How to make a customizable user survey in Django - django

I am making a system for a company which among other things must hold information about the satisfactory level about various things, I have made it work fine using a fixed model with fixed questions and answers, but I am sure that they will need to change or add questions.
So I want to make a system where users can make custom evaluation schemas that consists of custom questions defined by them. How do I go about making such a design?
Right now my model is this, but wrong:
RATING_CHOICES = ((0, u"Good"), (1, u"Bad"), (2, u"Dunno"),)
class EvaluationScheme(models.Model):
title = models.CharField(max_length=200)
class Evaluation(models.Model):
doctor = models.CharField(max_length=200)
agency = models.CharField(max_length=200)
scheme = models.ForeignKey(EvaluationScheme)
class EvaluationQuestion(models.Model):
question = models.CharField(max_length=200)
evaluation = models.ForeignKey(EvaluationScheme)
def __unicode__(self):
return self.question
class EvaluationAnswer(models.Model):
evaluation = models.ForeignKey(Evaluation)
question = models.ForeignKey(EvaluationQuestion)
answer = models.SmallIntegerField(choices=RATING_CHOICES)
This is sort of what I want, except that the EvaluationScheme is useless, since you still have to chose all questions and answers yourself - it does not display a list of only the questions related to the schema of choice.

I think your models are fine. I used the Django admin to create an EvaluationScheme with EvaluationQuestions, then I created an Evaluation and I was able to answer its questions. Here's the code I used to go with your models:
# forms.py:
from django.forms.models import inlineformset_factory
import models
AnswerFormSet = inlineformset_factory(models.Evaluation,
models.EvaluationAnswer, exclude=('question',),
extra=0, can_delete=False)
# views.py
from django.http import HttpResponse
from django.shortcuts import render_to_response, get_object_or_404
import models, forms
def prepare_blank_answers(evaluation):
for question in evaluation.scheme.evaluationquestion_set.all():
answer = models.EvaluationAnswer(evaluation=evaluation,
question=question)
answer.save()
def answer_form(request, id):
evaluation = get_object_or_404(models.Evaluation, id=id)
if len(evaluation.evaluationanswer_set.all()) == 0:
prepare_blank_answers(evaluation)
if request.method == 'POST':
formset = forms.AnswerFormSet(request.POST, instance=evaluation)
if formset.is_valid():
formset.save()
return HttpResponse('Thank you!')
else:
formset = forms.AnswerFormSet(instance=evaluation)
return render_to_response('answer_form.html',
{'formset':formset, 'evaluation':evaluation})
# answer_form.html:
<html><head></head><body>
Doctor: {{ evaluation.doctor }} <br>
Agency: {{ evaluation.agency }}
<form method="POST">
{{ formset.management_form }}
<table>
{% for form in formset.forms %}
<tr><th colspan="2">{{ form.instance.question }}</th></tr>
{{ form }}
{% endfor %}
</table>
<input type="submit">
</form>
</body></html>

Have you checked django-survey? It's pretty neat.

Django-crowdsourcing is a fork of django-survey that is actively maintained as of 2012 and targets Django 1.2+.

Not a django expert so you might wish to wait for a more experience person to answer but you could try something like:
EvaluationQuestions.objects.filter(evaluationscheme__title="myscheme").select_related()
Could also put the relationships the other way around, depends how you need to access the data.
class EvaluationScheme(models.Model):
title = models.CharField(max_length=200)
evaluations = models.ManyToMany(Evaluation)
questions = models.ManyToMany(EvaluationQuestions)

