Related
I've two models:
First one:
class A(models.Model):
a_user = models.ForeignKey(User, unique=False, on_delete=models.CASCADE)
a_title = models.CharField("A title", max_length=500)
Second one:
class B(models.Model):
b_a = models.ForeignKey(A, verbose_name=('A'), unique=False, on_delete=models.CASCADE)
b_details = models.TextField()
Now, I'm using CreateView to create form for Value filling :
class B_Create(CreateView):
model = B
fields = ['b_a','b_details']
Then using this to render these field in templates.
Now, my problem is, while giving the field b_a ( which is the dropdown ), it list downs all the values of model A, but the need is to list only the values of model A which belongs to the particular logged in user, in the dropdown.
I've seen all the answers, but still not able to solve the problem.
The things I've tried:
limit_choices_to in models : Not able to pass the value of A in the limit_choices
form_valid : Don't have the model A in the CreateView, as only B is reffered model in B_Create
passing primary key of A in templates via url : Then there is no instance of A in the template so can't access. Also, don't want to handle it in templates.
I'm new to Django and still learning, so don't know to override admin form.
Please suggest the implemented way, if possible to the problem. I've researched and tried most of the similar questions with no result for my particular problem. I feel like, this is a dumb question to ask, but I'm stuck here, so need help.
Thanks..
(Please feel free to suggest corrections.)
You have access to self.request.user in the form_valid of the view. But in order to limit the choices in the form you have to customize the form before it is served initially. You best override the view's get_form and set the form field's queryset:
class B_Create(CreateView):
model = B
fields = ['b_a','b_details']
def get_form(self, *args, **kwargs):
form = super(B_Create, self).get_form(*args, **kwargs)
form.fields['b_a'].queryset = self.request.user.a_set.all()
# form.fields['b_a'].queryset = A.objects.filter(a_user=self.request.user)
return form
Generally, there are three places where you can influence the choices of a ModelChoiceField:
If the choices need no runtime knowledge of your data, user, or form instance, and are the same in every context where a modelform might be used, you can set limit_choices_to on the ForeignKey field itself; as module level code, this is evaluated once at module import time. The according query will be built and executed every time a form is rendered.
If the choices need no runtime knowledge, but might be different in different forms, you can use custom ModelForms and set the queryset in the field definition of the respective form field.
If the queryset needs any runtime information, you can either override the __init__ of a custom form and pass it any information it needs to set the field's queryset or you just modify the queryset on the form after it is created which often is a quicker fix and django's default views provide nice hooks to do that (see the code above).
The #schwobaseggl answer is excellent.
Here is a Python 3 version. I needed to limit the projects dropdown input based on the logged-in user.
class ProductCreateView(LoginRequiredMixin, CreateView):
model = Product
template_name = 'brand/product-create.html'
fields = '__all__'
def get_form(self, form_class=None):
form = super().get_form(form_class=None)
form.fields['project'].queryset = form.fields['project'].queryset.filter(owner_id=self.request.user.id)
return form
I am trying to practice a simple project: A user registers (using Django registration-redux), uploads some file, and then s/he is provided with a list of her files, being downloadable. Here are my models.py, forms.py, and views respectively:
models.py
class UserProfile(models.Model):
user = models.ForeignKey(User, related_name='uploaded_by')
names = models.CharField(max_length=40)
lastname = models.CharField(max_length=50)
email = models.EmailField()
uploads = models.FileField(upload_to= 'blablabla')
def __str__(self):
return self.email
forms.py
class UserProfileForm(forms.ModelForm):
class Meta:
model = UserProfile
fields = ['names', 'uploads']
view.py
from .forms import UserProfileForm
from .models import UserProfile
#login_required()
def new(request):
form = UserProfileForm(request.POST or None, request.FILES or None)
if form.is_valid():
form.save()
context = {'title': 'welcome', 'form': form}
return render(requst, 'upform.html', context)
however when I login with a user and try to upload a file I get the error: IntegrityError at /new
NOT NULL constraint failed: _userprofile.user_id
After digging a lot I noticed someone suggested the reason of the error is because the user is not included anyhow in the process of posting the form, so I tried whatever came to my mind and the case in which I added the user field to the forms.py worked:
forms.py
class UserProfileForm(forms.ModelForm):
class Meta:
model = UserProfile
fields = ['names', 'uploads']
the problem however is that the form shown in the browser now includes a drop-down list containing all the registered users. I tried to associate the logged-in user with the form in the views by I kept seeing different errors.
