I suggest Use JSONObject keys() to get the key and then iterate each key to get to the dynamic value. according to your json string code must be something look like this:- //refers to the current element in the array “data” JSONObject mainObj = new JSONObject(yourString); JSONObject dynamicValue1 = mainObj.getJSONObject(“dynamicValue1”); Iterator key = dynamicValue1.keys(); while(key.hasNext()) … Read more
Introduction Parsing JSON strings in Android can be a tricky task, but with the right tools and techniques, it can be done quickly and easily. This post will provide an overview of the different methods available for parsing JSON strings in Android, as well as provide some tips and tricks for getting the most out … Read more
var str=”{“id”:[1],”latitude”:[12.45787],”longitude”:[11.77878]}”; var obj = JSON.parse(str); for (var key in obj) { obj[key] = obj[key][0]; } console.log(obj); 0 solved how to remove square bracket in json formate string?
According to data flile you provide some of elements doesn’t include ‘platform’ property (see examples downhere). So use techniques I sugessted before: $platform = (isset($locations[‘platform’]))?$locations[‘platform’]:”; You should verify if property exists before use it. code: foreach($result_arr[‘locations’] as $locations){ $platform = (isset($locations[‘platform’]))?$locations[‘platform’]:”; echo ‘== PLATFORM : ‘.$platform.’ <br />’; } outputs: == PLATFORM : 7 == … Read more
response is an object. result is an object. full is an object. Array1 and Array2 are, obviously enough, arrays. For mytext1: response.result.full.Array1[0] For mytext2: response.result.full.Array1[1] For mytext3: response.result.full.Array2[0] For mytext4: response.result.full.Array2[1] If you want to log everything in the array, use a simple for…loop: var arr = response.result.full.Array1; for (var i = 0, l = … Read more
Try this var items= { “people”: [ { “name”: “Person A”, “score”: 100 }, { “name”: “Person B”, “score”: 101 }, { “name”: “Person C”, “score”: 100000 }, { “name”: “Person D”, “score”: 555 } ] }; items.people.sort(function(a, b){ return a.score – b.score; }); console.log(items); it works as expected! 0 solved I want to sort … Read more
There is a native API for this called NSJSONSerialization. This will serialise and deserialise JSON. 3 solved iOS library serialize/deserialize JSON [closed]
For better understanding you can call ajax method as below $.ajax({ url: ‘https://prod.api.pvp.net/api/lol/euw/v1.1/summoner/by-name/kerrie?api_key=[key]’, type: ‘GET’, dataType: ‘JSON’, async: false, error: function(){}, success: function(resp){ // resp variable will be your JSON responce } } }); 1 solved jQuery AJAX not receiving JSON from http request
You can use a dictionary to parse a JSON object with custom property names: public class Test { public string col1 { get; set; } public string col2 { get; set; } } public class DataValue { public Test test { get; set; } } public class RootObject { public RootObject() { this.data = new … Read more
This is less a question targeted towards REST then it is actually targeting HTTP as the default transport protocol used by applications following a REST architecture approach. According to RFC 7231 4.3.4 a server SHOULD verify that the PUT representation is consistent with any constraints the server has for the target resource that cannot or … Read more
Try mysql_fetch_assoc: $json = array(); while ($row = mysql_fetch_assoc($result)) { $json[$row[‘uid’]] = array( ‘lat’ => $row[‘lat’], ‘lon’ => $row[‘lon’], ‘loc’ => $row[‘loc’] ); } echo json_encode($json); You should use MySQLi or PDO_MySQL instead of mysql_. 1 solved How to create json response
GetYouTubeUserCommentsTask task = new GetYouTubeUserCommentsTask(null, viewCount); // passing null. And you have public GetYouTubeUserCommentsTask(Handler replyTo, String username) { this.replyTo = replyTo; // replyTo is null this.username = username; } replyTo is null. You need to Initialize the handler replyTo 10 solved TextView isn’t updated with JSON Response
I’m guessing you mean by not writing your own bean for GSON to deserialize to. GSON provides support by having certain JSON types that you can make use of: Gson gson = new Gson(); final JsonArray jsonElements = gson.fromJson(JSON, JsonArray.class); 0 solved Could anyone help me to convert the below JSON to java object (List), … Read more
The first rule of debugging is to assume that the error message is telling you the truth. In this case it is telling you keyerror u’True. This means that somewhere in your code you’re doing the equivalent of some_dictionary[u’True’]. In the code you posted, there is nothing that is using the literal value True as … Read more
do Directly as let Media_list = jsonData[“staffData”] as? NSDictionary let dataurl = Media_list[“url”] as? String 2 solved Json data is not parsing in swift3 [duplicate]