Python: Django: How to run django app using a python script

Question

I've a python file named "config.py" which actually checks the ip of user and run the django server on that port accordingly.
The method I used there was to create a batch file using python and execute it later.

import socket
import subprocess
x = socket.gethostbyname(socket.gethostname())
x = str(x)
with open("run.bat", "w") as f:
    f.write(f'manage.py runserver {x}:0027')
subprocess.call([r'run.bat'])

But this method is not very effective. I want a way like:
import something something.run("manage.py")

or something accordingly
Kindly Help me doing this

Edit:
I've seen Django's manage.py.
Can I edit this file (the sys.argv section) in such a way that it automatically runs a server on 192.168.x.x:8000 just by clicking on manage.py?
Please help

not sure what you actually want to do, but if you want to check the IP, then use middleware for that. — Alex
– Alex, Commented Sep 29, 2021 at 15:43

Bernardo Duarte · Accepted Answer · 2021-09-28 13:29:40Z

2

I think the method you're trying to reach is to run a shell command inside a python script, which can be achieved by using the os.system() function.

You can use it in your code as the following:

import socket
import subprocess
import os
x = socket.gethostbyname(socket.gethostname())
x = str(x)
os.system(f'manage.py runserver {x}:0027')

answered Sep 28, 2021 at 13:29

Bernardo Duarte

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5 Comments

Ojas_Gupta Over a year ago

That did it. Thank you very much

Bernardo Duarte Over a year ago

@Ojas_Gupta Great! Don't forget to accept as the answer ;D

Ojas_Gupta Over a year ago

Hi! @Bernardo Duarte , would you like to please edit your answer according to edited question please? Thank you very much

Bernardo Duarte Over a year ago

@Ojas_Gupta I believe it is a whole new question, I think you should create a new one due to community guidelines and for better visibility so that I or someone else can help you ;D

Flimm Over a year ago

You can also do subprocess.call(['manage.py', 'runserver', x + ':0027']) instead of the os.system line.

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