如何在Django-form下拉列表中刷新值？

Question

In my django project I have a form with dropdown list of options from database. When server starts up all options are there, but if there is an option added while it runs, the new option won't show up in dropdown list.

forms.py

class filter(forms.Form): 
    filterOption=forms.CharField(widget=forms.Select(choices=getOptions())

functions.getOptions() returns a list of options from mySQL (database is not the problem here). It is then rendered in views.py and put into html form.

from .forms import filter

def index(request):
    filter = filter()
    return render(request, 'homePage/home.html',{'filter':filter})

I understand that the form object is created when the server starts, but if a new option is added and getOptions() should return one more option, how can I re-initialise that form object without restarting the server. Sometimes it is added after ~10min, but I need it instantly.

Answer 1

You can use ChoiceField with a callable:

class filter(forms.Form): 
    filterOption = forms.ChoiceField(choices=getOptions)

Answer 2

Well, if you want to show the new checkbox possibilities after you reload the form html page in your browser, it should be there immediately.

Have you really looked in the database that this option was added successfully?

Do you have some kind of customized cache active?

If you want to show the new checkbox option without a page reload, you'd need to implement a pretty complicated custom form rendering which reloads the form every X seconds via an AJAX request.

Answer 3

Solved it by following: https://modwsgi.readthedocs.io/en/develop/user-guides/reloading-source-code.html

I basicly kill the process and make it reload every time new value is added.

import signal, os
os.kill(os.getpid(), signal.SIGINT)

In linked documentation it says 'touch'-ing the wsgi.py file should make it reload. And it does, but only when you do it manualy, not if you use os.system('touch .../wsgi.py'). To make it reload with code I had to kill the process.