My question is: How can I associate the uploads with the logged-in user in a transparent manner.
sorry if the question is too newbie-liked
Keep the user out of the form and add it on save:
if form.is_valid():
profile = form.save(commit=False)
profile.user = request.user
profile.save()
I must say your model looks a bit odd; you have multiple profiles for each user, each with a single upload. Seems more likely you want a single profile, with a OneToOne relationship to User, than a separate Uploads model with a ForeignKey to UserProfile.
I added the user field to the forms.py worked:
This probably also opens up a security hole, because you could then set the user from outside of your application, overriding the logged-in user.
After digging a lot I noticed someone suggested the reason for the error is because the user is not included anyhow in the process of posting the form.
You figured that quite right. If the user is not a field, the form never knows how to fill the user field of UserProfile. Since you cannot save a UserProfile without a filled in user foreign key, you get the error.
There are various ways around this:
One way to solve the problem is to use save(commit=False) on the form, patch the user into the unsaved instance and manually saving the object to the database:
if form.is_valid():
profile = form.save(commit=False)
profile.user = request.user
profile.save()
This slightly violates the encapsulation because you now handle the database save outside of the form.
You are also able to provide a initial "template" instance to the form:
form = UserProfileForm(
request.POST,
request.FILES,
instance=UserProfile(user=self.request.user)
)
You probably want to do so, anyway because the form also allows to edit an existing userprofile. Currently you are saving a new UserProfile each time and since user is not unique=True in your model class, you will get multiple profiles per user.
If you do not want this to be possible, check Daniel Roseman's answer, because then you probably want UserProfile.user to be a OneToOne field.
In this case, we can simplify to
profile, created = UserProfile.objects.get_or_create(user=request.user)
form = UserProfileForm(
request.POST,
request.FILES,
instance=profile
)
Note also that I removed the or None as it is not necessary. BaseForm (which ModelForm derives from) does this check for you (it actually does self.data = data or {}) which essentially undoes the prior none-ification)
You can do following the advice of Daniel.
class UserProfile(models.Model):
user = models.OneToOneField(User, related_name='uploaded_by')
names = models.CharField(max_length=40)
lastname = models.CharField(max_length=50)
email = models.EmailField()
class UserFiles(models.Model):
user = models.ForeignKey(UserProfile)
file = models.FileField(upload_to= 'blablabla')
Remember OneToOneField is similar a ForeignKey with the attribute unique=True.
I am trying to create custom fields for users to enter on signup with django-allauth. I have referred to several posts about this, but I am not able to get my custom form to save to my database. I do get a combined form on my signup.html page with username, password1 and 2, email and my extra fields of city and school, but I am not able to save the extra fields to the database. I have run syncdb and can see my User Profile table in the admin area.
This advice is the closest I have come to the answer but I do not understand how to implement it: "You can't use UserProfileForm to the allauth.SIGNUP_FORM_CLASS. You need to extend it from SignUpForm and write a save method which will accept the newly created user as the only parameter," from this post:
Custom registration form for use with django-allauth
I have also attempted to integrate advice on this form these posts:
Django Allauth not saving custom form
How to customize user profile when using django-allauth
This is my code:
Models.py
from django.db import models
from django.contrib.auth.models import User
class UserProfile(models.Model):
# A required line - links a UserProfile to User.
user = models.OneToOneField(User)
# The additional attributes we wish to include.
school = models.CharField(max_length=128)
city = models.CharField(max_length=128)
def __unicode__(self):
return self.user.username
Forms.py
from django import forms
from django.contrib.auth.models import User
from myapp.models import UserProfile
from django.forms.widgets import HiddenInput
class UserProfileForm(forms.ModelForm):
class Meta:
model = UserProfile
fields = ('city', 'school')
def signup(self, request, user):
user=User.objects.get(email=request.email)
city=request.POST.get('city','')
school=request.POST.get('school','')
userprofile_obj = UserProfile(user=user,city=city,school=school)
userprofile_obj.save()
Settings.py
ACCOUNT_SIGNUP_FORM_CLASS = 'myapp.forms.UserProfileForm'
My template is the basic Signup.html from the django-allauth templates and I do not have a view made for this, although I attempted to make one from the tangowithdjango user authentication section register view, and this gave similar behavior (not saving to the database).
