imagecreatefromjpeg not giving me a resource?

Question

I am trying to code a script for file resizing.In the script, I use getimagesize() to get the mime information from the image and then test it's value using a switch statement, assigning different values to variables in different cases.

However after running the script, I'm getting the error

PHP Warning:  imagecopyresampled() expects parameter 2 to be resource, string given in ..........

Why exactly am I getting this error when all I'm doing is passing it the variable from createimagefromjpeg?

My Code extract:

list($width,$height,$type)=getimagesize($_FILES["BusinessCreateFileUpload"]["tmp_name"]);
        //checks the mime type and gets the extension
        //Declares the $uploadedfile,$src and $ext variables.
        $uploadedfile='';
        $src='';
        $ext='';
        switch($type)
        {
            case"image/jpeg":
                $uploadedfile=$_FILES["BusinessCreateFileUpload"]["tmp_name"];
                $src=imagecreatefromjpeg($uploadedfile);
                $ext=".jpg";
                break;
            case"image/png":
                $uploadedfile=$_FILES["BusinessCreateFileUpload"]["tmp_name"];
                $src=imagecreatefrompng($uploadedfile);
                $ext=".png";
                break;
            default:
                echo"An error has occurred.Please follow the link to try again";
        }
        //Creates thumbnail
        $newwidth=100;
        $newheight=100;
        $thumbimg=imagecreatetruecolor($newwidth, $newheight);
        imagecopyresampled($thumbimg,$src,0,0,0,0,$newwidth,$newheight,$width,$height);
        //Creates displayimage
        $newwidth1=250;
        $newheight1=250;
        $displayimg=imagecreatetruecolor($newwidth1, $newheight1);
        imagecopyresampled($displayimg,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height);

I'm editing bits and pieces of a script I found on the internet as I'm still rather new to programming so I might have inadvertently screwed something up, though I can't seem to figure out where exactly did I go wrong.

Stoic · Accepted Answer

Your code sets a default value of $src to an empty string. The following lines are then accessing this empty string of $src, which is why it is giving the error:

imagecopyresampled($thumbimg,$src,0,0,0,0,$newwidth,$newheight,$width,$height);
imagecopyresampled($displayimg,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height);

As per your code, this would occur when the switch case for $type is assuming a default value, and therefore your code must be outputting the following line as well:

An error has occurred.Please follow the link to try again

Regarding OP's comment (the answer seemed long, so including it here):

First, make your script use die instead of echo in your default case of the switch statement. So, if the value of $type is not as desired, then the script will give an error and stop.
Secondly, the function getimagesize returns mimetype as an INT value, which is what your variable $type will be set as. Instead, you are comparing your $type variable to strings which is wrong, and would always make use of the default case. Therefore, change your switch statement and compare the values to INT values.
Also, make sure your case statements and echo statements have a space after keywords, i.e. after case and echo

Considering the above, you can replace your switch statement with this:

$src = 0;
switch($type)
    {
        case 2:
            $uploadedfile=$_FILES["BusinessCreateFileUpload"]["tmp_name"];
            $src=imagecreatefromjpeg($uploadedfile);
            $ext=".jpg";
            break;
        case 3:
            $uploadedfile=$_FILES["BusinessCreateFileUpload"]["tmp_name"];
            $src=imagecreatefrompng($uploadedfile);
            $ext=".png";
            break;
        default:
            die "An error has occurred.Please follow the link to try again";
    }

imagecreatefromjpeg not giving me a resource?

Answers (2)

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