I have this copy function
$srcfile='C:\xampp\htdocs\imagenes\M1001.jpg';
$dstfile='C:\xampp\htdocs\imagenes-copiadas\M1001.jpg';
mkdir(dirname($dstfile), 0777, true);
copy($srcfile, $dstfile);
I need the name M1001.jpg to be a variable call...
asked by
26.05.2018 / 10:49