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Home/ Questions/Q 5964281
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T19:24:11+00:00 2026-05-22T19:24:11+00:00

Here is my code to json decode picasa api results, but it always Warning:

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Here is my code to json decode picasa api results, but it always Warning: Invalid argument supplied for foreach(), Where is the problem? Thanks.

<?php
header('Content-type:text/html; charset=utf-8');
$url="http://picasaweb.google.com/data/feed/base/all?alt=json&kind=photo&access=public&filter=1&q=usa&imgor=landscape&max-results=50&hl=en";
$json = file_get_contents($url);
$data = json_decode($body, true);
foreach ($data['feed']['entry'] as $result){         
echo html_entity_decode($result->content->src, ENT_QUOTES, 'UTF-8');
echo html_entity_decode($result['updated']['$t'], ENT_QUOTES, 'UTF-8');
}
?>

json tree here

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  1. Editorial Team
    Editorial Team
    2026-05-22T19:24:12+00:00Added an answer on May 22, 2026 at 7:24 pm

    You are using a wrong variable:

    $json = file_get_contents($url);
    $data = json_decode($body, true);
    

    It should be $json instead of $body.

    You used json_decode(, true) so it converts all objects to associative arrays. But here you use object-notation to access your data which might result in another error:

    echo html_entity_decode($result->content->src, ENT_QUOTES, 'UTF-8');
    
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