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How do you make a Django Form for a model and all of its children following a particular foreign key?
原标题:

For example, lets say you want to create a ModelForm for Supervisor that also allows you to create or update 3 or more Underlings in the same form.

from django.db import models
class Supervisor(models.Model):
    name = models.CharField(max_length=100)

class Underling(models.Model):
    supervisor = models.ForeignKey(Superisor, related_name="underlings")
    name = models.CharField(max_length=100)

This should be pretty standard, right? Just make a FormSet for the underlings, and... then what? The Django Admin interface does it, so how do I do it?

问题回答

First you would define your ModelForm for Underlings:

class UnderlingForm(forms.ModelForm):
    class Meta:
        model = Underling

Then you would create a FormSet of UnderlingForms:

UnderlingFormSet = formset_factory(UnderlingForm, extra=3) # 3 Underlings.

Then you instantiate and loop over them (or pass them to a template context):

formset = UnderlingFormSet()
for form in formset.forms:
    print form.as_table()

Since you ll also have a SupervisorForm in the same view, I d suggest having a glance at the prefix option too. If you are still left with questions about Formsets, here is the official docs.





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