Django - Accessing relational database in view - django

This is my forms.py:
class RegistrationForm(forms.Form):
PET_CHOICES = (
('none','---'), ('dog', 'Dog'),
)
username = forms.CharField()
email = forms.EmailField()
password1 = forms.CharField()
birthdate = forms.DateField(widget=extras.SelectDateWidget)
pet = forms.ChoiceField(choices=PET_CHOICES)
This is my models.py:
from django.db import models
from django.contrib.auth.models import User
# Create your models here.
class UserExtended(models.Model):
user = models.ForeignKey(User)
pet = models.CharField(max_length=40)
age = models.PositiveSmallIntegerField()
As you can see, this model extends the default Django User model. Now, this is my view which registers users:
def registerView(request):
form = RegistrationForm()
if request.method == 'POST':
form = RegistrationForm(request.POST)
if form.is_valid():
user = User.objects.create_user(
username=form.cleaned_data['username'],
password=form.cleaned_data['password1'],
email=form.cleaned_data['email']
)
user.userextended_set.pet=form.cleaned_data['pet']
user.userextended_set.age=form.cleaned_data['birthdate']
return HttpResponseRedirect('/')
I'm wondering, is this part of my code correct?
user.userextended_set.pet=form.cleaned_data['pet']
user.userextended_set.age=form.cleaned_data['birthdate']
With the two lines above, I am trying to add to the UserExtended model which has a ForiegnKey with User. I think I am doing something wrong because after the user is created and I authenticate and log the user in using the functions
authenticate()
and
login()
provided by django, and if I place this in my template which Django redirects to after logging in
<p>your pet is {{ user.userextended_set.pet }}</p>
it does not display another for
{{ user.userextended_set.pet }}
. It just says:
your pet is
. Any idea why it does not display anything? {{ user.userextended_set.pet }}

userextended_set is actually a manager, not a UserExtended object, because you have a ForeignKey in UserExtended, which indicates that a User could have multiple UserExtended objects. This probably isn't what you want — you probably want to use a OneToOneField instead.

The problem here is,
user.userextended_set returns a RelatedManager type. So, you cannot do a simple .pet would not work.
Now, looking at your models, i am assuming you intend UserExtended to a OneToOneField, rather than a ForeignKey. As long as you can manage the (non) creation of duplicates, you are just fine.
Now, coming back to your issue:
You can do
user_extended = user.userextended_set.all()[0]
#Or if you are sure it is a unique relationship,
#user_extended = user.userextended_set.get()
#Note - this would raise error if MultipleObjectsReturned, or DoesNotExist
user_extended.pet = form.cleaned_data['pet']
user_extended.age = form.cleaned_data['age']
Also, in the template,
{{ user.userextended_set.0.pet }}
Or even this:
{% for u_e in user.userextended_set.all %}
{{u_e.pet}}
{% empty %}
Not found
{% endfor %}

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

How do I catch IntegrityError in Django when using UniqueConstraint across multiple fields?

