JSON Extracting Data Into Variable Using PHP

Question

I'm using a YouTube data API request to grab the channel id, but i'm not too sure why it isn't working:

The returned JSON request i'm getting is:

{
     "kind": "youtube#channelListResponse",
     "etag": "\"PSjn-HSKiX6orvNhGZvglLI2lvk/k5qSWj-xcF96jAN3p1uQH1amSRc\"",
     "pageInfo": {
      "totalResults": 1,
      "resultsPerPage": 5
     },
     "items": [
      {
       "kind": "youtube#channel",
       "etag": "\"PSjn-HSKiX6orvNhGZvglLI2lvk/e1xTbLf6JLhwwzeWbdMfWdPfcwg\"",
       "id": "UC-lHJZR3Gqxm24_Vd_AJ5Yw"
      }
     ]
    }

To extract the JSON data i'm using the a few lines of code and a function in php:

$banner_data = file_get_contents('https://www.googleapis.com/youtube/v3/channels?part=brandingSettings&forUsername=pewdiepie&key=AIzaSyDTxvTLWXStUrhzgCDptVUG4dGBCpyL9MY');
$banner_data = json_decode($banner_data, true);
$YTid = $banner_data['items']['id'];

When i :

echo "YouTube Channel Id Of pewdiepie is " . $YTid . ".<br />";

I don't get the channel id? What's my problem?

vrijdenker · Accepted Answer · 2014-10-20 07:35:25Z

1

Items is an array containing one or more objects. So it has to be:

 $YTid = $banner_data['items'][0]->id;

This way you grab 'id' from the first item in the items-array.

BTW: learning to debug is crucial to learning to code. If you decode the json and then print the outcome you can see the structure of the array, which could have helped you to find the problem, like:

$banner_data = json_decode($banner_data, true);
var_dump($banner_data);

edited Oct 20, 2014 at 7:35

answered Oct 19, 2014 at 20:04

vrijdenker

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3 Comments

Conor Reid Over a year ago

This has been a helpful answer overall but still hasn't fixed the problem! Thanks and i never thought of using a var_dump. This has really helped! In my other parts of code i seem to be able to use $stats_data = $data['entry']['yt$statistics'];

Jake Ball Over a year ago

@ConorReid not resolved? Working just fine on my end. You have basically just provided a variable with absolutely no context there

vrijdenker Over a year ago

So does it work now? I think johnh10 is right in his answer BTW, it has to be ->id since it's an object. I edited my answer.

johnh10 · Accepted Answer · 2014-10-21 18:06:22Z

1

Try this instead:

    $YTid = $banner_data['items'][0]['id'];

edited Oct 21, 2014 at 18:06

answered Oct 20, 2014 at 2:35

johnh10

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2 Comments

johnh10 Over a year ago

I just noticed you used true for the 2nd argument so json returns assoc arrays. Above code will work.

Conor Reid Over a year ago

I ended up with conflicting code that was stopping it working. Sorry, it does work.

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