CODEWITHSUNDEEP
phpimagecurl
salmanhijazi
salmanhijazi

Reputation: 825

php/curl - get extension of image on remote server using headers

There is a form on my website that allows the upload of images from a remote server. When the user enters the link and submits, I want to check the file and make sure it's the right extension before I copy it off to my server.

I tried to directly use exif_imagetype, but allow_url_fopen is not allowed on the server so need help. I think using curl will solve the problem, but I'm not sure how to get the image extension from the header.

Thanks in advanced!

solution

$ch = curl_init ($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_exec ($ch);

$content_type = curl_getinfo($ch, CURLINFO_CONTENT_TYPE);
echo $content_type;

Thanks to Mario!

Upvotes: 2

Views: 3715

Answers (3)

stardust4891
stardust4891

Reputation: 2550

You can also do it this way:

$src = 'https://example.com/image.jpg';
$size = getimagesize($src);
$extension = image_type_to_extension($size[2]);

Upvotes: 0

Troy Tabor
Troy Tabor

Reputation: 1

Based on a few responses, I was trying to display an image from a curl request, and dynamically figure out the headers Content-Type. Using what Mario did, and some other things, here is a function which will correctly display an image from a standard url or file, and correctly put in the header. All PHP.

function displayImage( $filename ) {
    $ch = curl_init();
    $options =  array(

        CURLOPT_URL => $filename,
        CURLOPT_ENCODING => "",
        CURLOPT_RETURNTRANSFER => true,
        CURLOPT_HTTPGET => true,
        CURLOPT_CONNECTTIMEOUT => 60,
        CURLOPT_TIMEOUT => 60

    );

    curl_setopt_array($ch, $options);
    $response = curl_exec($ch);
    header('Content-Type: ' . curl_getinfo($ch, CURLINFO_CONTENT_TYPE));

    if(!curl_errno($ch)){
        curl_close($ch);
        $img = imagecreatefromstring($response);
        imagejpeg($img);
        imagedestroy($img);
        return true;
    }
    else{
        curl_close($ch);
        return curl_error($ch);
    }
}

Upvotes: 0

salmanhijazi
salmanhijazi

Reputation: 825

solution

$ch = curl_init ($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_exec ($ch);

$content_type = curl_getinfo($ch, CURLINFO_CONTENT_TYPE);
echo $content_type;

Thanks to Mario!

Upvotes: 6

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