I want that the path from which the file is uploaded C:\\xampp\\htdocs\\cmd is coming from
dirname(__FILE__), but when i do that the script throws an error.
Working code
exec( dirname(__FILE__)."\\xyz.exe C:\\xampp\\htdocs\\cmd\\blue.png D:\\1.png\", $output);
But I want:
exec( dirname(__FILE__)."\\xyz.exe" dirname(__FILE__)."\\blue.png D:\\1.png\", $output);
Why not chdir before executing all: