How to access image field in PHP methods

Hi guys we are sitting with a bit of a problem, we have a form with some drop downs and image (database) fields. If the drop-down is a certain value the user has to upload an image too. Everytime we try to access the image field in a method it comes back as empty. Below is an example of what we are trying to achieve.

if (({check_1} == ‘Not Updated’ && {notes_1}==’’ && {img_up1}==’’) || ({check_1} == ‘Not Updated’ && {notes_1}!=’’ && {img_up1}==’’) || ({check_1} == ‘Not Updated’ && {notes_1}==’’ && {img_up1}!=’’))

            {

            sc_error_message('Pictures and Notes are required for "Update Service Control Board by means of paint coding - Green *Monthly Check (MC)"');

            sc_label('img_up1') = "Take Photo <b style='color:red'>*</b>";

            sc_label('notes_1') = "Add Note <b style='color:red'>*</b>";

            $alert1 = true;

            }else{

            sc_label('img_up1') = "Take Photo ";

            sc_label('notes_1') = "Add Note";

            $alert1 = false;

            }`

From using a print_r({img_up1}); we see the img_up1 field is empty therefore always entering the if even if an image is uploaded. Is there a method of doing this that we are missing? or are we somewhat on the right track?

Any advice would be greatly appreciated. TIA