android,json,android-studio,jsonobject
I´ll respond to the future, if someone has the same error: I did this (Thaks to Bigdestroyer): File -> Invalidate caches/Restart -> Invalidate and Restart You can try to delete ~/.gradle/caches directory to force re-download the dependencies. Now its works perfect. Thanks !!!! ...
android,json,arrays,jsonobject
Do this way Replace these two lines LessonTemplateArrayList.setVideo(customFieldsObject.getString("wpcf-video").replace("\\", "")); LessonTemplateArrayList.setIconImage(customFieldsObject.getString("wpcf-icon").replace("\\", "")); With LessonTemplateArrayList.setVideo(customFieldsObject.getJSONArray("wpcf-video").getString(0)); LessonTemplateArrayList.setIconImage(customFieldsObject.getJSONArray("wpcf-icon").getString(0)); ...
java,android,stringbuilder,jsonobject
Use this: try { JSONObject job = new JSONObject(sb.toString()); int Status = job.getInt("Status"); String SessionId = job.getString("SessionId"); } catch (JSONException e) { // Handle error } Reference: http://developer.android.com/reference/org/json/JSONObject.html...
java,javascript,json,unicode,jsonobject
There are two different things happening: Unicode encoding and JSON string escaping. Per 2.5 Strings of the JSON RFC: .. All Unicode characters may be placed within the quotation marks except for the characters that must be escaped .. Any character may be escaped. If the character is in the...
javascript,angularjs,angularjs-directive,jsonobject
At the moment, you aren't referencing an actual object, only a property on the attrs construct that will return a string. There are a couple of ways you could pull in an actual reference. Using scope.$eval //Target object will be pulled in if it exists on the scope // this...
If it starts with a square bracket then it's not a JSON object, it's a JSON array.
android,json,android-asynctask,android-5.0-lollipop,jsonobject
Okay, so after looking for an answer it seems the problem isn't with my code, the problem is with android 5. I tried many tutorials of JsonArray/JsonObject, downloaded their samples etc but nothing worked. Also, I've witnessed a growing number of developer who suffer from the same problem. Thanks to...
java,arrays,json,parsing,jsonobject
The JSON you've got is for an array, not an object. You probably want JSONArray array = new JSONArray(jsonString); Full sample code: import org.json.*; public class Test { public static void main(String[] args) { String json = "[[\"1\",\"100\"],[\"2\",\"200\"],[\"3\",\"300\"]]"; JSONArray array = new JSONArray(json); JSONArray first = array.getJSONArray(0); System.out.println(first.getString(1)); // Prints...
json,maven,hadoop,noclassdeffounderror,jsonobject
You can use jackson to process json. <repositories> <repository> <id>codehaus</id> <url>http://repository.codehaus.org/org/codehaus</url> </repository> </repositories> <dependencies> <dependency> <groupId>org.codehaus.jackson</groupId> <artifactId>jackson-mapper-asl</artifactId> <version>1.8.5</version> </dependency> </dependencies> or google's gson <dependency>...
Refer to this link http://developer.android.com/reference/org/json/JSONObject.html JSONObject put(String name, Object value) Maps name to value, clobbering any existing name/value mapping with the same name. Datatype Object is the root class of the Java class hierarchy. All non-primitive types (including arrays) inherit either directly or indirectly from this class. Therefore jsonObject.put("name", "myname");...
In your code "doc list" log could be truncated. "total records" is an Integer so it's truncated. To log "jsonObj" the only way is to store it in a file. You have to add permission to write external storage: <manifest ...> <uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" /> ... </manifest> And then put your...
java,json,string,equals,jsonobject
Jackson Json parser has a nice feature that it can parse a Json String into a Map. You can then query the entries or simply ask on equality: import com.fasterxml.jackson.databind.ObjectMapper; import java.util.*; public class Test { public static void main(String... args) { String input1 = "{\"state\":1,\"cmd\":1}"; String input2 = "{\"cmd\":1,\"state\":1}";...
