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Home/ Questions/Q 7616137
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T02:49:00+00:00 2026-05-31T02:49:00+00:00

<?php header(‘Cache-Control: no-cache, must-revalidate’); header(‘Content-type: application/json’); $jsonData = json_decode(file_get_contents(urlencode(‘https://chart.googleapis.com/chart?cht=p3&chs=250×100&chd=t:60,40&chl=Hello|World&chof=json’))); echo $jsonData; ?> Error: failed

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<?php

header('Cache-Control: no-cache, must-revalidate');
header('Content-type: application/json'); 

$jsonData = json_decode(file_get_contents(urlencode('https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json')));

echo $jsonData;

?>

Error: failed to open stream: No such file or directory in <b>C:\wamp\www\file.php

I want to print the result as a json string so i can handle it with jquery ajax. What am i missing? Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-31T02:49:02+00:00Added an answer on May 31, 2026 at 2:49 am

    You realise you have a space in your URL (results in a 400 from Google). Also, you don’t want to use urlencode() here.

    I’d also hazard a guess that you don’t want to use json_decode() as this will return an object or array.

    So, try this instead

    readfile('https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json');
    exit;
    

    To do what you’re attempting, please pay attention to this note in the manual

    A URL can be used as a filename with this function if the fopen wrappers have been enabled. See fopen() for more details on how to specify the filename. See the Supported Protocols and Wrappers for links to information about what abilities the various wrappers have, notes on their usage, and information on any predefined variables they may provide.

    If you cannot configure your servers as such, you’ll want to check out cURL or the sockets extension

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