I have a model with user as 1 field (Foreign Key) and one other field skill_group. I need to make sure the user does not add duplicate skill groups so I added a UniqueConstraint. This is working as the system errors out with IntegrityError at /skillgroup/create/
duplicate key value violates unique constraint "unique_skillgroup" - How do I catch this exception and notify user if duplicate; otherwise save it?
New to Django/Python/Postgres and I thought I could handle it by overriding the save() function, but there is no access to user which is part of the check and I have read this should not be handled here. Is there a try/save catch/message I should be employing? I have tried a few things with no luck. I have seen similar questions on here, but they have not helped. Any help is appreciated.
models.py
class SkillGroup(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
skill_group = models.CharField(max_length=35)
sequence = models.IntegerField(default=999)
class Meta:
constraints = [
models.UniqueConstraint(fields=['user', 'skill_group'], name='unique_skillgroup'),
]
def __str__(self):
return self.skill_group
def get_absolute_url(self):
return reverse('skillgroup-list')
views.py
class SkillGroupCreateView(LoginRequiredMixin, CreateView):
model = SkillGroup
fields = ['skill_group']
def form_valid(self, form):
form.instance.user = self.request.user
form.instance.sequence = SkillGroup.objects.filter(user=self.request.user).order_by('sequence').last().sequence + 1
return super().form_valid(form)
skillgroup_form.html
{% extends "recruiter/baseskills.html" %}
{% load crispy_forms_tags %}
{% block content%}
<div class="content-section">
<form method="post">
{% csrf_token %}
<fieldset class="form-group">
<legend class="border-bottom mb-4">Skill Group</legend>
{{ form|crispy }}
</fieldset>
<div class="form-group">
<button class="btn btn-outline-info" type="submit">Add Skill Group</button>
</div>
</form>
</div>
{% endblock content%}
I want to either catch the exception and save the record if not a duplicate or put message on screen saying "Skill Group already exists" and leave user on create page. Also, I could remove the UniqueConstraint and handle with code if that is the best solution.
You are inadvertently bypassing Django's form validation here and then trying to save invalid input to the database, which is why Django is feeding back an ugly IntegrityError from the database instead of handling the error gracefully.
If you submit a duplicate User and SkillGroup in your form, your CreateView will helpfully return the error message back into your form template:
"Skill group with this User and Skill group already exists."
But it can only do this if you include a User field in your form. I assume you have excluded User to keep the form template tidy, but that prevents Django's form validation from checking if the combination already exists.
To get around this, add User to your form field as a hidden input. I don't think that's possible using CreateView's behind-the-scenes magic, so you'll need to create a SkillGroupForm to handle that.
# forms.py
from django import forms
from .models import SkillGroup
class SkillGroupForm(forms.ModelForm):
class Meta:
model = SkillGroup
fields = ('user', 'skill_group')
widgets = {
'user': forms.HiddenInput,
}
# views.py
from .forms import SkillGroupForm
class SkillGroupCreateView(LoginRequiredMixin, CreateView):
model = SkillGroup
form_class = SkillGroupForm
def get_initial(self):
return {'user': self.request.user}
def form_valid(self, form):
form.instance.sequence = SkillGroup.objects.filter(user=self.request.user).order_by('sequence').last().sequence + 1
return super().form_valid(form)
The get_initial method passes the request.user as an initial value into the hidden form field, so no user input is needed.
Try to validate skill_group field inside your form class. Define clean_skill_group method like in docs.There you can get queryset of all SkillGroup objects related to your User (like here) and then compare skills. But you need to push somehow your User object or user_id (to get then User object) to the form before form.is_valid() will be called (or call one more time form.is_valid() after). Then show form errors in your html template.
class YourForm(forms.ModelForm):
....some fields, Meta....
def clean_skill_group(self):
your_user_object = ....
previously_created_skills = your_user_object.skill_group_set.all()
skill_input = self.cleaned_data["skill_group"]
if skill_input in previously_created_skills:
raise forms.ValidationError(("This skill group is already exist"), code="invalid") # I suppose you are using model form

Create a User and ExtendedUser object in one form using an Inline formset?

I have a model Sachbearbeiter which extends my User model.
from django.contrib.auth.models import User
class Sachbearbeiter(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
verein = models.ForeignKey(Verein)
Now I need CreateView form to create a new Sachbearbeiter instance that is immediately linked to a new User instance.
The form should have the following fields:
First Name: [textfield]
Last Name: [textfield]
Email Address: [textfield]
Verein: [Dropdown that lists `Vereine`]
What I tried is this:
class SachbearbeiterForm(ModelForm):
class Meta:
model = Sachbearbeiter
fields = '__all__' # Go into detail once it's working
class UserForm(ModelForm):
class Meta:
model = User
fields = '__all__' # Go into detail once it's working
SachbearbeiterFormSet = inlineformset_factory(User, Sachbearbeiter, fields='__all__')
class SachbearbeiterCreate(CreateView):
form_class = SachbearbeiterFormSet
But this gives me a form that looks like this:
Verein: [Dropdown that lists `Vereine`]
Delete? [checkbox]
Is this even possible, or do I have to create a custom form for that?
This is not the solution I was looking for, but works well. Because of that, I won't mark it as an answer, though.
I had to create a custom view and was able to simply use both forms inside it. (In the code example I omitted the unimportant stuff.
In the ModelForms I only changed the fields attribute to have only the fields I want to show.
def my_view(request):
...
if request.method == 'POST':
if user_form.is_valid() and sachbearbeiter_form.is_valid():
user = user_form.save(commit=False)
user.username = user.email
user.save()
sachbearbeiter = sachbearbeiter_form.save(commit=False)
sachbearbeiter.user = user
sachbearbeiter.save()
...
return redirect('sachbearbeiter_create_or_edit', username=user.username)
The template looks something like that:
{% if user_form.errors or sachbearbeiter_form.errors %}
<div class="alert alert-danger" role="alert">
Please correct the errors indicated by red color.
</div>
{% endif %}
<form method="post" action="#">
{{ user_form.as_p }}
{{ sachbearbeiter_form.as_p }}
</form>