look closely at data[0]: posts contains an array of two strings that say "[object Object]", not two objects. How are you parsing the json? oh, and your syntax is correct for extracting the username from the nested objects...
json,extjs,textfield,jsonobject
The scope within your select function no longer contains datajson. To continue to use this.datajson you can pass the required scope into the function using bind Ext.application({ name: 'Fiddle', datajson: { 'name': 'Hello' }, displayName: function (combo, record, index) { Ext.getCmp('name').setValue((this.datajson.name)); }, launch: function () { Ext.create('Ext.form.Panel', { title: 'Contact...
thanks for your response I've manage to solve this problem as follow: JSONObject outerObject = new JSONObject(); JSONArray outerArray = new JSONArray(); JSONObject [] innerObject = new JSONObject[4]; for(int i =0; i<3; i++) { innerObject[i]=new JSONObject(); innerObject[i].put("id", "id of "+i); innerObject[i].put("url", "url of "+i); innerObject[i].put("Definition", "Definition of "+i); outerArray.put(innerObject[i]); }...
php,android,json,security,jsonobject
Server response is came in JSONArray so no need to convert in JSONObject just build JSONArray from server response and return as JSONArray instead JSONObject from makeHttpRequest : JSONArray jsonArray = jsonParser.makeHttpRequest(EVENTS_URL, "POST", params); ...
That package is discontinued and shouldn't be used any more. If you have a console application, I could recommend two options: JSON.NET. Not a part of the .NET Framework itself, but much faster than Microsofts implementation; Microsofts implementation of the JavaScriptSerializer. Both are capable of reading and writing JSON. I...
android,arrays,json,jsonobject
If this is the end of your code, JSONObject solutions = new JSONObject(); solutions.put("solutions", 20); JSONObject refundable = new JSONObject(); refundable.put("refundable","false"); then the issue is you're never actually adding the objects to your request object. Also, those are values and not JSONObjects, so you should replace it with: requests.put("solutions", 20);...
spring,resttemplate,jsonobject
After some research on how MappingJackson2HttpMessageConverter actually works and taking a closer look at Camunda REST API's request content format, I found that the problem lies in Jackson converter not able to parse anything that doesn't starts with the ArrayList start token "[". Since I have to pass a request...
json,java-ee,gson,java-ee-7,jsonobject
JAX-RS will convert your objects to JSON, no need to do it manually. I.e., the following code: @GET @Path("/books") @Produces(MediaType.APPLICATION_JSON) public Response getBooks() { List<Book> listOfBooks = getMiscService().getbooks(); return Response.status(Response.Status.OK).entity(listOfBooks).build(); } ...will produce a JSON like: [ { "title": "Book1", "author": "Foo", ... }, { "title": "Book2", "author": "Bar", ......
The problem is in your method that calls addCounters(). The reference you're passing in the parameter long[] counters is being initialized once to new long[2] and mistakenly reused. You must initialize the variable to a new long array each time you execute the sorting algorithm.
Try this.. JSONObject jsonObject = new JSONObject(response.toString()); JSONObject jsonOb=jsonObject.getJSONObject("LocationMatrix"); Log.e(TAG,"jsonOb :"+jsonOb.toString()); Iterator i = jsonOb.keys(); while (i.hasNext()) { try { String key = i.next().toString(); String j = jsonOb.getString(key); locationList.add(new Location(Integer.parseInt(key), j)); } catch (JSONException e) { e.printStackTrace(); } } ...
java,javascript,json,jsp,jsonobject
Yes, that's right JSON.parse(data) resulted in an error. Also wrap your script in a document.ready function $(function(){ ... }); to be sure that the DOM is ready before you access it. The rest of your code is working. I just commented the parameters to the server because I have no...
headers should be a JSONArray, not JSONObject. An object is a set of name-to-value mappings, an array is a linear collection. JSONArray headers = new JSONObject(); headers.add("ID"); headers.add("Organization Name"); headers.add("Submission Date"); headers.add("Status"); JSONObject organizationsHJ = new JSONObject(); organizationsHJ.put("headers", headers); array.add(organizationsHJ ); ...