Related

multi step form and model inheritance in django

I have seen this approach in many web applications (e.g. when you subscribe for an insurance), but I can't find a good way to implement it in django. I have several classes in my model which inherit from a base class, and so they have several fields in common. In the create-view I want to use that inheritance, so first ask for the common fields and then ask for the specific fields, depending on the choices of the user.
Naive example, suppose I want to fill a database of places
class Place(Model):
name = models.CharField(max_length=40)
address = models.CharField(max_length=100)
class Restaurant(Place):
cuisine = models.CharField(max_length=40)
website = models.CharField(max_length=40)
class SportField(Place):
sport = models.CharField(max_length=40)
Now I would like to have a create view when there are the common fields (name and address) and then the possibility to choose the type of place (Restaurant / SportField). Once the kind of place is selected (or the user press a "Continue" button) new fields appear (I guess to make it simple the page need to reload) and the old one are still visible, already filled.
I have seen this approach many times, so I am surprised there is no standard way, or some extensions already helping with that (I have looked at Form Wizard from django-formtools, but not really linked to inheritance), also doing more complicated stuff, as having more depth in inheritance.
models.py
class Place(models.Model):
name = models.CharField(max_length=40)
address = models.CharField(max_length=100)
class Restaurant(Place):
cuisine = models.CharField(max_length=40)
website = models.CharField(max_length=40)
class SportField(Place):
sport = models.CharField(max_length=40)
forms.py
from django.db import models
from django import forms
class CustomForm(forms.Form):
CHOICES = (('restaurant', 'Restaurant'), ('sport', 'Sport'),)
name = forms.CharField(widget=forms.TextInput(attrs={'placeholder': 'Name'}))
address = forms.CharField(widget=forms.TextInput(attrs={'placeholder': 'Address'}))
type = forms.ChoiceField(
choices=CHOICES,
widget=forms.Select(attrs={'onChange':'renderForm();'}))
cuisine = forms.CharField(required=False, widget=forms.TextInput(attrs={'placeholder': 'Cuisine'}))
website = forms.CharField(required=False, widget=forms.TextInput(attrs={'placeholder': 'Website'}))
sport = forms.CharField(required=False, widget=forms.TextInput(attrs={'placeholder': 'Sport'}))
views.py
from django.http.response import HttpResponse
from .models import Restaurant, SportField
from .forms import CustomForm
from django.shortcuts import render
from django.views import View
class CustomView(View):
def get(self, request,):
form = CustomForm()
return render(request, 'home.html', {'form':form})
def post(self, request,):
data = request.POST
name = data['name']
address = data['address']
type = data['type']
if(type == 'restaurant'):
website = data['website']
cuisine = data['cuisine']
Restaurant.objects.create(
name=name, address=address, website=website, cuisine=cuisine
)
else:
sport = data['sport']
SportField.objects.create(name=name, address=address, sport=sport)
return HttpResponse("Success")
templates/home.html
<html>
<head>
<script type="text/javascript">
function renderForm() {
var type =
document.getElementById("{{form.type.auto_id}}").value;
if (type == 'restaurant') {
document.getElementById("{{form.website.auto_id}}").style.display = 'block';
document.getElementById("{{form.cuisine.auto_id}}").style.display = 'block';
document.getElementById("{{form.sport.auto_id}}").style.display = 'none';
} else {
document.getElementById("{{form.website.auto_id}}").style.display = 'none';
document.getElementById("{{form.cuisine.auto_id}}").style.display = 'none';
document.getElementById("{{form.sport.auto_id}}").style.display = 'block';
}
}
</script>
</head>
<body onload="renderForm()">
<form method="post" action="/">
{% csrf_token %}
{{form.name}}<br>
{{form.address}}<br>
{{form.type}}<br>
{{form.website}}
{{form.cuisine}}
{{form.sport}}
<input type="submit">
</form>
</body>
</html>
Add templates folder in settings.py
TEMPLATES = [
{
...
'DIRS': [os.path.join(BASE_DIR, 'templates')],
...
]
I've created a 2-page working example using modified Class Based Views.
When the form is submitted on the first page, an object of place_type is created. The user is then redirected to the second page where they can update existing details and add additional information.
No separate ModelForms are needed because the CreateView and UpdateView automatically generate the forms from the relevant object's model class.
A single template named place_form.html is required. It should render the {{ form }} tag.
# models.py
from django.db import models
from django.urls import reverse
class Place(models.Model):
"""
Each tuple in TYPE_CHOICES contains a child class name
as the first element.
"""
TYPE_CHOICES = (
('Restaurant', 'Restaurant'),
('SportField', 'Sport Field'),
)
name = models.CharField(max_length=40)
address = models.CharField(max_length=100)
place_type = models.CharField(max_length=40, blank=True, choices=TYPE_CHOICES)
def __str__(self):
return self.name
def get_absolute_url(self):
return reverse('place_update', args=[self.pk])
# Child models go here...
# urls.py
from django.urls import path
from . import views
urlpatterns = [
path('create/', views.PlaceCreateView.as_view(), name='place_create'),
path('<pk>/', views.PlaceUpdateView.as_view(), name='place_update'),
]
# views.py
from django.http import HttpResponseRedirect
from django.forms.models import construct_instance, modelform_factory
from django.views.generic.edit import CreateView, UpdateView
from django.urls import reverse_lazy
from . import models
class PlaceCreateView(CreateView):
model = models.Place
fields = '__all__'
def form_valid(self, form):
"""
If a `place_type` is selected, it is used to create an
instance of that Model and return the url.
"""
place_type = form.cleaned_data['place_type']
if place_type:
