Parsing JSON object in PHP and get each value with json_decode

Question

Hi tried to request each src value from wordpress media playlist , so i return data in JSON

My request Code which did not succeed was :

$decode_tracks = json_decode($track , true);

foreach ($decode_tracks[0][0] as $item) {
 print $item['src'];
}

This is some of the data that was returned.

[
      {
        "src": "http:\/\/localhost\/wordpress\/wp-content\/uploads\/2015\/08\/output.mkv",
        "type": "video\/x-matroska",
        "title": "output",
        "caption": "",
        "description": "",
        "meta": {
          "length_formatted": "0:16"
        }
      },
      {
        "src": "http:\/\/localhost\/wordpress\/wp-content\/uploads\/2015\/08\/christmas.mp4",
        "type": "video\/mp4",
        "title": "christmas",
        "caption": "tytj",
        "description": "tyjtyj",
        "meta": {
          "length_formatted": "1:21"
        }
      }
    ]

You've got an array of items inside your json object, meaning that even after you run json_decode($track, true) you've still got numerically indexed arrays to go through. — Zarathuztra, Aug 29 '15 at 19:07

score 1 · Answer 1 · answered Aug 29 '15 at 19:10

1

Try this:

foreach ($decode_tracks as $item) {
     print $item['src'];
}

There is no need to convert json to array, you can easily do like this:

foreach ($track as $item) {
     print $item->src;
}

answered Aug 29 '15 at 19:10

Yaroslav Tkachenko

137
1
10

0xdenishdev · Answer 2 · 2015-08-29T19:44:07.983

1

$decode_tracks = json_decode($track);

for ($i = 0; $i < count($decode_tracks); $i++) {
  print $decode_tracks[$i]->src;
}

Or if you are using array: try this one

$decode_tracks = json_decode($track, true);

for ($i = 0; $i < count($decode_tracks); $i++) {
  print $decode_tracks[$i]['src'];
}

edited Aug 29 '15 at 19:44

answered Aug 29 '15 at 19:34

0xdenishdev

186
4
10

Parsing JSON object in PHP and get each value with json_decode

2 Answers2