hashmap,put,jsonobject,linkedhashmap
Luca, please describe more clearly what you have in your structure at the end of your test, {"Key" : { } } doesn't mean much, what kind of Object is the {} value? Looking at this series.put(serieObject.getSerie_instance(), serieObject); serieObject seems to have at least that weirdly named getSerie_instance key and...
java,android,json,escaping,jsonobject
Problem caused because of = sign near by actionResult as well as actionResult not surrounded with double quotes and you didn't close json string properly. Replace Json String With: { "responseData": { "emotion":"", "lastinput":{"actionResult":{"value":{"label":"green","key":"1"},"result":"success","action":"displayClickableList"}, "answer":"Sorry, I did not understand.", "link": { "href":"", "target":"" }, "extraData": { }, "responseSession": { "id":"c4a5ef257851a991eb32c69132c9",...
android,json,parsing,jsonobject
The way to do it is: JSONArray cast = obj.getJSONArray("data3"); String[] string_array = new String[](); for (int i=0; i<cast.length(); i++) { string_array[i] = cast.getString(i); retVal.data3.set(i, string_array[i]); } Try this. This will work....
I have made changes in such a way to build the JSONArray from SQL output. JSONObject jo = new JSONObject(); Connection conn = DB.getConnection(); ResultSet rs; PreparedStatement ps = conn.prepareStatement("select city,state from zips limit 2"); rs = ps.executeQuery(); JSONArray ja = new JSONArray(); while (rs.next()) { jo.put("city", rs.getString("city")); jo.put("state", rs.getString("state"));...
android,android-fragments,serializable,jsonobject
Well I managed to figure a way out of this, it may not be the most standard of way of doing things but it actually solved my problem and might help others to some extent. What I did was instead of converting the JSONArray that was received from the API...
Finally got the solution, at the time of debugging, string shows all data doesn't matter how much it is, but in JSONObject all data is available but it doesn't shows while debugging.When retrieved the data from JSONObject i got all fields which are not shown in debugger.
json,java-8,optional,jsonobject
It is not refactorable in a nicely way to Optionals in this case. Why? getJSONObject() method will never return null. If there is no value associated with provided key, it will throw JSONException. Optional monad was designed for handling nulls and not exceptions. Those checks are necessary in this case....
json,swift,parsing,jsonobject,nsjsonserialization
& is required for in-out parameters. I hope you are aware that your app will crash if it doesn't receive a JSON dictionary. Which is very likely to happen if you are behind a paywall, for example. So your app is likely to crash if I try it in a...
Change this section : for(int x = 0; x < retorno.length(); x++){ JSONObject jsoObject = retorno.getJSONObject(x); JSONObject jsoLocal = jsoObject.getJSONObject("Local"); JSONObject jsoTipoLocal =jsoObject.getJSONObject("TipoLocal"); } To for(int x = 0; x < retorno.length(); x++){ JSONObject jsoObject = retorno.getJSONObject(x); if(jsoObject.has("Local")) { JSONObject jsoLocal = jsoObject.getJSONObject("Local"); } if(jsoObject.has("TipoLocal")) { JSONObject jsoTipoLocal = jsoObject.getJSONObject("TipoLocal");...
I was able to resolve this. The problem was message.put("message", tempObject.toString()); The method call tempObject.toString() leads to an iteration through the list (presumption) and the message.put(...) leads to accessing the resultant string. Sometimes this leads to ConcurrentModificationException. I was able to resolve it by doing it in two lines and...
