发送复杂的JSON对象
我想与Web服务器通信并交换JSON信息。
我的webservice URL看起来像以下格式：http： http://46.157.263.140/EngineTestingWCF/DPMobileBookingService.svc/SearchOnlyCus
这是我的JSON请求格式。
{ "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" }
我尝试使用以下代码发送请求：
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"; }
然后使用Gson我创建JSON对象，如：
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方法是
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exception：
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.(JSONObject.java:158) at org.json.JSONObject.(JSONObject.java:171)
在我发出请求之前打印出JSON字符串如下：
{ "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" }
我该如何解决这个exception？ 提前致谢！
要创建一个附加了JSON对象的请求，您应该执行以下操作：
public static String sendComment (String commentString, int taskId, String sessionId, int displayType, String url) throws Exception { Map jsonValues = new HashMap(); jsonValues.put("sessionID", sessionId); jsonValues.put("NewTaskComment", commentString); jsonValues.put("TaskID" , taskId); jsonValues.put("DisplayType" , displayType); JSONObject json = new JSONObject(jsonValues); DefaultHttpClient client = new DefaultHttpClient(); HttpPost post = new HttpPost(url + SEND_COMMENT_ACTION); AbstractHttpEntity entity = new ByteArrayEntity(json.toString().getBytes("UTF8")); entity.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json")); post.setEntity(entity); HttpResponse response = client.execute(post); return getContent(response); }
我对Json不是很熟悉，但我知道它现在很常用，你的代码似乎没问题。
如何将此JSON字符串转换为JSON对象？
好吧，你几乎到达那里，只需将JSON字符串发送到您的服务器，然后在您的服务器中再次使用Gson：
Gson gson = new Gson(); Fpack f = gson.fromJSON(json, Fpack.class);
http://google-gson.googlecode.com/svn/trunk/gson/docs/javadocs/index.html
关于例外：
您应该删除此行，因为您正在发送请求，而不是响应一个请求：
httpPostRequest.setHeader("Accept", "application/json");
我会改变这一行：
httpPostRequest.setHeader("Content-type", "application/json");
至
se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
如果这没有任何区别，请在发送请求之前打印出您的JSON字符串，让我们看看那里有什么。
根据我的理解，您希望使用您创建的JSON向服务器发出请求，您可以执行以下操作：
URL url; HttpURLConnection connection = null; String urlParameters ="json="+ jsonSend; try { url = new URL(targetURL); connection = (HttpURLConnection)url.openConnection(); connection.setRequestMethod("POST"); connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded"); connection.setRequestProperty("Content-Language", "en-US"); DataOutputStream wr = new DataOutputStream ( connection.getOutputStream ()); wr.writeBytes (urlParameters); wr.flush (); wr.close (); InputStream is = connection.getInputStream(); BufferedReader rd = new BufferedReader(new InputStreamReader(is)); String line; StringBuffer response = new StringBuffer(); while((line = rd.readLine()) != null) { response.append(line); response.append('\r'); } rd.close(); return response.toString(); } catch (Exception e) { e.printStackTrace(); return null; } finally { if(connection != null) { connection.disconnect(); } } }
实际上这是一个不好的请求。 这就是服务器以XML格式返回响应的原因。 问题是将非原始数据（DATE）转换为JSON对象..因此它将是Bad Request ..我解决了自己以理解GSON适配器..这是我使用的代码：
try { JsonSerializer ser = new JsonSerializer () { @Override public JsonElement serialize(Date src, Type typeOfSrc, JsonSerializationContext comtext) { return src == null ? null : new JsonPrimitive("/Date("+src.getTime()+"+05300)/"); } }; JsonDeserializer deser = new JsonDeserializer () { @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)); } };