问题描述
I have a model, Foo
. It has several database properties, and several properties that are calculated based on a combination of factors. I would like to present these calculated properties to the user as if they were database properties. (The backing factors would be changed to reflect user input.) Is there a way to do this with the Django admin interface?
I would suggest you subclass a modelform for Foo
(FooAdminForm) to add your own fields not backed by the database. Your custom validation can reside in the clean_*
methods of ModelForm.
Inside the save_model
method of FooAdmin
you get the request, an instance of Foo
and the form data, so you could do all processing of the data before/after saving the instance.
Here is an example for a model with a custom form registered with django admin:
from django import forms
from django.db import models
from django.contrib import admin
class Foo(models.Model):
name = models.CharField(max_length=30)
class FooAdminForm(forms.ModelForm):
# custom field not backed by database
calculated = forms.IntegerField()
class Meta:
model = Foo
class FooAdmin(admin.ModelAdmin):
# use the custom form instead of a generic modelform
form = FooAdminForm
# your own processing
def save_model(self, request, obj, form, change):
# for example:
obj.name = 'Foo #%d' % form.cleaned_data['calculated']
obj.save()
admin.site.register(Foo, FooAdmin)
Providing initial values for custom fields based on instance data
(I'm not sure if this is the best solution, but it should work.)
When a modelform for a existing model instance in the database is constructed, it gets passed this instance. So in FooAdminForm's __init__
one can change the fields attributes based on instance data.
def __init__(self, *args, **kwargs):
super(FooAdminForm, self).__init__(*args, **kwargs)
# only change attributes if an instance is passed
instance = kwargs.get('instance')
if instance:
self.fields['calculated'].initial = (instance.bar == 42)
这篇关于Django:在管理界面中伪造一个字段?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!