我正在面对Django的一个有趣的情况,我希望有人会看到解决方案,或者至少可以给我一个暗示.
我正在尝试制作ModelForm模型 - 通用.我不知道这是否应该做,但在这里.
这很好用:
元组引用模型
# settings.py
SPECIES = (
('TIG', 'Tiger'),
('SHR', 'Shark'),
)
用于创建Animal对象的URL
# urls.py
from django.conf.urls.defaults import patterns, include, url
urlpatterns = patterns('species.views',
url(r'^add/$', 'add_animal', name='add_animal'),
)
动物模型和它的两个孩子
# models.py
from django.db import models
from django.conf import settings
class Animal(models.Model):
name = models.CharField(max_length=100)
nickname = models.CharField(max_length=100)
species = models.CharField(max_length=3, choices=settings.SPECIES)
class Tiger(Animal):
fangs_size = models.IntegerField()
class Shark(Animal):
color = models.CharField(max_length=20)
显示表单的视图
通过GET参数选择正确的模型.
# views.py
def add_animal(request):
if request.method == "GET":
if request.GET['model_name']:
model_name = request.GET['model_name']
else:
model_name = 'Animal'
print "Model name is: %s" % model_name
model = get_model("species", model_name)
form = AnimalForm(model=model)
return create_object(
request,
model=model,
post_save_redirect=reverse('index'),
template_name='species/my_form.html',
)
一个模板
# my_form.html
<!doctype html>
<html>
<head>
<title>Adding an animal</title>
</head>
<body>
<h1>Add an animal to the farm</h1>
<form>
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit" />
</form>
</body>
</html>
当我访问/添加？model_name = tiger时,我会显示正确的表单.
现在,假设我要隐藏昵称字段.然后我需要使用自定义的ModelForm.如何用正确的模型实现它？那是我的问题.
这是表格？
# forms.py
from species.models import Animal
from django import forms
class AnimalForm(forms.ModelForm):
class Meta:
model = Animal
def __init__(self, *args, **kwargs):
model = kwargs.pop('model')
super(AnimalForm, self).__init__(*args, **kwargs)
self.Meta.model = model
观点变为:
# views.py
...
model = get_model("species", model_name)
form = AnimalForm(model=model)
return create_object(
request,
# model=model, # Need for customization
# form_class=AnimalForm, # With the class name, how to pass the argument?
form_class=form, # Not sure this can be done, I get this error: 'AnimalForm' object is not callable
post_save_redirect=reverse('index'),
template_name='species/my_form.html',
)
...
我的目标是能够创建继承自Animal的新模型,将它们添加到SPECIES元组并完成.这可以实现吗？
谢谢你的帮助!
您可能想要使用django.forms.models.modelform_factory- 它需要模型类和exclude其参数中的元组.您可以在视图中使用它来动态创建表单类.
form_class = modelform_factory(model, exclude=('nickname',))