What put does is set the key (the first parameter) to the value (the second parameter). So, first you set json.Country to Ireland and json.Capital to Dublin. Next, you overwrite json.Country and set it to Spain, and continue to do so until finally, the Country and Capital are overwritten to...
arrays,json,variables,jmeter,jsonobject
You could trt with this snippet: if(${i}%(${bufferSize}*${sample}-1)==0 && ${i}!=0 || ${i}==${totalNumber}-${endOfDb}){ vars.put("json",vars.get("json")+part.toString()); JSONArray records= new JSONArray("["+vars.get("json")+"]"); log.info(records.toString()); //records.put(tempArray); JSONObject kinesis = new JSONObject(); kinesis.put("records",records); kinesis.put("streamName","kinesis"); vars.put("jsonBuffer",kinesis.toString()); vars.put("json",""); } ...
javascript,json,arrays,jsonobject
This should populate the cities array: var data = { China: ["Guangzhou","Fax"], Majorette: ["Fungous","Godzilla"], Bhutan: ["Thimphu","Parr","Photofinishing"] } var cities = []; for(var i in data){ if(data.hasOwnProperty(i)){ for(var j in data[i] ){ if(data[i].hasOwnProperty(j)){ cities.push(data[i][j]); } } } } console.log(cities); JSFIDDLE Demo...
android,json,android-volley,jsonobject,postman
The problem was in Postman caching It's better to disable the parameter "Auto save request" in the settings....
android,arrays,json,jsonobject
You should really look into How to parse JsonObject / JsonArray in android. Please put your code of also what you have tried because people are here to help solve error/problem not to do coding. Here is code from which you can Parse your json JSONObject jsonObject = new JSONObject();...
android,listview,android-listview,android-arrayadapter,jsonobject
As you are using just a single text field in your code you need not make a custom Adapter just for that as you can use Android's ArrayAdapter. ArrayAdapter<String> adapter = new ArrayAdapter<String>(this, android.R.layout.simple_list_item_1, android.R.id.text1, listdata.toArray()); menu_list.setAdapter(adapter); This snippet uses default Android List layout and a TextView which will take...
android,json,facebook-graph-api,jsonobject
Since data is inside of friends. Try this JSONObject friends= object.getJSONObject("friends"); JSONArray data = friends.getJSONArray("data"); ...
javascript,google-maps-api-3,latitude-longitude,jsonobject
var lat = this.getPosition().lat(); var lng = this.getPosition().lng(); Generally there are getters and setters for many things in GM, and you should look for those in the docs before going into the inspector....
javascript,json,google-maps,marker,jsonobject
From your current code you would then need to loop through the markers with a for loop like so and add a marker to the map: var data = '{"markers":[{"k":48.80686346108517,"B":1.494140625},{"k":50.28933925329177,"B":14.326171875},{"k":43.70759350405294,"B":21.357421875},{"k":30.977609093348686,"B":11.337890625},{"k":40.58058466412761,"B":-0.87890625},{"k":48.45835188280866,"B":1.318359375}]}'; function createMarkerBasedOnFetchedPosition(fetchedMarkersPosition) { var jsonData = JSON.parse(fetchedMarkersPosition); for (var i = 0; i <...
java,android,json,arrays,jsonobject
You are getting response in JSONObject and you are trying to get it in JSONArray.. thats why you are getting error.. Try this way... try { JSONObject result = new JSONObject(response); if(data.has("ValidateLoginResult"){ JSONArray array = result.getJSONArray("ValidateLoginResult"); for (int i = 0; i < array.length(); i++) { JSONObject obj = array.getJSONObject(i);...
json,datetime,gson,jodatime,jsonobject
In the JS/Jquery, I used var endTimestamp = new Date(data[i].endTimestamp.iMillis); ...
android,json,android-edittext,jsonobject
Try this: ArrayList<Integer> minqtyList = new ArrayList<Integer>(); JSONArray array = ck.getJSONArray(BULKPRICE_LIST); for (int i = 0; i < array.length(); i++) { JSONObject row = array.getJSONObject(i); int minqty = row.getInt("minqty"); minqtyList.add(minqty); } In your code, you need to edit this portion like that: bulkprice = ck.getJSONArray(BULKPRICE_LIST); allqtys=new ArrayList<String>(); // initialisation must...
