running a python script with values from views.py in Django

Question

I am trying to pass in a value i got in a views.py file in Django into another python script i wrote, but i have no idea how to do so. Here is my code from views.py file:

    def get_alarm_settings(request):
      if request.method == 'POST':
        username = request.POST.get('username')
        password = request.POST.get('password')
        alarm = Alarm.objects.all()[0].alarm_setting
        response = HttpResponse(alarm, content_type='text/plain')
        response_a = execfile('alarm_file.py')
        return response_a

I am trying to pass response into alarm_file.py, does anyone know how to do that?

If the script is written in Python, why don't you just import the relevant function and call it? — Blender
– Blender, Commented Jul 25, 2014 at 0:21
I am fully with Blender, you should implement function inside alarm_file and use import alarm_file to have access to it — Max Markov
– Max Markov, Commented Jul 25, 2014 at 0:33
How would i make it so it has the same effect as python alarm_file.py?? — Rjzheng
– Rjzheng, Commented Jul 25, 2014 at 4:15

pavel_form · Accepted Answer · 2014-07-25 09:48:19Z

As per comments under the question, make function in alarm_file.py and import it. To make python alarm_file.py call do the same, use if __name__ == '__main__': an alarm_file.py. It will contain logic to run when called python alarm_file.py. Additionally, you can use sys.argv to get argument, passed in command line. For example:

alarm_file.py:

import sys

def do_something(val):
    # do something
    print val
    # return something
    return val

if __name__ == '__main__':
    try:
        arg = sys.argv[1]
    except IndexError:
        arg = None

    return_val = do_something(arg)

This will ensure that you can run simply: python alarm_file.py some_argument

In your view just import function from alarm_file and use it:

from alarm_file import do_something
...
        response = HttpResponse(alarm, content_type='text/plain')
        response_a = do_something(response)
        return response_a

Nijil · Accepted Answer · 2018-11-15 13:04:50Z

I would suggest you to use Popen method of Subprocess module.

For Example:

process = Subprocess.Popen(['python','alarm_file.py','arg1','arg2'], stdout=PIPE, stderr=STDOUT)
output = process.stdout.read()
exitstatus = process.poll()

In this example, stderr and stdout of the script is saved in variable 'output', And exit code of the script is saved in variable 'exitstatus'.

You can refer my blog for detailed explanation >>

http://www.easyaslinux.com/tutorials/devops/how-to-run-execute-any-script-python-perl-ruby-bash-etc-from-django-views/

alphwe · Accepted Answer · 2014-07-25 04:09:19Z

0

You may use the process-response middleware to handle the response returned from the view method, more details could be found here https://docs.djangoproject.com/en/dev/topics/http/middleware/#process-response

answered Jul 25, 2014 at 4:09

alphwe

111 bronze badge

Comments

DeeGautam · Accepted Answer · 2014-07-25 04:44:16Z

0

You cannot pass arguments using execfile(), if you want to call another instance of the interpreter and run the script, you should probably use subprocess.popen (suprocess docs)

Something like this: subprocess.Popen(['alarm_file.py', 'arg1', ...], ..., shell=True)

answered Jul 25, 2014 at 4:44

DeeGautam

415 bronze badges

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