PHP JSON not decoding properly

PHP JSON not decoding properly

By : matlabrookie
Date : November 18 2020, 01:01 AM
fixed the issue. Will look into that further The second argument to json_decode forces all objects to be parsed as associative arrays so you need to access it with array notation. Additionally result is an array always so you need to loop over it or access it by index:
code :
$price = json_decode($pricejson, true);
// print the first price
echo $price['result'][0]['Last'];

// print all prices:
foreach ($price['result'] as $data) {
   echo $data['Last'];
$price = json_decode($pricejson);
echo $price->result[0]->Last;

// print all prices:
foreach ($price->result as $data) {
   echo $data->Last;

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JSON not decoding properly

JSON not decoding properly

By : Chandu Rajan
Date : March 29 2020, 07:55 AM
To fix the issue you can do The $.getJSON method already parses the result from the server as JSON, so you don't need to call $.parseJSON once again. The data parameter that is passed to your callback will already represent a JSON object that you could use directly:
code :
$.getJSON(url, function(data) {
Apple JSON API not decoding properly in PHP

Apple JSON API not decoding properly in PHP

By : user2061470
Date : March 29 2020, 07:55 AM
I wish this helpful for you I find out about Apple lookup API and wanted to try something. , You need to understand the structure of the object. What you need is:
code :
$appname = $data['results'][0]['artistName'];
PHP json encode not properly decoding in javascript

PHP json encode not properly decoding in javascript

By : xarshaal
Date : March 29 2020, 07:55 AM
To fix this issue XMLHttpRequest works asynchronously. So your function requestData ends after the xhttp.send(); and does not return anything. The event handler for readystatechange (which is called a bit later, once XMLHttpRequest actually receives the response from the server) does return a value, but nothing is done with it.
Solution: move parsing and interpretation of the response in that event handler.
Swift 4 Not decoding JSON optional properly

Swift 4 Not decoding JSON optional properly

By : Aportes Educativos G
Date : March 29 2020, 07:55 AM
Hope this helps Swift noob here. , If copyright is optional , then you can make use of decodeIfPresent.
code :
self.copyright = try valueContainer.decodeIfPresent(String.self, forKey: CodingKeys.copyright)
Decoding json object to get its value not working properly?

Decoding json object to get its value not working properly?

By : user3439161
Date : March 29 2020, 07:55 AM
Hope this helps You had done most of the hardwork. The result.Name is of Type Newtonsoft.Json.Linq.JValue. You only had to take the Value from it using the JValue.Value property
code :
var result = JsonConvert.DeserializeObject<dynamic>(statusText.text);
var gettingTheName = (string)result.name.Value; // Change here
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