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..

Answer 2

do expects a callable and any arguments it make take.

Therefore your call to do should look like:

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

The do implementation can be found here.

def do(self, job_func, *args, **kwargs):
    """Specifies the job_func that should be called every time the
    job runs.

    Any additional arguments are passed on to job_func when
    the job runs.
    """
    self.job_func = functools.partial(job_func, *args, **kwargs)
    functools.update_wrapper(self.job_func, job_func)
    self._schedule_next_run()
    return self