Update Django Choice Field With Database Results
Solution 1:
classMyForm(forms.Form):
my_choice_field = forms.ChoiceField(choices=MY_CHOICES)
So if you want the values to be dynamic(or dependent of some logic) you can simply modify your code to something like this:
either
defget_my_choices():
# you place some logic here return choices_list
classMyForm(forms.Form):
my_choice_field = forms.ChoiceField(choices=get_my_choices())
or
User_list = [ #place logic here]classMyForm(forms.Form):
my_choice_field = forms.ChoiceField(choices=get_my_choices())
but once database value is updated, new data value will be popoulated only on restart of server. So write a function like this in forms:
classMyForm(forms.Form):def__init__(self, *args, **kwargs):
super(MyForm, self).__init__(*args, **kwargs)
self.fields['my_choice_field'] = forms.ChoiceField( choices=get_my_choices() )
or in place of the get_my_choices u can ad the USER_LIST too.
Solution 2:
If you have models for location
and quantity
, a ModelChoiceField should work:
classdform(forms.Form):
location = forms.ModelChoiceField(queryset = Location.objects.all())
Otherwise, you'll need to query the database directly, for example:
classdform(forms.Form):
location = forms.ChoiceField(choices = get_location_choices())
# elsewhere
from django.db import connection
def get_location_choices():
cursor = connection.cursor()
cursor.execute("select location_id, name from location_table")
return cursor.fetchall()
The SQL query to use here depends on your database engine and table schema.
Solution 3:
I think that, based on my understanding of your question, the best solution would be to include JSON objects with your form and load these using jQuery instead of submitting the form over and over. Included in your form, you should add something like:
classMyForm(forms.Form):
CHOICE_DICT = {
'choice_1': [
'option_1',
'option_2',
],
etc...
Then you should include form.CHOICE_DICT in your context, load that with jQuery, and render it depending on changes to other fields.
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