klass = getattr(models, place_type)
instance = klass()
obj = construct_instance(form, instance)
obj.save()
return HttpResponseRedirect(obj.get_absolute_url())
return super().form_valid(form)
class PlaceUpdateView(UpdateView):
fields = '__all__'
success_url = reverse_lazy('place_create')
template_name = 'place_form.html'
def get_object(self, queryset=None):
"""
If the place has a `place_type`, get that object instead.
"""
pk = self.kwargs.get(self.pk_url_kwarg)
if pk is not None:
obj = models.Place.objects.get(pk=pk)
if obj.place_type:
klass = getattr(models, obj.place_type)
obj = klass.objects.get(pk=pk)
else:
raise AttributeError(
"PlaceUpdateView must be called with an object pk in the URLconf."
)
return obj
def get_form_class(self):
"""
Remove the `place_type` field.
"""
model = self.object.__class__
return modelform_factory(model, exclude=['place_type',])
We did something similar manually, we created the views and forms based on design and did the linkage based on if conditions.
I think a nice solution would be to dynamically access subclasses of the main class and then do the necessary filtering/lists building.
UPD: I've spent some more time today on this question and made a "less raw" solution that allows to use the inheritance.
You can also check the code below deployed here. It has only one level of inheritance (as in example), though, the approach is generic enough to have multiple levels
views.py
def inheritance_view(request):
all_forms = {form.Meta.model: form for form in forms.PlaceForm.__subclasses__()}
all_forms[models.Place] = forms.PlaceForm
places = {cls._meta.verbose_name: cls for cls in models.Place.__subclasses__()}
# initiate forms with the first one
context = {
'forms': [forms.PlaceForm(request.POST)],
}
# check sub-forms selected on the forms and include their sub-forms (if any)
for f in context['forms']:
f.sub_selected = request.POST.get('{}_sub_selected'.format(f.Meta.model._meta.model_name))
if f.sub_selected:
sub_form = all_forms.get(places.get(f.sub_selected))
if sub_form not in context['forms']:
context['forms'].append(sub_form(request.POST))
# update some fields on forms to render them on the template
for f in context['forms']:
f.model_name = f.Meta.model._meta.model_name
f.sub_forms = {x.Meta.model._meta.verbose_name: x for x in f.__class__.__subclasses__()}
f.sub_options = f.sub_forms.keys() # this is for rendering selector on the form for the follow-up forms
page = loader.get_template(template)
response = HttpResponse(page.render(context, request))
return response
forms.py
class PlaceForm(forms.ModelForm):
class Meta:
model = models.Place
fields = ('name', 'address',)
class RestaurantForm(PlaceForm):
class Meta:
model = models.Restaurant
fields = ('cuisine', 'website',)
class SportFieldForm(PlaceForm):
class Meta:
model = models.SportField
fields = ('sport',)
templates/inheritance.html
<body>
{% for form in forms %}
<form method="post">
{% csrf_token %}
{{ form.as_p }}
{% if form.sub_options %}
<select class="change-place" name="{{ form.model_name }}_sub_selected">
{% for option in form.sub_options %}
<option value="{{ option }}" {% if option == form.sub_selected %}selected{% endif %}>{{ option }}</option>
{% endfor %}
</select>
{% endif %}
<button type="submit">Next</button>
</form>
{% endfor %}
</body>
What I didn't make here is saving the form to the database. But it should be rather trivial using the similar snippet:
for f in context['forms']:
if f.is_valid():
f.save()
Add a PlaceType table, and a FK, e.g. type_of_place, to the Place table:
class PlaceType(Model):
types = models.CharField(max_length=40) # sportsfield, restaurants, bodega, etc.
class Place(Model):
name = models.CharField(max_length=40)
address = models.CharField(max_length=100)
type_of_place = models.ForeignKey('PlaceType', on_delete=models.SET_NULL, null=True)
class Restaurant(Place):
cuisine = models.CharField(max_length=40)
website = models.CharField(max_length=40)
This allows you to create a new Place as either SportsField, restaurant or some other type which you can easily add in the future.
When a new place is created, you'll use the standard CreateView and Model Form. Then, you can display a second form which also uses a standard CreateView that is based on the type_of_place value. These forms can be on the same page (and with javascript on the browser side, you'll hide the second form until the first one is saved) or on separate pages--which may be more practical if you intend to have lots of extra columns. The two key points are as follows:
type_of_place determines which form, view, and model to use. For
example, if user chooses a "Sports Field" for type_of_place, then
you know to route the user off to the SportsField model form;
CreateViews are designed for creating just one object/model. When
used as intended, they are simple and easy to maintain.
There are lot of way you can handle multiple froms in django. The easiest way to use inlineformset_factory.
in your froms.py:
forms .models import your model
class ParentFrom(froms.From):
# add fields from your parent model
Restaurant = inlineformset_factory(your parent model name,Your Child model name,fields=('cuisine',# add fields from your child model),extra=1,can_delete=False,)
SportField = inlineformset_factory(your parent model name,Your Child model name,fields=('sport',# add fields from your child model),extra=1,can_delete=False,)
in your views.py
if ParentFrom.is_valid():
ParentFrom = ParentFrom.save(commit=False)
Restaurant = Restaurant(request.POST, request.FILES,) #if you want to add images or files then use request.FILES.
SportField = SportField(request.POST)
if Restaurant.is_valid() and SportField.is_valid():
ParentFrom.save()
Restaurant.save()
SportField.save()
return HttpResponseRedirect(#your redirect url)
#html
<form method="POST" enctype="multipart/form-data">
{% csrf_token %}
#{{ Restaurant.errors}} #if you want to show error
{{ Restaurant}}
{{ SportField}}
{{form}}
</form>
you can use simple JavaScript in your html for hide and show your any froms fields

