PHP Detecting if source image url link leads to a "broken" image?

Question

Suppose you have a thumbnail generator script that accepts source images in the form of a URL. Is there a way to detect if the source URL is "broken" - whether nonexistent or leads to an non-image file?

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Just brute force using getimagesize() or another PHP GD function is not a solution, since spoofed stray URL's that might not be images at all (http://example.com/malicious.exe or the same file, but renamed as http://example.com/malicious.jpg) could be input - such cases could easily be detected by PHP before having to invoke GD. I'm looking for GD pre-sanitizing before having GD try its battalion at parsing the file.

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as a first step, the following regular expression checks if the URL is an image extension: preg_match('@(https?://([-\w\.]+)+(:\d+)?(/([\w/_\.]*(\?\S+)?)?)?)([^\s]+(\.(?i)(jpg|png|gif|bmp))$)@', $txt,$url);

Answer 1

use file_exists function in php, you can check urls with it.

See documentation below, shows how to check img... exactly what you need

FILE EXISTS - http://www.php.net/manual/en/function.file-exists.php#93572

URL EXISTS - http://www.php.net/manual/en/function.file-exists.php#85246

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Here is alternative code for checking the url. If you will test in browser replace \n with <br/>

<?php

$urls = array('http://www.google.com/images/logos/ps_logo2.png', 'http://www.google.com/images/logos/ps_logo2_not_exists.png');

foreach($urls as $url){
   echo "$url - ";
   echo url_exists($url) ? "Exists" : 'Not Exists';
   echo "\n\n";
}


function url_exists($url) {
    $hdrs = @get_headers($url);

    echo @$hdrs[1]."\n";

    return is_array($hdrs) ? preg_match('/^HTTP\\/\\d+\\.\\d+\\s+2\\d\\d\\s+.*$/',$hdrs[0]) : false;
}
?>

Output is as follows

http://www.google.com/images/logos/ps_logo2.png - Content-Type: image/png
Exists

http://www.google.com/images/logos/ps_logo2_not_exists.png - Content-Type: text/html; charset=UTF-8
Not Exists

Answer 2

I have used the following to detect attributes for remote images

$src='http://example.com/image.jpg';
list($width, $height, $type, $attr) = @getimagesize($src);

example (checking stackoverflows "Careers 2.0" image)

$src='http://sstatic.net/ads/img/careers2-ad-header-so.png';
list($width, $height, $type, $attr) = @getimagesize($src);

echo '<pre>';
echo $width.'<br>';
echo $height.'<br>';
echo $type.'<br>';
echo $attr.'<br>';
echo '</pre>';

If $height, $width etc is null the image is obvious not an image or the file does not exists. Using cURL is overkill and slower (even with CURLOPT_HEADER)

Answer 3

The only really reliable way is to request the image using file_get_contents(), and finding out its image type using getimagesize().

Only if getimagesize() returns a valid file type, can you rely that it is in fact a valid image.

This is quite resource heavy, though.

You could consider not doing any server-side checks at all, and adding an onerror JavaScript event to the finished image resource:

<img src="..." onerror="this.style.display = 'none'">

Answer 4

try for local files

<?php 
if(file_exists($filename))
{
//do what you want
}
else
{
//give error that file does not exists
}
?>

for external domains

$headers = @get_headers($url);
if (preg_match("|200|", $headers[0])) {
// file exists
} else {
// file doesn't exist
}

Also you can use curl request for the same.

Answer 5

Fast Solution for broken or not found images link
i suggest you that don't use getimagesize() because it will 1st download image then it will check images size+if this will not image then it will throw exception so use below code

if(checkRemoteFile($imgurl))
{
//found url, its mean
echo "this is image";
}

function checkRemoteFile($url)
{
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL,$url);
    // don't download content
    curl_setopt($ch, CURLOPT_NOBODY, 1);
    curl_setopt($ch, CURLOPT_FAILONERROR, 1);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    if(curl_exec($ch)!==FALSE)
    {
        return true;
    }
    else
    {
        return false;
    }
}

Note: this current code help you to identify broken or not found url image this will not help you to identify image type or headers

Answer 6

You could check the HTTP status code (it should be 200) and the Content-type header (image/png etc.) of the HTTP response before you put the actual image through the generator.

If these two preconditions are ok, after retrieving the image you can call getimagesize() on it and see if it breaks, what MIME type it returns etc.

Answer 7

did you try file_get_contents() method?

http://php.net/manual/en/function.file-get-contents.php

Answer 8

onerror Event

Execute a JavaScript if an error occurs when loading an image:

The onerror event is triggered if an error occurs while loading an external file (e.g. a document or an image).

Example:

<!DOCTYPE html>
<html>
<body>

<img src="image.gif" onerror="myFunction()">

<p>A function is triggered if an error occurs when loading the image. The function shows an alert box with a text.
In this example we refer to an image that does not exist, therefore the onerror event occurs.</p>

<script>
function myFunction() {
  alert('The image could not be loaded.');
}
</script>

</body>
</html>

Answer 9

Although not detecting broken links, might be useful for someone else...

onerror=""

<img src="PATH" onerror="this.src='NEW PATH'" />