How to get a object from JSON to PHP

Question

I need to take a img link (https://upload.wikimedia.org/wikipedia/en/5/51/Minecraft_cover.png) from this api https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&piprop=original&titles=Minecraft&pilicense=any. How to do it?

I wrote code like this, but I can print :

$img_url = "https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&piprop=original&titles=Minecraft&pilicense=any";
$img_url = str_replace(" ", "%20", $img_url);
$img = json_decode(file_get_contents($img_url));
print_r ($img);

But how to print only img source?

score 1 · Answer 1 · answered Feb 23 '20 at 15:34

1

The simplest way would be to use the following.

echo $img->query->pages->{'27815578'}->original->source;

Where 27815578 is the Page ID

answered Feb 23 '20 at 15:34

Uzair Hayat

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THANKS! I knew that i should do something with that pageid, but i didn't know how to write it in code. Thank you!! – Stawikendo Feb 23 '20 at 15:39
No worries. Haha. Glad I could help. – Uzair Hayat Feb 23 '20 at 15:41

How to get a object from JSON to PHP

1 Answers1