WP7 - 将json数据发布到Web Service

Question

我想知道将小型json字符串/数据发送到Web服务的最佳方法。我可以使用WebClient或httpwebrequest这不是问题，但我主要关注如何将字符串转换为json格式并发布请求。

正如所建议的，这里我使用的是json.net，以下是代码：

 Uri url = new Uri("http://example.com");




         //Create the web request object
        HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create(url);
        webRequest.Method = "POST";
        webRequest.ContentType = "application/json";

        // Start the request
        webRequest.BeginGetRequestStream(new AsyncCallback(GetRequestStreamCallback), webRequest);


void GetRequestStreamCallback(IAsyncResult asynchronousResult)
    {
        HttpWebRequest webRequest = (HttpWebRequest)asynchronousResult.AsyncState;
        // End the stream request operation
        Stream postStream = webRequest.EndGetRequestStream(asynchronousResult);



        //
        JObject json =
            new JObject(
            new JProperty("customer", new JObject
            (
            new JProperty("phoneNumber", "07700555555"),
            new JProperty("name", "John")
            ))
            ,

            new JProperty("pickupAddress", new JObject
            (
            new JProperty("street", "1 Seagull Lane"),
            new JProperty("city", "London"),
            new JProperty("county", "London"),
            new JProperty("postcode", "E000XX"),
            new JProperty("country", "England"),
            new JProperty("longitude", "10.18"),
            new JProperty("latitude", "12.214")
            ))
            ,
            new JProperty("destinationAddress", new JObject
            (
            new JProperty("county", "London"),
            new JProperty("street", "1 Snow Lane"),
            new JProperty("longitude", "1.79"),
            new JProperty("latitude", "1.294"),
            new JProperty("postcode", "E00XX"),
            new JProperty("country", "England"),
            new JProperty("city", "London")
            ))
            ,


           new JProperty("pickupTime", "1311467121460"),
           new JProperty("notes", "some notes"),
           new JProperty("accountNumber", "account1"),
          new JProperty("accountPassword", "account password")


        );

        //



        // Create the post data
        // Demo POST data 

        //byte[] byteArray = Encoding.UTF8.GetBytes(json.ToString());
        //byte[] byteArray = Encoding.UTF8.GetBytes("1233456");


        //
        JsonSerializer serializer = new JsonSerializer();
        serializer.NullValueHandling = NullValueHandling.Ignore;

        using (StreamWriter sw = new StreamWriter(postStream))
        using (JsonWriter writer = new JsonTextWriter(sw))
        {
            //serializer.Serialize(writer, JsonConvert.SerializeObject(json));
             json.WriteTo(writer,null);

        }

        //



        // Add the post data to the web request
        //postStream.Write(byteArray, 0, byteArray.Length);
        postStream.Close();

        // Start the web request
        webRequest.BeginGetResponse(new AsyncCallback(GetResponseCallback), webRequest);
    }

    void GetResponseCallback(IAsyncResult asynchronousResult)
    {
        try
        {
            HttpWebRequest webRequest = (HttpWebRequest)asynchronousResult.AsyncState;
            HttpWebResponse response;

            // End the get response operation
            response = (HttpWebResponse)webRequest.EndGetResponse(asynchronousResult);
            Stream streamResponse = response.GetResponseStream();
            StreamReader streamReader = new StreamReader(streamResponse);
            var Response = streamReader.ReadToEnd();
            MessageBox.Show(Response);
            streamResponse.Close();
            streamReader.Close();
            response.Close();

        }
        catch (WebException e)
        {
            // Error treatment
            // ...
        }
    }

现在代码似乎没问题，json数据创建正常，但服务器响应显示：“服务器响应e = {”远程服务器返回错误：NotFound。“}”

我不知道出了什么问题，请各位朋友帮忙！有人告诉我，我必须在请求的标题中添加一些内容，即标题应显示：booking = {... JSON ...}

Answer 1

您可以使用内置的DataContractJsonSerializer Class。

或者，您可以使用json.net获得更好的效果。