Calling main from another file in python

Question

I have a file named client_simulator.py and it contains a Class named Client_simulator which simulates clients behavior.

I have another file named pool_manager.py and it has no class, it has one __main__ and lot of functions and basically, I want to call a method named generator of Client_simulator class from one of the methods of pool_manager.py.

the basic structure of client_simulator.py is as follows

class Client_simulator(object):
    def generator(self):

if __name__ == '__main__':
Client_simulator().generator()

the basic structure of file pool manager.py is as follows

def start_client_simulator():

     client_simulator.Client_simulator().generator()

if __name__ == "__main__":
   start_client_simulator()

I am getting the following error

'module' object is not callable

P.S: I want to call __main __ instead of `generator()', how to do that?

I am recently moving from java to python that's why having these basic doubts. Thanks in advance

I don't understand . Even when you call the main , it had generator call , so it will be called right ? Is that what you want to call the main directly ? — Saranya Sridharan
– Saranya Sridharan, Commented Nov 28, 2017 at 4:38
can you please let me know, how to call main. For example, I tried client_simulator.main() but its giving me same error — Amarjit Dhillon
– Amarjit Dhillon, Commented Nov 28, 2017 at 4:40
using client_simulator.main() gives 'module' object has no attribute 'main' — Amarjit Dhillon
– Amarjit Dhillon, Commented Nov 28, 2017 at 4:43
Yes. if name == "main": this statement checks if the invoking call is from the same program or not. — Saranya Sridharan
– Saranya Sridharan, Commented Nov 28, 2017 at 4:44
When you ask a question about an exception, include the traceback. — Terry Jan Reedy
– Terry Jan Reedy, Commented Nov 28, 2017 at 6:54

Blckknght · Accepted Answer · 2017-11-28 06:51:01Z

I think you're confused a bit, at least in terminology if maybe not in code.

When you guard a section of code with if __name__ == "__main__":, you're not defining a __main__ function. It's just a normal if statement that reads the global variable __name__ (which is set up by the Python interpreter automatically) and compares its value to a string.

So there's no __main__ to be called from another module. If you want the contents of that block to be callable from elsewhere, you should put it in a function, which you can call from both the if __name__ == "__main__": block and from the other module, as necessary.

So try this in client_simulator.py:

class Client_simulator(object):
    def generator(self):

def main():    # you can name this whatever you want, it doesn't need to be main()
    Client_simulator().generator()

if __name__ == '__main__':
    main()

The pool_manager.py file can call client_simulator.main() too.

Exis Zhang · Accepted Answer · 2019-04-01 23:28:17Z

9

@blckknght is right, but if you absolutely do need to do that, use below antipattern

import subprocess
subprocess.run('python D:/your_script.py')

answered Apr 1, 2019 at 23:28

Exis Zhang

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