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Home/ Questions/Q 8758433
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T14:32:48+00:00 2026-06-13T14:32:48+00:00

I’m using this structure for Instagram: <?php // Supply a user id and an

  • 0

I’m using this structure for Instagram:

<?php
        // Supply a user id and an access token
        $userid = "--user--";
        $accessToken = "--token--";

        // Gets our data
        function fetchData($url){
             $ch = curl_init();
             curl_setopt($ch, CURLOPT_URL, $url);
             curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
             curl_setopt($ch, CURLOPT_TIMEOUT, 20);
             $result = curl_exec($ch);
             curl_close($ch); 
             return $result;
        }

        // Pulls and parses data.
        $result = fetchData("https://api.instagram.com/v1/users/--user--/media/recent/?access_token=--token--");
        $result = json_decode($result);
    ?>


    <?php foreach ($result->data as $post): ?>
        <!-- Renders images. @Options (thumbnail,low_resoulution, high_resolution) -->
        <a class="group" rel="group1" href="<?= $post->images->standard_resolution->url ?>"><img src="<?= $post->images->thumbnail->url ?>"></a>
    <?php endforeach ?>

But this print:

Warning: Invalid argument supplied for foreach() in /home/mysite/index.php on line 40

40.line:

<?php foreach ($result->data as $post): ?>

Whats wrong in this structure?

This structure is http://www.blueprintinteractive.com/blog/how-instagram-api-fancybox-simplified

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T14:32:50+00:00Added an answer on June 13, 2026 at 2:32 pm

    Ok, just to check – in your code, you have $userid = "--user--" and $accessToken = "--token--", then in the URL, you have --user-- and --token-- again – you are replacing them with $userid and $accessToken? If not, the issue could simply be that Instagram is returning an error, which in turn caused your code to fail.

    If you have taken the above in to account and are still getting nothing, then print the value of $result->data before the loop (var_dump($result->data)) and see what your variable contains.

    If the above shows nothing, then try var_dump($result) – It is possible that the json_decode() has failed and will show false.

    If $result->data is an Array, and it is just that you have no data (I’m not sure what Instagram would retrun if there were no results), add a check to see if the Array is empty before the foreach() loop.

    <!-- This will output the contents of '$result->data' - let us know what it is -->
    <?php var_dump($result->data)`) ?>
    
    <?php if(!empty($result->data)): ?>
        <?php foreach ($result->data as $post): ?>
            <!-- Renders images. @Options (thumbnail,low_resoulution, high_resolution) -->
            <a class="group" rel="group1" href="<?= $post->images->standard_resolution->url ?>"><img src="<?= $post->images->thumbnail->url ?>"></a>
        <?php endforeach ?>
    <?php endif ?>
    
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