java,android,arrays,json,jsonobject
If you want to get "Yuliem" you can do this: String json = "{\"id\":[{\"id\":\"2\",\"name\":\"Yuliem\"}],\"success\":1,\"message\":\"Product successfully created.\"}"; JSONObject obj = new JSONObject(json); JSONArray array = (JSONArray) obj.get("id"); JSONObject innerObj = (JSONObject) array.get(0); Log.d("User's name: ",innerObj.get("name")); ...
java,android,jsonobject,keyset,missingmethod
See the @hide annotation * @hide. */ public Set<String> keySet() { return nameValuePairs.keySet(); } ...
java,android,json,arrays,jsonobject
You are trying to get JSONArray of "questionAnswers" from "data" instead of "survey" String yourJSONString = "{"code":1,"data":{"survey":{"id":null,"questionAnswers":[{"id":null,"min":"1","max":"5","step":"0.1","last":"4.7","text":"I am happy at work today","key":"Mood","testQuestionId":"1","datetime":"2014-04-29 15:47:02","answer":"4.7","testAnswerId":null}]}}}"; JSONObject jObj_main = new JSONObject(yourJSONString); JSONObject jOb_data = jObj_main.getJSONObject("data"); JSONObject jOb_survey = jOb_data.getJSONObject("survey");...
android,json,parse.com,arrays,jsonobject
whats the mimetype of the data you stored as file on parse? Normally when you upload to parse you provide a file name to the parse api (including ".typ" ) that will indicate a mimetype to any subsequent GET requests. Those gets can just wrap http protocol with the parse...
android,arrays,json,parsing,jsonobject
There are several ways to do this, but I'll post the way I feel like makes the most sense. If you don't mind putting this in separate arrays you can do this: String stringOfJSONCode = <read in the JSON here>; JSONObject json = new JSONObject(stringOfJSONCode); Iterator<String> iter = json.keys(); while...
android,json,arrays,jsonobject
How do i convert it into jsonobject. e.g "name":["The name field is required."] You dont need to convert these JSONarrays to JSONobjects. when you found "[]" means its array and "{}" means its object. In your case first convert your total json to JSONobject then retrive its value from its...
android,json,arrays,jsonobject,android-json
Hope this may help you JSONObject jsonObject=new JSONObject(result); jsonObject=jsonObject.getJSONObject("1"); Double lat=jsonObject.getDouble("latitude"); Toast.makeText(getBaseContext(),""+ lat, Toast.LENGTH_LONG).show(); ...
android,parsing,retrofit,jsonobject
You need to create a response object public class UserResponse { private User responseData; private String message; private boolean success; } And change your callback to return a UserResponse public interface RetrofitService { @FormUrlEncoded @POST("/login") public void login(@Field("email_id") String emailId, @Field ("password") String password, Callback <UserResponse> callback); } NOTE: You...
The problem is that you are trying to parse the next level also. When you parse from JSON to an object, it will parse all levels. You get an object containing objects, not an object containing JSON strings that needs to be parsed. Just get the object from the property:...
javascript,iteration,jsonobject
Try this code. Ensure you have parsed the json string. And use array map function to get all names. var jsonString = '{"data":{"pkg":[{"name":"ThisIsName1","FRONTPAGE":"n","IP":"n","MAXSQL":"1","MAXPOP":"unlimited"},{"name":"ThisIsName2","FRONTPAGE":"n","IP":"n","MAXSQL":"0","MAXPOP":"unlimited"},{"name":"ThisIsName3","FRONTPAGE":"n","IP":"n","MAXSQL":"0","MAXPOP":"unlimited"}]},"metadata":{"version":1,"reason":"OK","result":1,"command":"listpkgs"}}'; var jsonData = JSON.parse(jsonString); var names =...
There are millions of implementation and usecase about JSON+Android. I use gson in order to de/serialize an object as below. It is more reliable than sequence-sensitive Parcelable conversion for me. public static String toJson(Object obj){ Gson gson = new Gson(); return gson.toJson(obj); } public static Object fromJson(String json, Class clazz)...