Beginners: Dynamic filtering of foreign key field in Django CreateView form

Models:
class Instructional_Cycle(models.Model):
date_started = models.DateField()
date_finished = models.DateField()
standard_tested = models.OneToOneField(Standard, on_delete=models.CASCADE)
class Standard(models.Model):
subject = models.CharField(max_length=14, choices=subjects)
grade_level = models.IntegerField(choices=gradeLevels)
descriptor = models.CharField(max_length=15)
description = models.TextField()
essential_status = models.BooleanField(default=False)
View:
class CycleCreateView(CreateView):
model = Instructional_Cycle
template_name = 'cycle_new.html'
fields = '__all__'
success_url = reverse_lazy('student_progress:cycles')
Template:
<!-- student_progress/cycle_new.html -->
{% extends 'base.html' %}
{% block content %}
<h1>Add a new instructional cycle:</h1>
<form action="{% url 'student_progress:cycle_new' %}" method="post">
{% csrf_token %}
{{ form.as_p }}
<button name="submit">add cycle</button>
</form>
{% endblock content %}
The problem I'm having with this form is that the dropdown to select Instructional_Cycle.standard_tested has literally 1000 records from Standard. There's no way that the user can scroll through all of those and find the one record they want.
What I need is some way to click a link and filter the dropdown list by subject or grade_level and/or a search box, similar to what's achieved on the admin side by creating a custom admin model in admin.py like so:
class StandardAdmin(admin.ModelAdmin):
list_display = ('descriptor', 'description', 'essential_status')
list_filter = ('subject', 'grade_level', 'essential_status')
search_fields = ('descriptor',)
inlines = [MilestoneInLine]
def get_search_results(self, request, queryset, search_term):
queryset, use_distinct = super().get_search_results(request, queryset, search_term)
try:
search_term_as_int = int(search_term)
except ValueError:
pass
else:
queryset |= self.model.objects.filter(age=search_term_as_int)
return queryset, use_distinct
Please "dumb it down" for this newbie. I just finished working through Django for Beginners, and my conceptual model of how this all fits together is still full of holes. Please assume that I know hardly anything. Thanks!
That amount of reactive work on one page will require you to be comfortable with Javascript, Ajax, etc. If that is the case, there are a number of approaches you could take that let you refresh the form with the desired options.
Alternatively, you could ask the user for the necessary data one step earlier in the process and let Django build the correct form for you in the first place by overriding the form's default queryset.
You should look into using something like django-ajax-select. https://github.com/crucialfelix/django-ajax-selects

