发送复杂的JSON对象 [英] Sending Complex JSON Object

问题描述

我想用一个Web服务器和交换JSON的信息沟通。

I want to communicate with a web server and exchange JSON information.

我的web服务的URL看起来像如下格式： http://46.157.263.140/EngineTestingWCF/DPMobileBookingService.svc/SearchOnlyCus ​​

my webservice URL looking like following format: http://46.157.263.140/EngineTestingWCF/DPMobileBookingService.svc/SearchOnlyCus

下面是我的JSON请求格式。

Here is my JSON Request format.

{
    "f": {
        "Adults": 1,
        "CabinClass": 0,
        "ChildAge": [
            7
        ],
        "Children": 1,
        "CustomerId": 0,
        "CustomerType": 0,
        "CustomerUserId": 81,
        "DepartureDate": "/Date(1358965800000+0530)/",
        "DepartureDateGap": 0,
        "Infants": 1,
        "IsPackageUpsell": false,
        "JourneyType": 2,
        "PreferredCurrency": "INR",
        "ReturnDate": "/Date(1359138600000+0530)/",
        "ReturnDateGap": 0,
        "SearchOption": 1
    },
    "fsc": "0"
}

我试着用下面的code发送一个请求：

I tried with the following code to send a request:

public class Fdetails {
    private String Adults = "1";
    private String CabinClass = "0";
    private String[] ChildAge = { "7" };
    private String Children = "1";
    private String CustomerId = "0";
    private String CustomerType = "0";
    private String CustomerUserId = "0";
    private Date DepartureDate = new Date();
    private String DepartureDateGap = "0";
    private String Infants = "1";
    private String IsPackageUpsell = "false";
    private String JourneyType = "1";
    private String PreferredCurrency = "MYR";
    private String ReturnDate = "";
    private String ReturnDateGap = "0";
    private String SearchOption = "1";
}

public class Fpack {
    private Fdetails f = new Fdetails();
    private String fsc = "0";
}

然后用GSON我创建 JSON对象这样的：

public static String getJSONString(String url) {
String jsonResponse = null;
String jsonReq = null;
Fpack fReq = new Fpack();

try {                                                
    Gson gson = new Gson();
    jsonReq = gson.toJson(fReq);                        
    JSONObject json = new JSONObject(jsonReq);
    JSONObject jsonObjRecv = HttpClient.SendHttpPost(url, json);
    jsonResponse = jsonObjRecv.toString();
} 
catch (JSONException e) {
                           e.printStackTrace();
                }               
return jsonResponse;
    }

和我的 HttpClient.SendHttpPost 方法

public static JSONObject SendHttpPost(String URL, JSONObject json) {

        try {
            DefaultHttpClient httpclient = new DefaultHttpClient();
            HttpPost httpPostRequest = new HttpPost(URL);

            StringEntity se;
            se = new StringEntity(json.toString());
            httpPostRequest.setEntity(se);
            se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));          
            HttpResponse response = (HttpResponse) httpclient.execute(httpPostRequest);
            HttpEntity entity = response.getEntity();
            if (entity != null) {
                // Read the content stream
                InputStream instream = entity.getContent();
                Header contentEncoding = response.getFirstHeader("Content-Encoding");
                if (contentEncoding != null && contentEncoding.getValue().equalsIgnoreCase("gzip")) {
                    instream = new GZIPInputStream(instream);
                }

                // convert content stream to a String
                String resultString= convertStreamToString(instream);
                instream.close();
                resultString = resultString.substring(1,resultString.length()-1); // remove wrapping "[" and "]"

                // Transform the String into a JSONObject
                JSONObject jsonObjRecv = new JSONObject(resultString);
            return jsonObjRecv;
            } 
    catch (Exception e)
        {
            e.printStackTrace();
        }
        return null;
    }

现在我得到以下异常：

org.json.JSONException: Value !DOCTYPE of type java.lang.String cannot be converted to JSONObject
at org.json.JSON.typeMismatch(JSON.java:111)
at org.json.JSONObject.<init>(JSONObject.java:158)
at org.json.JSONObject.<init>(JSONObject.java:171)

和JSON字符串的打印输出之前，我提出请求如下：

and the printout of JSON string right before I make the request is as follows:

{
    "f": {
        "PreferredCurrency": "MYR",
        "ReturnDate": "",            
        "ChildAge": [
            7
        ],
        "DepartureDate": "Mar 2, 2013 1:17:06 PM",
        "CustomerUserId": 0,
        "CustomerType": 0,
        "CustomerId": 0,
        "Children": 1,
        "DepartureDateGap": 0,
        "Infants": 1,
        "IsPackageUpsell": false,
        "JourneyType": 1,
        "CabinClass": 0,
        "Adults": 1,
        "ReturnDateGap": 0,
        "SearchOption": 1

    },
    "fsc": "0"
}

我该怎么解决这个异常？在此先感谢！

How do I solve this exception? Thanks in advance!

推荐答案

其实这是一个错误的请求。这就是为什么服务器返回的响应为XML格式。 现在的问题是转换的非原始数据（DATE），以JSON对象..所以这将是错误的请求...... 我解决了自己了解GSON适配器。这里是code我用：

Actually it was a BAD REQUEST. Thats why server returns response as XML format. The problem is to convert the non primitive data(DATE) to JSON object.. so it would be Bad Request.. I solved myself to understand the GSON adapters.. Here is the code I used:

try {                                                       
                        JsonSerializer<Date> ser = new JsonSerializer<Date>() {

                            @Override
                            public JsonElement serialize(Date src, Type typeOfSrc,
                                    JsonSerializationContext comtext) {

                                return src == null ? null : new JsonPrimitive("/Date("+src.getTime()+"+05300)/");
                            }
                        };
                        JsonDeserializer<Date> deser = new JsonDeserializer<Date>() {                           
                            @Override
                            public Date deserialize(JsonElement json, Type typeOfT,
                                    JsonDeserializationContext jsonContext) throws JsonParseException {

                                String tmpDate = json.getAsString();

                                  Pattern pattern = Pattern.compile("\\d+");
                                  Matcher matcher = pattern.matcher(tmpDate);
                                  boolean found = false;

                                  while (matcher.find() && !found) {
                                       found = true;
                                        tmpDate = matcher.group();
                                  }
                                return json == null ? null : new Date(Long.parseLong(tmpDate));                             
                            }
                        };

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