Question

I have encountered a problem with creating a thumbnail from an uploaded image file, then to upload it to the server.

As of now, I have a function that creates the thumbnail, saves it as a temp and returns it to the caller.

Then what I try to do, is upload that created thumb image with move_uploaded_file(tempthumb, path);

Here is the createThumb function and the caller:

function createThumb( $image, $thumbWidth )
{
  // load image and get image size
  $img = imagecreatefromjpeg( "{$image}" );
  $width = imagesx( $img );
  $height = imagesy( $img );

  // calculate thumbnail size
  $new_width = $thumbWidth;
  $new_height = floor( $height * ( $thumbWidth / $width ) );

  // create a new temporary image
  $tmp_img = imagecreatetruecolor( $new_width, $new_height );

  // copy and resize old image into new image 
  imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

  return $tmp_img;

}

return $tmp_img; // Here I return the new image. Is this the proper way to get a binary image back??

Here is the caller:

$thumb = createThumb($_FILES['propform-previmg']['tmp_name'], $max_previmg_width); 
        $filenamepath = $src_dir . '/thumb/' . $_FILES['propform-previmg']['name'];
        if ( !move_uploaded_file($thumb, $filenamepath ))
            echo "Error moving file {$filenamepath}";

I tried to just upload the uploaded file directly without trying to make a thumbnail first, and that worked fine. So I guess there is some error with the variable I return from the createThumb function, but I can’t figure out exactly what.

Also, I need to do the upload from the caller code, and not inside the createThumb function with imagejpeg(file, path).

Thank you!

Answer 1

First line from the manual:

This function checks to ensure that the file designated by filename is a valid upload file (meaning that it was uploaded via PHP’s HTTP POST upload mechanism). If the file is valid, it will be moved to the filename given by destination.

Your thumbnail wasn’t uploaded, it even isn’t a file yet, but just an image resource. To write the image, call something like imagejpeg($resource, $filename) to write it to the path specified in $filename.