How to get value from onetoone relation using request user

im setting the request user in templates using this code, and this code work really good.
{% if user.is_authenticated %}
{{ user.name }}
{% else %}
xxxxxxx
{% endif %}
tables:
user(model.Models):
id
...
Profile(model.Models):
user onetoone(user)
name
last name
my user table have a onetoone relation with Profile and i want to know if i can get a field of Profile table using the request.user or other way to make it, thanks for all.
If I rightly understand you are using built-in user model. So we can write:
models.py
from django.contrib.auth.models import User
class Profile(models.Model):
***user*** = models.OneToOneField(User)
name = models.CharField(max_length=50)
User.*profile* = property(lambda u: Profile.objects.get_or_create(***user***=u)[0])
views.py
def profile(request):
user = request.user
profile = user.*profile*
after you can use profile fields like profile.name or other

Django and users entering data

I am building a webapp which will be used by a company to carry out their daily operations. Things like sending invoices, tracking accounts receivable, tracking inventory (and therefore products). I have several models set up in my various apps to handle the different parts of the web-app. I will also be setting up permissions so that managers can edit more fields than, say, an office assistant.
This brings me to my question. How can I show all fields of a model and have some that can be edited and some that cannot be edited, and still save the model instance?
For example, I have a systems model for tracking systems (we install irrigation systems). The system ID is the primary key, and it is important for the user to see. However, they cannot change that ID since it would mess things up. Now, I have a view for displaying my models via a form using the "form.as_table". This is efficient, but merely spits out all the model fields with input fields filling in the values stored for that model instance. This includes the systemID field which should not be editable.
Because I don't want the user to edit the systemID field, I tried making it just a label within the html form, but django complains. Here's some code:
my model (not all of it, but some of it):
class System(models.Model):
systemID = models.CharField(max_length=10, primary_key=True, verbose_name = 'System ID')
systemOwner = models.ForeignKey (System_Owner)
installDate = models.DateField()
projectManager = models.ForeignKey(Employee, blank=True, null=True)
#more fields....
Then, my view for a specific model instance:
def system_details(request, systemID):
if request.method == 'POST':
sysEdit = System.objects.get(pk=systemID)
form = System_Form(request.POST, instance=sysEdit)
if form.is_valid():
form.save()
return HttpResponseRedirect('/systems/')
else:
sysView = System.objects.get(pk=systemID)
form = System_Form(instance=sysView)
return render_to_response('pages/systems/system_details.html', {'form': form}, context_instance=RequestContext(request))
Now the html page which displays the form:
<form action="" method="POST">
{% csrf_token %}
<table>
{{ form.as_table }}
</table>
<input type="submit" value="Save Changes">
<input type="button" value="Cancel Changes" onclick="window.location.href='/systems/'">
</form>
So, what I am thinking of doing is having two functions for the html. One is a form for displaying only those fields the user can edit, and the other is for just displaying the content of the field (the systemID). Then, in the view, when I want to save the changes the user made, I would do:
sysValues = System.objects.get(pk=SystemID)
form.save(commit = false)
form.pk = sysValues.sysValues.pk (or whatever the code is to assign the sysValues.pk to form.pk)
Is there an easier way to do this or would this be the best?
Thanks
One thing you can do is exclude the field you don't need in your form:
class System_Form(forms.ModelForm):
class Meta:
exclude = ('systemID',)
The other is to use read-only fields: http://docs.djangoproject.com/en/1.3/ref/contrib/admin/#django.contrib.admin.ModelAdmin.readonly_fields as #DTing suggessted
To make a field read only you can set the widget readonly attribute to True.
using your example:
class System_Form(ModelForm):
def __init__(self, *args, **kwargs):
super(System_Form, self).__init__(*args, **kwargs)
self.fields['systemID'].widget.attrs['readonly'] = True
class Meta:
model = System
or exclude the fields using exclude or fields in the class Meta of your form and display it in your template if desired like so:
forms.py
class System_Form(ModelForms):
class Meta:
model = System
exclude = ('systemID',)
views.py
def some_view(request, system_id):
system = System.objects.get(pk=system_id)
if request.method == 'POST':
form = System_Form(request.POST, instance=system)
if form.is_valid():
form.save()
return HttpResponse('Success')
else:
form = System_Form(instance=system)
context = { 'system':system,
'form':form, }
return render_to_response('some_template.html', context,
context_instance=RequestContext(request))
some_template.html
<p>make changes for {{ system }} with ID {{ system.systemID }}</p>
<form method='post'>
{{ form.as_p }}
<input type='submit' value='Submit'>
</form>