Django 1.9 forms not displaying in template but CSRF token is

I'm working on a Django project and I'm having issues with my Model Forms rendering in my template. I've read through the documentation, tutorials as well as searched here; particularly this post here someone had with their input fields not showing. I thought that my issue was similar but I'm still not having any success. Basically, when I call {{ form.as_p }} nothing shows in the DOM. This occurs despite my csrf token working. Here's the code for my Model:
class Workout(models.Model):
upper_body = models.CharField(max_length=30, help_text="ex: Bench Press, Shoulder Press, Shrugs")
lower_body = models.CharField(max_length=30, help_text="Squats, Lunges, Deadlifts")
explosive = models.CharField(max_length=30, help_text="Hang Cleans, Box Jumps, Hang Snatch")
cardio = models.CharField(max_length=30, help_text="Jogging, Cycling, 110s")
speed = models.CharField(max_length=30, help_text="Sprints, 10 yard starts, Speed Sled")
skill = models.CharField(max_length=30, help_text="Boxing, Jump Shooting, Route Running")
athlete = models.ForeignKey(User, related_name='athlete')
def __unicode__(self):
return u"{}".format(self.athlete)
Here is the forms.py for that Model:
from django import forms
from .models import User, Workout, Meal
from django.forms import ModelForm
# Beginning of Workout Form
class WorkoutForm(forms.ModelForm):
class Meta:
model = Workout
fields = ["upper_body",
"lower_body",
"explosive",
"cardio",
"speed",
"skill"]
# End of Workout Form
Here is my views.py for the form:
from .models import User, Workout, Meal
from Spartan.forms import UserForm, WorkoutForm, MealForm
from django.shortcuts import render, redirect
from django.http import HttpResponse, HttpResponseForbidden, HttpResponseRedirect
def workout_form(request):
form = WorkoutForm(request.POST or None)
if form.is_valid():
workout = form.save(commit=False)
workout.athlete = request.user
workout.save()
return redirect('user/workouts.html')
Here is the actual the form that should be rendered on the template:
<div class="container">
<h3>Create your workout</h3>
<form method="POST" action=".">
{% csrf_token %}
{{ form.as_p }}
<button class="btn btn-primary" type="submit" value="Create Your Workout">CREATE WORKOUT</button>
</form>
</div>
As I've stated, the token is there, along with the form and button just no input fields. If anyone can assist me in pointing out what I'm not doing or what I'm overlooking it'll be GREATLY appreciated.
Thanks!
The problem is in your view. The else block doesn't matter because in your GET request is_valid is always returning False and you're redirecting to user/workouts.html without context.
One solution is to use an if/else statement to segment how you handle GET and POST:
def workout_form(request):
form = WorkoutForm() # Will be overridden if `is_valid` returns `False`
if request.method == "POST":
form = WorkoutForm(data=request.POST)
if form.is_valid():
workout = form.save(commit=False)
workout.athlete = request.user
workout.save()
return redirect('user/workouts.html')
return render(request, 'user/workouts.html',
{'form': form})
Figured it out. Silly mistake on my part. Basically had a view written for the page that was rendering while I styled it. Completely forgot it was being called and the function for the workout_form wasn't even being used. That's also why the pdb.set_trace() wasn't running either. All is well now. Thanks to everyone who assisted.

Django filter ModelFormSet field choices... different from limiting the Formset's queryset

