I want to start a python file that containing imports from django library in the OS shell

Question

I have a very simple case in which I want to run a script with python

myscript.py:

from models import Company
c = Company(number="1234567890", name="Company name")

models.py:

from django.db import models

class Company(models.Model):
    number = models.CharField("company number", max_length=20, unique=True)
    name = models.CharField("company name", max_length=30, unique=True)

    class Meta:
        verbose_name = "Company"
        verbose_name_plural = "Companies"

I want to run python myscript.py or something like python manage.py execute_file myscript.py

I know my question is trivial and I can import a python manage.py shell environment, but that's not the point. Can I run it in a much simple manner?

Define a command: https://docs.djangoproject.com/en/2.0/howto/custom-management-commands/ — Willem Van Onsem, Jul 26 '18 at 10:08

score 1 · Answer 1 · answered Jul 26 '18 at 10:13

1

Yes, Django has support for that: you define a custom command. It has some support for example to parse command parameters in a more convenient way.

You first need to define a file with the same name as the command you want (so here myscript.py). This should be placed in the app/management/commands/ directory (where app is an app of your system). So then the file tree looks like:

app/
    __init__.py
    management/
        __init__.py
        commands/
            __init__.py
            myscript.py

If you have not constructed any custom management commands before, you probably need to construct the files in boldface. The __init__.py files are simply empty files.

Then your command can look like:

# app/management/commands/myscript.py

from django.core.management.base import BaseCommand
from models import Company

class Command(BaseCommand):

    def handle(self, *args, **options):
        c = Company(number="1234567890", name="Company name")

Note that this script is actually not doing much: you construct a Company, but you do not store it in the database, so that means that usually the program will just stop, or raise an error in case the input is invalid.

You can then call your script with:

python3 manage.py myscript

and the command will also be listed in case you ask for help (with python3 manage.py help).

The Django tutorial on custom commands explains in more depth what you can do to parse commands, add helptext, etc.

answered Jul 26 '18 at 10:13

Willem Van Onsem

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the solution is working, but is still very complex. If nothing new comes will accept your answer, th you for your time – Искрен Станиславов Jul 26 '18 at 10:17
@ИскренСтаниславов So what could be easier than run shell and execute the code? `cat script.py | ./manage.py shell` – Sergei Zherevchuk Jul 26 '18 at 13:58
@WillemVanOnsem do you mind reworking your example to work like `python3 manage.py run myscript.py` – Искрен Станиславов Jul 27 '18 at 13:33
1

@ИскренСтаниславов: that would not only be a major anti-pattern (for which employees should get fired :) ), but also have terrible security consequences. It would imply one should use `eval`, also known as *evil*. – Willem Van Onsem Jul 27 '18 at 13:38
how about the usage of `execfile(filename)` in python27 instead of `eval` and `exec(compile(open(filename).read(), filename, 'exec'))` in python3 ? is it `eval` as well ? – Искрен Станиславов Jul 30 '18 at 07:19
1

@ИскренСтаниславов: all functions like `eval`, `exec`, and friends are problematic and insecure: see https://stackoverflow.com/questions/1832940/why-is-using-eval-a-bad-practice#1832957 – Willem Van Onsem Jul 30 '18 at 07:53
I totally agree that using the functions you listed are evil for usage in production. but for local usage such as running a test file from the current one that doesn't do any harm to the testing environment, is that so? – Искрен Станиславов Jul 30 '18 at 09:06

I want to start a python file that containing imports from django library in the OS shell

1 Answers1