PHP file_get_contents() throws error "failed to open stream: HTTP request failed! <!doctype html>"

Question

I'm trying to return JSON results from a page but I get the following error using file_get_contents(): " failed to open stream: HTTP request failed! "

Can someone tell me why I get this error?

<?

$newURL = 'https://api.qwant.com/api/search/images?count=10&offset=1&q=Spicy 
Cranberry Cavolo Nero (Kale)';



$returnedData = file_get_contents($newURL);

?>

Any reason not to have "api.qwant.com/api/search/…" instead of what you have right now? With %20 for each space — RAZERZ
– RAZERZ, Commented Jan 5, 2018 at 22:07
Possible duplicate of PHP file_get_contents() returns "failed to open stream: HTTP request failed!" — Liora Haydont
– Liora Haydont, Commented Jan 5, 2018 at 22:09
have your tried to escape the parameters correctly? for example, replace the spaces with %20 in the url. — Ibu
– Ibu, Commented Jan 5, 2018 at 22:24
@RAZERZ I tried doing a urlencode on the query itself but it gives me a different error (Too many request): $baseURL = 'api.qwant.com/api/search/images?'; $query = urlencode("count=10&offset=1&q=Spicy Cranberry Cavolo Nero (Kale)"); $composedURL = $baseURL.$query; $returnedData = file_get_contents($composedURL); — user1493588
– user1493588, Commented Jan 5, 2018 at 22:39

Ivan Ponomarev · Accepted Answer · 2018-01-06 00:14:23Z

1

Never use <? ?>. Use only <?php ?> or <?= ?>
urlencode() use only on values of your parameters, not on the whole parameter's line.

file_get_contents() is not a really good method to receive data from the outer servers. Better use CURL.

<?php

// Your URL with encoded parts
$newURL = 'https://api.qwant.com/api/search/images?count=10&offset=1&q='.urlencode('Spicy Cranberry Cavolo Nero (Kale)');

// This is related to the specifics of this api server, 
// it requires you to provide useragent header
$options = array(
    CURLOPT_RETURNTRANSFER => true,
    CURLOPT_USERAGENT => 'any-user-agent-you-want',
);

// Preparing CURL
$curl_handle = curl_init($newURL);

// Setting array of options
curl_setopt_array( $curl_handle, $options );

// Getting the content
$content = curl_exec($curl_handle);

// Closing conenction
curl_close($curl_handle);

// From this point you have $content with your JSON data received from API server

?>

edited Jan 6, 2018 at 0:14

answered Jan 6, 2018 at 0:08

Ivan Ponomarev

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1 Comment

ivanivan Over a year ago

and you can send POST requests with curl as well

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