; 
      var serializer1 = new DataContractJsonSerializer(typeof(string)); 
      var ms1 = new MemoryStream(); 
      serializer1.WriteObject(ms1, requestBody); 
      ms1.Position = 0; 
      var reader = new StreamReader(ms1); 
      bytes = ms1.ToArray(); 
      return bytes; 
     }

我在客户端使用UseHttpWebApproach的呼叫如下：

string serviceBaseUrl = <<your service base url>>; 
string resourceUrl = "/TestPost"; 
string method = "POST"; 
string jsonText = "{\"value\":{\"name\":\"value\",\"name1\":\"value\"}}"; 
UseHttpWebApproach(serviceBaseUrl, resourceUrl, method, jsonText);

下面是我的小提琴手请求：

POST http://localhost/VDName/AppName/TestPost HTTP/1.1 
Content-Type: application/json 
Content-Length: 54 
Connection: Keep-Alive 

"{\"value\":{\"name\":\"value\",\"name1\":\"value\"}}"

来源

2012-03-15 10:17:09 Rajesh

WCF WebInvoke方法POST

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