I understand that it is possible to override the default queryset 'used' by the modelformset. This just limits the objects for which a form is created.
I also found a Stack Overflow question about filtering ForeignKey choices in a Django ModelForm, but not a ModelForm Set and about limiting available choices in a Django formset, but not a Model FormSet. I have included my version of this code below.
What I want to do is render a ModelFormSet, for a school class ('teachinggroup' or 'theclass' to avoid clashing with the 'class' keyword) with one field limited by a queryset. This is for a teacher's class-editing form, to be able to reassign pupils to a different class, but limited to classes within the same cohort.
My models.py
class YearGroup(models.Model):
intake_year = models.IntegerField(unique=True)
year_group = models.IntegerField(unique=True, default=7)
def __unicode__(self):
return u'%s (%s intake)' % (self.year_group, self.intake_year)
class Meta:
ordering = ['year_group']
class TeachingGroup(models.Model):
year = models.ForeignKey(YearGroup)
teachers = models.ManyToManyField(Teacher)
name = models.CharField(max_length=10)
targetlevel = models.IntegerField()
def __unicode__(self):
return u'Y%s %s' % (self.year.year_group, self.name)
class Meta:
ordering = ['year', 'name']
My views.py
def edit_pupils(request, teachinggroup):
theclass = TeachingGroup.objects.get(name__iexact = teachinggroup)
pupils = theclass.pupil_set.all()
PupilModelFormSet = modelformset_factory(Pupil)
classes_by_year = theclass.year.teachinggroup_set.all()
choices = [t for t in classes_by_year]
# choices = [t.name for t in classes_by_year] #### I also tried this
if request.method == 'POST':
formset = PupilModelFormSet(request.POST,queryset=pupils)
if formset.is_valid():
formset.save()
return redirect(display_class_list, teachinggroup = teachinggroup)
else:
formset = PupilModelFormSet(queryset=pupils)
for form in formset:
for field in form:
if 'Teaching group' == field.label:
field.choices = choices
return render_to_response('reassign_pupils.html', locals())
As you can see, I am limiting the choices to the queryset classes_by_year, which is only classes which belong to the same year group. This queryset comes out correctly, as you can see in the rendered page below, but it doesn't affect the form field at all.
My template
{% for form in formset %}
<tr>
{% for field in form.visible_fields %}
<td> {# Include the hidden fields in the form #}
{% if forloop.first %}
{% for hidden in form.hidden_fields %}
{{ hidden }}
{% endfor %}
{% endif %}
<p><span class="bigtable">{{ field }}</span>
{% if field.errors %}
<p><div class="alert-message error">
{{field.errors|striptags}}</p>
</div>
{% endif %}
</td>
{% endfor %}
</tr>
{% endfor %}
</table>
<input type="submit" value="Submit changes"></p>
</form>
{{ choices }} <!-- included for debugging -->
The page renders with all teaching groups (classes) visible in the select widget, but the tag at the bottom of the page renders as: [<TeachingGroup: Y8 82Ma2>, <TeachingGroup: Y8 82Ma3>], accurately showing only the two classes in Year 8.
Note that I've also read through James Bennett's post So you want a dynamic form as recommended by How can I limit the available choices for a foreign key field in a django modelformset?, but that involves modifying the __init__ method in forms.py, and yet the only way I know how to create a ModelFormSet is with modelformset_factory, which doesn't involve defining any classes in forms.py.
Further to help from Luke Sneeringer, here is my new forms.py entry. After reading Why do I get an object is not iterable error? I realised that some of my problems came from giving a tuple to the field.choices method, when it was expecting a dictionary. I used the .queryset approach instead, and it works fine:
class PupilForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
super(PupilForm, self).__init__(*args, **kwargs)
thepupil = self.instance
classes_by_year = thepupil.teaching_group.year.teachinggroup_set.all()
self.fields['teaching_group'].queryset = classes_by_year
class Meta:
model = Pupil
As best as I can tell, you've actually put all the pieces together except one. Here's the final link.
You said you read the dynamic form post, which involves overriding the __init__ method in a forms.Form subclass, which you don't have. But, nothing stops you from having one, and that's where you can override your choices.
Even though modelformset_factory doesn't require an explicit Form class (it constructs one from the model if none is provided), it can take one. Use the form keyword argument:
PupilModelFormset = modelformset_factory(Pupil, form=PupilForm)
Obviously, this requires defining the PupilForm class. I get the impression you already know how to do this, but it should be something like:
from django import forms
class PupilForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
super(PupilForm, self).__init__(*args, **kwargs)
self.fields['teaching_group'].choices = ______ # code to generate choices here
class Meta:
model = Pupil
The last problem you might have is that a modelformset_factory just takes the class, which means that the constructor will be called with no arguments. If you need to send an argument dynamically, the way to do it is to make a metaclass that generates the form class itself, and call that metaclass in your modelformset_factory call.
You can accomplish this by setting field choices of a form in a formset is in the forms init and overwriting the self.fields['field_name'].choices. This worked fine for me but I needed more logic in my view after the formset was initialized. Here is what works for me in Django 1.6.5:
from django.forms.models import modelformset_factory
user_choices = [(1, 'something'), (2, 'something_else')] # some basic choices
PurchaserChoiceFormSet = modelformset_factory(PurchaserChoice, form=PurchaserChoiceForm, extra=5, max_num=5)
my_formset = PurchaserChoiceFormSet(self.request.POST or None, queryset=worksheet_choices)
# and now for the magical for loop and override each desired fields choices
for choice_form in my_formset:
choice_form.fields['model'].choices = user_choices
I wasn't able to find the answer for this but tried it out and it works in Django 1.6.5. I figured it out since formsets and for loops seem to go so well together :)