Thanks,
Kelly
Not sure if this is still an active question/issue for the original poster: if so, and for anyone else who comes across this, a few things to correct to at least move in the right direction:
I don't see an __init__() method that calls the superclass? E.g.:
def __init__(self, *args, **kwargs):
super(SignupForm, self).__init__(*args, **kwargs)
use the user parameter to the signup method. It should be populated; don't reload it.
Ensure the two objects are linking correctly (I didn't use Django to build my profile table so YMMV but I set user.profile = Profile(...); then execute user.profile.save() at the end of my signup() method.
get the values to place into the profile from the form cleaned_data (e.g. self.cleaned_data['city'] not the POST.
Then start debugging: is your signup() method firing? What does it get? What happens when you execute the profile.save() method?
I have a model registered on the admin site. One of its fields is a long string expression. I'd like to add custom form fields to the add/update pages of this model in the admin. Based on the values of these fields I will build the long string expression and save it in the relevant model field.
How can I do this?
I'm building a mathematical or string expression from symbols. The user chooses symbols (these are the custom fields that are not part of the model) and when they click save then I create a string expression representation from the list of symbols and store it in the DB. I don't want the symbols to be part of the model and DB, only the final expression.
Either in your admin.py or in a separate forms.py you can add a ModelForm class and then declare your extra fields inside that as you normally would. I've also given an example of how you might use these values in form.save():
from django import forms
from yourapp.models import YourModel
class YourModelForm(forms.ModelForm):
extra_field = forms.CharField()
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# ...do something with extra_field here...
return super(YourModelForm, self).save(commit=commit)
class Meta:
model = YourModel
To have the extra fields appearing in the admin just:
Edit your admin.py and set the form property to refer to the form you created above.
Include your new fields in your fields or fieldsets declaration.
Like this:
class YourModelAdmin(admin.ModelAdmin):
form = YourModelForm
fieldsets = (
(None, {
'fields': ('name', 'description', 'extra_field',),
}),
)
UPDATE:
In Django 1.8 you need to add fields = '__all__' to the metaclass of YourModelForm.
It it possible to do in the admin, but there is not a very straightforward way to it. Also, I would like to advice to keep most business logic in your models, so you won't be dependent on the Django Admin.
Maybe it would be easier (and maybe even better) if you have the two seperate fields on your model. Then add a method on your model that combines them.
For example:
class MyModel(models.model):
field1 = models.CharField(max_length=10)
field2 = models.CharField(max_length=10)
def combined_fields(self):
return '{} {}'.format(self.field1, self.field2)
Then in the admin you can add the combined_fields() as a readonly field:
class MyModelAdmin(models.ModelAdmin):
list_display = ('field1', 'field2', 'combined_fields')
readonly_fields = ('combined_fields',)
def combined_fields(self, obj):
return obj.combined_fields()
If you want to store the combined_fields in the database you could also save it when you save the model:
def save(self, *args, **kwargs):
self.field3 = self.combined_fields()
super(MyModel, self).save(*args, **kwargs)
Django 2.1.1
The primary answer got me halfway to answering my question. It did not help me save the result to a field in my actual model. In my case I wanted a textfield that a user could enter data into, then when a save occurred the data would be processed and the result put into a field in the model and saved. While the original answer showed how to get the value from the extra field, it did not show how to save it back to the model at least in Django 2.1.1
This takes the value from an unbound custom field, processes, and saves it into my real description field:
class WidgetForm(forms.ModelForm):
extra_field = forms.CharField(required=False)
def processData(self, input):
# example of error handling
if False:
raise forms.ValidationError('Processing failed!')
return input + " has been processed"
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# self.description = "my result" note that this does not work
# Get the form instance so I can write to its fields
instance = super(WidgetForm, self).save(commit=commit)
# this writes the processed data to the description field
instance.description = self.processData(extra_field)
if commit:
instance.save()
return instance
class Meta:
model = Widget
fields = "__all__"
You can always create new admin template, and do what you need in your admin_view (override the admin add URL to your admin_view):
url(r'^admin/mymodel/mymodel/add/$','admin_views.add_my_special_model')
If you absolutely only want to store the combined field on the model and not the two seperate fields, you could do something like this:
Create a custom form using the form attribute on your ModelAdmin. ModelAdmin.form
Parse the custom fields in the save_formset method on your ModelAdmin. ModelAdmin.save_model(request, obj, form, change)
I never done something like this so I'm not completely sure how it will work out.
