Using Model Form

Introduction#

Django ModelForm enables the creation of a Form class from a Django model.

Using Django Model Form with Django Class Based View.

Django Model Form with Django Class Based view is a classic way of building pages to do create/update operations in django application quickly. Within the form we can put methods to execute tasks. Its a cleaner way to put tasks in forms rather than putting in views/models.

To give an example using Django Model Form, first we need to define our Model.

class MyModel(models.Model):
   name = models.CharField(
          verbose_name = 'Name',
          max_length = 255)

Now let us make a form using this model:

class MyModelForm(forms.ModelForm):

    class Meta:
        model = MyModel
        fields = '__all__'

Lets add a method to print hello world in it.

class MyModelForm(forms.ModelForm):
    class Meta:
        model = MyModel
        fields = '__all__'
    
    def print_hello_world(self):
         print('Hello World')

Lets make a template to display the form:

<form method="post" action="">
    {% csrf_token %}
<ul>
    {{ form.as_p }}
</ul>
<input type="submit" value="Submit Form"/>
</form>

Now we will use this form in three different views which will respectively Create and Update tasks.

from django.views.generic.edit import CreateView, UpdateView
from myapp.models import MyModel

class MyModelCreate(CreateView):
    model = MyModel
    fields = ['name']
    form_class = MyModelForm
    template_name = 'my_template.html'

    def form_valid(self, form):
        # This method is called when valid form data has been POSTed.
        # It should return an HttpResponse.
        form.print_hello_world() # This method will print hello world in console
        return super(MyModelCreate, self).form_valid(form)
    

class MyModelUpdate(UpdateView):
    model = MyModel
    fields = ['name']
    form_class = MyModelForm
    template_name = 'my_template.html'

Now lets create a urls for accessing those views.

from django.conf.urls import url
from myapp.views import MyModelCreate, MyModelUpdate

urlpatterns = [
    # ...
    url(r'mymodel/add/$', MyModelCreate.as_view(), name='author-add'),
    url(r'mymodel/(?P<pk>[0-9]+)/$', MyModelUpdate.as_view(), name='author-update')
]

Okay, our work has been done. We can access url: localhost:8000/mymodel/add for creating entry in the model. Also access localhost:8000/mymodel/1 to update that entry.

Making fields not editable

Django 1.9 added the Field.disabled attribute:

The disabled boolean argument, when set to True, disables a form field using the disabled HTML attribute so that it won’t be editable by users. Even if a user tampers with the field’s value submitted to the server, it will be ignored in favor of the value from the form’s initial data.

And so you only need to do:

MyChangeForm(ModelForm): 

    def __init__(self, *args, **kwargs): 
        super(MyChangeForm, self).__init__(*args, **kwargs)                       
        self.fields['<field_to_disable>'].disabled = True

And creating the form you need:

MyChangeForm(initial={'<field_to_disable>': "something"})

Before version 1.9 you had to:

class MyChangeForm(ModelForm):
    def __init__(self, *args, **kwargs):
        super(ItemForm, self).__init__(*args, **kwargs)
        instance = getattr(self, 'instance', None)
    if instance and instance.id:
        self.fields['<field_to_disable>'].required = False
        self.fields['<field_to_disable>'].widget.attrs['disabled'] = True

def clean_<field_to_disable>(self):
    # As shown in the above answer.
    instance = getattr(self, 'instance', None)
    if instance:
        return instance.<field_to_disable>
    else:
        return self.cleaned_data.get('<field_to_disable>', None)

And creating the form you need:

MyChangeForm(instance=MyChange.objects.get_or_create(<field_to_disable>="something"))

This example was based on this question.