imagecreatefromstring - catch fail - PHP

Question

I'm trying to detect if an image is an image when given a URL. To save an image from a url I used the following:

// create the image and save it
$imagefile = $URL;
$resource = imagecreatefromstring(file_get_contents($imagefile));
// X amount of quality is the last number
imagejpeg($resource, "images/covers/$imagepath.jpeg", 50);
imagedestroy($resource);

The $URL is simply an image link provided by the user.

I have tried the following:

$imagefile = $addBuildCover;
$resource = imagecreatefromstring(file_get_contents($imagefile));

if ($resource !== false) {
  $return['imageGood'] = true;  
} else {
  $return['imageBad'] = true;
}

I have tried that code and i returns the correct JSON return of 'imageBad' but it's giving me an error before which is:

Warning: file_get_contents(ewffw.png): failed to open stream: No such file or directory in /var/www/clients/client2/web3/web/process/addnewbuild.php on line 116

Warning: imagecreatefromstring(): Empty string or invalid image in /var/www/clients/client2/web3/web/process/addnewbuild.php on line 116

How do I try I catch the URL fail but not actually have an error return like above?

score 10 · Accepted Answer · answered Jun 17 '14 at 23:49

imagecreatefromstring will unfortunately warn if you pass it invalid data, so using it unnecessarily problematic.

You will have to shut it up with the error suppression operator @:

$resource = @imagecreatefromstring(file_get_contents($imagefile));

Using this operator is something usually frowned upon, but this case is one which you really do have a legitimate use case.

This approach will also take care of the warning given from file_get_contents if the file cannot be loaded. That function is not as badly behaved (there are ways to check with reasonable certainty if it will fail or not, e.g. is_readable) so you could check with code instead of using @, but since you need to suppress errors anyway and it makes no difference to you if the file read failed or not, just slapping a @ in front is IMHO fully warranted here.

Thankyou! the error was causing issues with the JSON return so i think using the @ is justified. — Lovelock, Jun 18 '14 at 07:05

score -5 · Answer 2 · edited Sep 09 '15 at 15:19

<?php
    $serverName = "server";

    $uid = "user";
    $pwd = "pass";
    $connectionInfo = array( "UID"=>$uid,
                             "PWD"=>$pwd,
                             "Database"=>"database");


    $conn = sqlsrv_connect( $serverName, $connectionInfo);
    if( $conn === false )
    {
         echo "Unable to connect.</br>";
         die( print_r( sqlsrv_errors(), true));
    }


    $fotoquery = "SELECT column_image FROM Table WHERE Column = 'xxxxxxxxxx'";

    $stmt = sqlsrv_query( $conn, $fotoquery);

    if( $stmt === false )
    {
       echo "Error in executing query.</br>";
       die( print_r( sqlsrv_errors(), true));
    }

    $dataImage = sqlsrv_fetch_array($stmt);
    $varimg = base64_encode($dataImage[0]);

    $data = base64_decode($varimg);

    $im = imagecreatefromstring($data);
    if ($im !== false) {
        header('Content-Type: image/png');
        imagepng($im);
        imagedestroy($im);
    }
    else {
        echo 'An error occurred.';
    }
?>

imagecreatefromstring - catch fail - PHP

2 Answers2