CODEWITHSUNDEEP
phpjson
kams
kams

Reputation: 128

How to access specific json data PHP

Good day,

I'm fetching my data directly from and API endpoint. I was able to get the id but how can I access "type":"participant", specifically the skillTier?

Here's my code but it's throwing an error: NOTICE: UNDEFINED INDEX: SKILLTIER IN C:\XAMPP\HTDOCS\VGAPP\INDEX.PHP ON LINE 219

if(is_array($finaljson) || is_object($finaljson)){
if(!empty($finaljson['included']) && $finaljson['included'] !== ""){
    foreach ($finaljson['included'] as $key => $value) {
        if(!empty($value['attributes']['name']) && $value['attributes']['name'] === $username){
            $userid = $value['id'];
            $skillTier = $value['skillTier'];
        }
    }

Here is the json data:

https://gist.github.com/kamaelxiii/18ef0c1600330b98717b96db00532d6a

Upvotes: 2

Views: 107

Answers (4)

mickmackusa
mickmackusa

Reputation: 47894

It is patient work to whittle down a big json / array to target and isolate your desired data. You just need to traverse the structure step-by-step, print each level's data to screen, and identify the keys that hold what you want.

Code:

$array = json_decode($json, true);
foreach ($array['included'] as $included) {  // iterate all children of "included"
    if ($included['type'] === 'participant') {  // only access "participant" subset data
        $result[$included['id']] = $included['attributes']['stats']['skillTier'];  // access the skillTier data which is beneath "stats" beneath "attributes"
    }
}
echo "<pre>";
var_export($result);
echo "</pre>";

Output:

array (
  '4cfe4736-f47b-11e6-b633-0242ac110030' => 16,
  '4cfe473f-f47b-11e6-b633-0242ac110030' => 10,
  '4cfe4727-f47b-11e6-b633-0242ac110030' => 27,
  'd4003883-f423-11e6-865e-0242ac110008' => 23,
  'd4003869-f423-11e6-865e-0242ac110008' => 24,
  'd400387a-f423-11e6-865e-0242ac110008' => 23,
  '4cfe471f-f47b-11e6-b633-0242ac110030' => 27,
  'd400386d-f423-11e6-865e-0242ac110008' => 24,
  'd4003887-f423-11e6-865e-0242ac110008' => 23,
  '4cfe4724-f47b-11e6-b633-0242ac110030' => 24,
  'd4003864-f423-11e6-865e-0242ac110008' => 18,
  '4cfe473b-f47b-11e6-b633-0242ac110030' => 17,
)

Upvotes: 1

Bram
Bram

Reputation: 2581

First of all you have 2 PHP syntax errors where you have two missing }'s to close off your first ifstatement and foreach loop.

Also you want to loop your included nodes and check if you have a participant object before continuing since the other items don't have the skillTier value. Also best practice to check if the skilltier isn't empty before trying to save it in a variable.

This is the updated version of your code:

if(is_array($finaljson) || is_object($finaljson)) {
    foreach($finaljson['included'] as $included) {
        if($included['type'] === 'participant') {
            if(!empty($included['attributes']['stats']['skillTier'])){
                $id = $included['id'];
                $skillTier = $included['attributes']['stats']['skillTier'];
            }

        }
    }
}

Here is a working example with your JSON that will var_dump the ID's and skilltier. http://ideone.com/DrXWe7 (click fork to see the code)

Upvotes: 1

Gourav
Gourav

Reputation: 71

foreach($finaljson['included'] as $included) {
  if($included['type'] == 'participant'){
    $userid = $included['id'];
    $skillTier = $included['attributes']['stats']['skillTier'];
   }
}

Upvotes: 1

hassan
hassan

Reputation: 8288

As the same way that you've accessed the $value['attributes']['name'] , it's multi-dimensional array , so think about it as a town , and you want to arrive to flat number 6 which is at building 9 , which is located at street number 4 , you will not be able to ask somebody to tell you where is flat 6 without telling the full address , so in your case , you simply can achieve your skillTier index when only you provide your variable with the full address of your index , as follows :

$value['attributes']['stats]['skillTier']

Upvotes: 1

Related Questions