You can use JsonParser to convert your Json into an intermediate structure which allow you to examine the json content. String yourJson = "{your json here}"; JsonParser parser = new JsonParser(); JsonElement element = parser.parse(yourJson); JsonObject obj = element.getAsJsonObject(); //since you know it's a JsonObject Set<Map.Entry<String, JsonElement>> entries = obj.entrySet();//will...
ios,objective-c,arrays,jsonobject
Depending on the JSON, the object returned from [NSJSONSerialization JSONObjectWithData] is either an NSArray or NSDictionary. If objectAtIndex doesn't work, chances are you are dealing with an NSDictionary. The safest way to go is to check which one it is in your code: if([strResult isKindOfClass:NSArray.class]) { // Do something that...
android,json,import,arrays,jsonobject
You can use JSONArray post_id = new JSONArray(jsonStr); instead of JSONObject jsonObj = new JSONObject(jsonStr); // Getting JSON Array node post_id = jsonObj.getJSONArray(TAG_POSTID); because jsonStr you get is a JSONArray...
android,string,http-post,jsonobject
Try using URL Connection : public JSONObject readJSONFromURL(String urlString, JSONObject param){ String response = null; JSONObject jObject = null; try { URL url = new URL(urlString); URLConnection conn = url.openConnection(); conn.setDoOutput(true); List<NameValuePair> params = new ArrayList<NameValuePair>(); params.add(new BasicNameValuePair("dmode", param.toString())); OutputStream os = conn.getOutputStream(); BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));...
javascript,json,arrays,jsonobject
You're close: var seriesValues = []; for (var abc in objJSON.yAxis) { seriesValues.push({name: abc, data: objJSON.yAxis[abc]}); } That will make a javascript array. If you need it as a JSON string then append: var myString = JSON.stringify(seriesValues); ...
java,android,json,android-intent,jsonobject
If you have to use org.json.simple.*, here is the solution to convert java Strings that look like JSONArrays into parseable JSONObjects String searchResponseJSON = API.search(term, location); JSONParser parser = new JSONParser(); JSONObject response = null; try { response = (JSONObject) parser.parse(searchResponseJSON); } catch (ParseException pe) { System.out.println("Error: could not parse...
The json string seem to be legit. Here's some code to parse it. I'm using json-simple-1.1.1.jar downloaded from here: https://code.google.com/p/json-simple/downloads/detail?name=json-simple-1.1.1.jar&can=2&q= import org.json.simple.JSONArray; import org.json.simple.JSONObject; import org.json.simple.parser.JSONParser; public class JsonExample { public static void main(String[] args) throws Exception { System.out.println("Starting example..."); String str =...
javascript,php,android,json,jsonobject
i've finally solve my problem, first i create a config.php for connection to the database <?php $username = "root"; $password = ""; $host = "localhost"; $dbname = "smkkimmanuel2"; $options = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'); try { $db = new PDO("mysql:host={$host};dbname={$dbname};charset=utf8", $username, $password, $options); } catch(PDOException $ex) { die("Failed to...
Your JSON consists of an object. That object contains an array named results. The array contains object elements. Each object in the array contains an object named geometry. That object contains an object named location. That object contains lat and lng floating-point values. So, your code should look something like...
android,spring-annotations,jsonobject
You should create on server side a class that looks like this public class UserDetails { private String id; private String password; private String username; public String getId() { return id; } public void setId(String id) { this.id = id; } ... // other two getter setter } And then...
I saw the Java Source Code for JSONObject and the difference between accumulate and put is that with accumulate(String key,Object Value), if there exists some value for "key" then the Object is checked for being an array, if it is an array then the "value" is added to the array...
I'm wondering how you ever came up with that JSON string. {"test":"testvalue", "test2":"{"boxcolor":"bg-white", "iconcolor":"bg-aqua"}"} ^ ^ Here's the fixed code: var testData = '{"test":"testvalue", "test2":{"boxcolor":"bg-white", "iconcolor":"bg-aqua"}}'; testFunction(testData); function testFunction(data) { var obj = JSON.parse(data); alert(obj.test2.boxcolor); // bg-white } ...
