TypeError: the first argument must be callable

Question

I am using python and schedule lib to create a cron-like job

class MyClass:

        def local(self, command):
                #return subprocess.call(command, s         


        

Answer 1

replace

schedule.every(1).minutes.do(self.local(script_path))

with this one:

schedule.every(1).minutes.do(self.local,script_path)

and it will work fine..

you should write the parameters of function after function name and comma separate them..