Question

I am trying to build a class that does many photo operations, one method will upload images from a user but I am also needing to build a method to grab a photo from a URL and run other methods on it just like if it were being uploaded with a POST form from user.

Below is my start of the function for getting image from URL, it works but needs work still. Below the code you can see a image that is the result of this function being ran. Also is the original image to see what it should look like. You can see that this function makes the image have a black background on this transparent image. How can I make it look better like it should look?

$url = 'http://a0.twimg.com/a/1262802780/images/twitter_logo_header.png';

//run our function
savePhotofromURL($url, 'no');



// photo function should grab an photo from a URL
function savePhotofromURL($photo_url, $saveimage = 'yes'){
    if(isset($photo_url) && $photo_url != '') {

        //get info about photo
        $photo_info = getimagesize($photo_url);
        $source_width = $photo_info['0'];
        $source_height = $photo_info['1'];
        $source_type = $photo_info['mime'];

        //grab the Photo from URL
        $photo = imagecreatefromstring(file_get_contents($photo_url));

        if (is_resource($photo) === true){
            if($saveimage === 'yes'){
                // TO DO: resize image and make the thumbs code would go here if we are saving image:
                // TO DO: resize source image if it is wider then 800 pixels
                // TO DO: make 1 thumbnail that is 150 pixels wide
            }else{
                // We are not saving the image show it in the user's browser
                // TO DO: we will add in correct photo type soon
                header('Content-Type: image/gif');
                imagejpeg($photo, null, 100);
                imagedestroy($photo); 
            }
        }else{
            // not a valid resource, show error
            echo 'error getting URL photo from ' .$photo_url;
        }
    }else{
        // url of image was empty
        echo 'The URL was not passed into our function';
    }
}

The result looks like this
alt text http://img2.pict.com/52/05/1f/2429493/0/screenshot2b181.png

Instead of like this

Answer 1

The following two calls will tell php to use the alpha blending present in the png image:

        ImageAlphaBlending($photo, false);
        ImageSaveAlpha($photo, true);

Edit:
I see you’re outputting the image as a JPEG also. JPEGs don’t support transparency, so no matter what you do you will end up with an incorrect background color. Also see this related question: PHP/GD ImageSaveAlpha and ImageAlphaBlending