无法将JSON参数从Android发布到WCF

Question

是的，我已经阅读了所有与此相关的提交，以寻找代码中的缺陷。 我的Android应用程序发布到WCF Web服务（4.0）。 该服务在UriTemplate UriTemplate =“ PostUpdate / {userIMEI}”中接收传递的参数，但是[DataContract]中的JSON参数不会反序列化。 在Windows 7 IIS和实时Web服务器上进行测试。

WCF IWebService.cs

    [OperationContract]
    [WebInvoke(Method = "POST", 
        RequestFormat=WebMessageFormat.Json, 
        ResponseFormat = WebMessageFormat.Json, 
        BodyStyle = WebMessageBodyStyle.Wrapped, 
        UriTemplate = "PostUpdate/{userIMEI}")]

    string PostUpdate(string userIMEI, PostUpdateContract UserInfo);

数据合同是...

   [DataContract]
public class PostUpdateContract
{
    [DataMember]
    public string UserIMEI { get; set; }
    [DataMember]
    public string UserEmail { get; set; }
    [DataMember]
    public string UserMobileNumber { get; set; }
    [DataMember]
    public string UserFirst { get; set; }
    [DataMember]
    public string UserLast { get; set; }
    [DataMember]
    public string UserPassword { get; set; }
}

在WebService.svc.cs中

    public string PostUpdate(string userIMEI, PostUpdateContract UserInfo)
    {   // Post User Information Update to database for this IMEI.

        string userEmail = UserInfo.UserEmail;
        string userMobileNumber = UserInfo.UserMobileNumber;
        string userFirst = UserInfo.UserFirst;
        string userLast = UserInfo.UserLast;
        string userPassword = UserInfo.UserPassword;

        // ** DEBUG **
           ClassLibrary.Functions.EventLogSave(userIMEI + ", " + userEmail + ", " + userFirst + ", " + userMobileNumber);

        // Insert new user or update existing
        //int thisUserId = ClassLibrary.Functions.SaveUserInformation(UserIMEI, UserInfo);

        return "IMEI:" + userIMEI + ", First: " + UserInfo.UserFirst;
    }

Android应用程式文章为：

     http://192.168.0.34/WebServices/WebService.svc/PostUpdate//355500000000000

Android应用程序的JSON数据如下所示：

    {"UserInfo":{"IMEI":"355500000000000","EMAIL":"myemail@cox.net","MOBILE":"19495551212","FIRST":"Myfirst","LAST":"Mylast","PW":"12345678"}}

函数PostUpdate接收userIMEI，但是PostUpdateContracl UserInfo中的值不可用。 除了数据，我什么都没看到？ 谢谢

最后是AsyncTask doInBackground内部的Android应用DoJsonPost。

    private String doJSONPost(String thisUrl){
        // Do a JSON Post called from inside of AsyncTask()
        //  Build JSON message to post.
        Integer intResult = 0;
        HttpURLConnection urlConnection;
        StringBuilder result = new StringBuilder();

        String returnString = "";

        IMEI = "/" + thisIMEI ;

        String jsonString = "";
        JSONObject jsonObject = new JSONObject();
        try {
            JSONStringer jsonStringer = new JSONStringer()
                    .object()
                    .key("UserInfo")
                    .object()
                    .key("IMEI").value(thisIMEI)
                    .key("EMAIL").value(thisUserEmail)
                    .key("MOBILE").value(thisUserMobileNumber)
                    .key("FIRST").value(thisUserFirstName)
                    .key("LAST").value(thisUserLastName)
                    .key("PW").value(thisUserPassword)
                    .endObject()
                    .endObject();
            jsonString = jsonStringer.toString();

        } catch (JSONException jex)
        {
            jsonString = "";
        }

        try{
            // Build URL web service
            StringBuilder sb = new StringBuilder();
            sb.append(SERVICE_URL);
            sb.append(WEB_SERVICE);
            sb.append(POST_UPDATE);
            sb.append(IMEI);

            String httpConnect = sb.toString();

            Log.d("TRAFFIC", "Post to: " + httpConnect);

            // Connect to web service
            URL url = new URL(httpConnect);

            urlConnection = (HttpURLConnection) url.openConnection();

            urlConnection.setDoOutput(true);

            urlConnection.setRequestMethod("POST");

            urlConnection.setRequestProperty("Content-Type", "application/json");
            urlConnection.setRequestProperty("Accept", "application/json");
            urlConnection.setRequestProperty("Content-Length", "" + Integer.toString(jsonString.getBytes().length));
            urlConnection.setUseCaches(false);
            urlConnection.setConnectTimeout(10000);
            urlConnection.setReadTimeout(10000);

            urlConnection.connect();

            // Post Json
            OutputStreamWriter outputStreamWriter = new OutputStreamWriter(urlConnection.getOutputStream());
            outputStreamWriter.write(jsonString);
            outputStreamWriter.close();

            // Receive Response from server
            int statusCode = urlConnection.getResponseCode();
            Log.d("TRAFFIC", "StatusCode: " + statusCode);

                /* 200 represents HTTP OK */
            if (statusCode == 200) {
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream(), "UTF-8"));
                String line;
                sb.setLength(0);    // clear stringbuilder
                while ( (line=bufferedReader.readLine()) != null){
                    sb.append(line + "\n");
                }
                returnString = sb.toString();

            }else{
                intResult = 0; //"Failed to fetch data!";
                returnString = "";
            }

        } catch (Exception ex){
            errorMessage = ex.toString();
            Log.d("TRAFFIC", "JSON Post ERROR: " + errorMessage);
            returnString = "";
        }
        return returnString;

    }   // end doJSONPost()

谢谢。

Answer 1

答案：JSON键值名称与DataContract中的名称不匹配。 这个问题（由我解决）。

已更改：

.object()
                .key("UserInfo")
                .object()
                .key("IMEI").value(thisIMEI) to "UserIMEI"
                .key("EMAIL").value(thisUserEmail) to "UserEmail"
                .key("MOBILE").value(thisUserMobileNumber) to "UserMobileNumber"
                .key("FIRST").value(thisUserFirstName) etc.
                .key("LAST").value(thisUserLastName)
                .key("PW").value(thisUserPassword)
                .endObject()
                .endObject();