Django inlineformsetfactory - What is it good for?

Sorry for a newbie question but...
Can someone shed some light on what is the use case for inlineformset_factory?
I have followed example from Django documentation:
#Models
class Author(models.Model):
name = models.CharField(max_length=100)
class Book(models.Model):
author = models.ForeignKey(Author)
title = models.CharField(max_length=100)
#View
def jojo(request):
BookFormSet = inlineformset_factory(Author, Book)
author = Author.objects.get(name=u'Mike Royko')
formset = BookFormSet(instance=author)
return render_to_response('jojo.html', {
'formset': formset,
})
#jojo.html
<form action="" method="POST">
<table>
{{ formset }}
</table>
<input type="submit" value="Submit" />
</form>
But it only displays book fields.
My understanding was that formset would display Book form with inline Author form just
like Django Admin. On top of that I can't easily pass initial values to formset?
Then how is it better then using two separate AuthorForm and BookForm?
Or am i missing something obvious?
inlineformset_factory only provides multiple forms for the nested elements, you need a separate form at the top if you want a form for the main model.
Here is an example of a working inlineformset_factory with the main form embedded at the top:
views.py
from django.shortcuts import get_object_or_404, render_to_response
from django.forms.models import inlineformset_factory
from django.http import HttpResponseRedirect
from django.template import RequestContext
from App_name.models import * #E.g. Main, Nested, MainForm, etc.
. . .
#login_required
def Some_view(request, main_id=None, redirect_notice=None):
#login stuff . . .
c = {}
c.update(csrf(request))
c.update({'redirect_notice':redirect_notice})#Redirect notice is an optional argument I use to send user certain notifications, unrelated to this inlineformset_factory example, but useful.
#Intialization --- The start of the view specific functions
NestedFormset = inlineformset_factory(Main, Nested, can_delete=False, )
main = None
if main_id :
main = Main.objects.get(id=main_id)#get_object_or_404 is also an option
# Save new/edited Forms
if request.method == 'POST':
main_form = MainForm(request.POST, instance=main, prefix='mains')
formset = NestedFormset(request.POST, request.FILES, instance=main, prefix='nesteds')
if main_form.is_valid() and formset.is_valid():
r = main_form.save(commit=False)
#do stuff, e.g. setting any values excluded in the MainForm
formset.save()
r.save()
return HttpResponseRedirect('/Home_url/')
else:
main_form = MainForm(instance=main, prefix='mains') #initial can be used in the MainForm here like normal.
formset = NestedFormset(instance=main, prefix='nesteds')
c.update({'main_form':main_form, 'formset':formset, 'realm':realm, 'main_id':main_id})
return render_to_response('App_name/Main_nesteds.html', c, context_instance=RequestContext(request))
template.html
{% if main_form %}
<form action="." method="POST">{% csrf_token %}
{{ formset.management_form }}
<table>
{{main_form.as_table}}
{% for form in formset.forms %}
<table>{{ form }}</table>
{% endfor %}
</table>
<p><input type="submit" name="submit" value="Submit" class="button"></p>
</form>
{% endif %}
The beauty of the inlineformset_factory (and modelformset_factory) is the ability to create multiple model instances from a single form. If you were to simply 'use two separate forms' the id's of the form's fields would trample each other.
The formset_factory functions know how many extra form(sets) you need (via the extra argument) and sets the id's of the fields accordingly.
inlineformset_factory creates a list of forms.
This can be used when the same form needs to be repeated at the page, for example:
Upload multiple photo's with a description.
Invite multiple members by email
Fill a calendar grid per hour
Fill a list of books for the author.
With some JavaScript code, you can add a "add another row" functionality as well.