How to check if remote image can be shown? - php

Question

Images from an remote server sometimes won't show on my site. They are broken. And getting 403 (Forbidden) when checking in console.
So I wanna find out a way to pre-check if they are forbidden and then not show them, instead of a broken image.
So, my thought was to check http status with get_headers(). But it shows 200 OK even for the broken images.
And then I found this solution after a google search, to use before get_headers():
$default_opts = array(
'http'=>array(
'method'=>"GET",
'header'=>"Referer: http://www.fakesite.com/hotlink-check/",
)
);
stream_context_set_default($default_opts);
But that doesn't work either. I have also tried file_get_contents() but that gave result for both non-broken and broken images.
So, how to check if a remote file has an 403 error?
EDIT:
Okay so, here are recent tests I've tried with:
$image = 'http://www.arabnews.com/sites/default/files/2018/02/28/1114386-582723106.jpg';
Test 1:
<?php
$default_opts = array(
'http' => array(
'method' => "GET",
'header' => "Referer: http://www.fakesite.com/hotlink-check/",
'user_agent' => 'Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.1.6) Gecko/20091201 Firefox/3.5.6',
)
);
stream_context_set_default($default_opts);
$output = #get_headers($image, 1);
echo '<pre>'.print_r($output, true).'</pre>';
?>
Output:
Array
(
[0] => HTTP/1.1 200 OK
[Date] => Wed, 28 Feb 2018 14:50:39 GMT
[Content-Type] => image/jpeg
[Content-Length] => 43896
[Connection] => close
[Set-Cookie] => __cfduid=d2a28d117e8a32d5755219bcd6e1892561519829439; expires=Thu, 28-Feb-19 14:50:39 GMT; path=/; domain=.arabnews.com; HttpOnly
[X-Content-Type-Options] => nosniff
[Last-Modified] => Tue, 27 Feb 2018 22:55:30 GMT
[Cache-Control] => public, max-age=1209600
[Expires] => Wed, 14 Mar 2018 14:50:39 GMT
[X-Request-ID] => v-e065c0be-1c11-11e8-9caa-22000a6508a2
[X-AH-Environment] => prod
[X-Varnish] => 175579289 178981578
[Via] => 1.1 varnish-v4
[X-Cache] => HIT
[X-Cache-Hits] => 8
[CF-Cache-Status] => HIT
[Accept-Ranges] => bytes
[Server] => cloudflare
[CF-RAY] => 3f44328fd0247179-ORD
)
Test 2:
<?php
function get_response_code($url) {
#file_get_contents($url);
list($version, $status, $text) = explode(' ', $http_response_header[0], 3);
return $status;
}
$output = get_response_code($image);
echo '<pre>'.print_r($output, true).'</pre>';
?>
Output:
200
Test 3:
<?php
$output = getimagesize($image);
echo '<pre>'.print_r($output, true).'</pre>';
?>
Output:
Array
(
[0] => 650
[1] => 395
[2] => 2
[3] => width="650" height="395"
[bits] => 8
[channels] => 3
[mime] => image/jpeg
)

Answer 1

Check the image size. This is the easiest and most effective solution:
list($width, $height, $type, $attr) = getimagesize("www.url.com/img.jpg");
If especially the first two; $width and $height are set, and $type is a valid value you are good to go.
And I would display them in this case.