Apply this to receive all the data types in string format occupation = firstResult.getJSONObject("job").get("occupation").toString(); ...
The JSONArray contains one JSONObject. When you jsonArray.put(jsonObject); you are adding it to the JSONArray, not to the JSONObject in the JSONArray. This will add the jsonObject to the first JSONObject in your JSONArray jsonArray.getJsonObject(0).put("events",jsonObject.get("events")); ...
javascript,html,json,jsonobject
JSFiddle here HTML <table align='center' > <tr> <td> <select id='listSelect' onChange="updateList()"> <option value='' selected >All</option> <option value='Admin'>Admin</option> <option value='Client'>Client</option> <option value='Regulator'>Regulator</option> </select></td> </tr> </table> <div id="result"></div> JS var jsonObj = { "headerContent": [ { "...": "..." } ], "bodyContent": [ { "userId": "user_1", "userName":...
javascript,html,json,angularjs,jsonobject
As per my understanding of your problem the following is the answer for your question . Your html code like this <div ng-app="myApp" ng-controller="myController"> <table> <tr> <th>Type</th> <th>Shape</th> </tr> <tr ng-repeat="(key ,value) in list.Kind"> <td>{{key}}</td><td>{{value.shape}}</td> </tr> </table> </div> After that your javascript like this var myApp = angular.module("myApp", []); myApp.controller("myController",...
This should do the trick: function extract(data){ return Object.keys(data).map(function(x){ return data[x].filter(function(z){return z!==null;}); }); } Here is the fiddle to play with. For explanation of map, take a look at MDN If you want to flatten the data, you could extend it to: function flatten(data){ return Object.keys(Object.keys(data).map(function(x){ return data[x].filter(function(z){return z!==null;}); }).reduce(function(o,n){...
java,php,android,httpurlconnection,jsonobject
I reckon the problem here is not at the Java side, If the parameters is of fixed type like in json in your case, the JSON Object as POST params method will work if collected this way on the php side: <?php $json = file_get_contents('php://input'); $obj = json_decode($json); print_r($obj); print_r("this...
Just get a String from the JSONObject and write it into the file like this: String content = json.toString(); FileWriter fw = new FileWriter("path"); BufferedWriter bw = new BufferedWriter(fw); bw.write(content); bw.close(); You will of course need to catch/throw exceptions and so on....
java,android,json,arraylist,jsonobject
If I understand you correctly, you want to create an ArrayList of your POJO? I assume you have getters and setters inside your POJO class. Initialize an ArrayList somewhere near the top like this private ArrayList<CholesterolInformation> mCholesterol; Now, parse through your json like this JSONobject data = new JSONObject(jsonStringData); JSONArray...
Maybe there are much better solutions, but at least you have two choices: 1.after building json object inside a method which calls JSONString.endObject(); you can use JSONObject class to add other key/values or modify existing ones: JSONStringer json = new JSONStringer(); json.object(); json.key("stuff").value("1") .key("name").value("2") .key("type_name").value("3") .key("type").value("4"); json.endObject(); //after calling the...
javascript,json,parsing,jsonobject
Something must have gone wrong in creating this string. Yes, it's valid JSON, but it has a different format than you think, because you escape control characters like " and [, ]. Try this string instead: [ { "obj":{ "markers":[ { "k":47.040182144806664, "B":0.52734375 }, { "k":50.90303283111257, "B":10.37109375 }, { "k":52.53627304145945,...
jsonResponse actually doesn't hold the "ratings" object. It is inside each object in the "movies" array. You need something like this: for (int i=0; i<movies.length(); i++) { try { JSONObject movie = movies.getJSONObject(i); JSONObject ratingsobj = movie.getJSONObject("ratings"); String audrating = ratingsobj.get("audience_score").toString(); audscore.add(audrating); } catch (Exception e) { audscore.add("Unavailable"); } }...
Just trim out the quotes: authorization = authorization.substring(1, authorization.length() - 1)