The first (highest score) solution (https://stackoverflow.com/a/23337009/10843740) was accurate, but I have more.
If you declare fields by code, that solution works perfectly, but what if you want to build those dynamically?
In this case, creating fields in the __init__ function for the ModelForm won't work. You will need to pass a custom metaclass and override the declared_fields in the __new__ function!
Here is a sample:
class YourCustomMetaClass(forms.models.ModelFormMetaclass):
"""
For dynamically creating fields in ModelForm to be shown on the admin panel,
you must override the `declared_fields` property of the metaclass.
"""
def __new__(mcs, name, bases, attrs):
new_class = super(NamedTimingMetaClass, mcs).__new__(
mcs, name, bases, attrs)
# Adding fields dynamically.
new_class.declared_fields.update(...)
return new_class
# don't forget to pass the metaclass
class YourModelForm(forms.ModelForm, metaclass=YourCustomMetaClass):
"""
`metaclass=YourCustomMetaClass` is where the magic happens!
"""
# delcare static fields here
class Meta:
model = YourModel
fields = '__all__'
This is what I did to add the custom form field "extra_field" which is not the part of the model "MyModel" as shown below:
# "admin.py"
from django.contrib import admin
from django import forms
from .models import MyModel
class MyModelForm(forms.ModelForm):
extra_field = forms.CharField()
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# Do something with extra_field here
return super().save(commit=commit)
#admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
form = MyModelForm
You might get help from my answer at :
my response previous on multicheckchoice custom field
You can also extend multiple forms having different custom fields and then assigning them to your inlines class like stackedinline or tabularinline:
form =
This way you can avoid formset complication where you need to add multiple custom fields from multiple models.
so your modeladmin looks like:
inlines = [form1inline, form2inline,...]
In my previous response to the link here, you will find init and save methods.
init will load when you view the page and save will send it to database.
in these two methods you can do your logic to add strings and then save thereafter view it back in Django admin change_form or change_list depending where you want.
list_display will show your fields on change_list.
Let me know if it helps ...
....
class CohortDetailInline3(admin.StackedInline):
model = CohortDetails
form = DisabilityTypesForm
...
class CohortDetailInline2(admin.StackedInline):
model = CohortDetails
form = StudentRPLForm
...
...
#admin.register(Cohort)
class CohortAdmin(admin.ModelAdmin):
form = CityInlineForm
inlines = [uploadInline, cohortDetailInline1,
CohortDetailInline2, CohortDetailInline3]
list_select_related = True
list_display = ['rto_student_code', 'first_name', 'family_name',]
...
I'm struggling to get my head round django forms.. I've been reading various documentation but just can't quite grasp the concepts. I have got to grips with models, views and templates. What I am trying to do is to create a form with various fields composing of dropdown lists and checkboxes which are populated by values in a database.
I have a working app called vms. Using the models.py I have a built a simple schema that holds size and type. Size consists of 'small', 'medium' & 'large'. Type is 'windows' & 'linux'. Using the admin site, I can add an extra size, for example 'Extra Large'.
What I would like to do is create a form that has a drop down list of the vm sizes. If an extra size gets added via the admin site, I would like that size to appear in the drop down list.
I would submit my attempts at the code, but actually am struggling with the concepts. Can anyone help guide me in how to accomplish the above?
Thanks
Oli
Forms are just a tool to simplify and speed-up (the development of) the process of fetching POST data from the request. A manual way would be to do request.POST.get('somefield') for all the fields there are in some HTML form. But Django can do better than that...
In its essence, a Form class holds a number of Fields and performs these tasks:
display HTML inputs,
collect and validate data when user submits it,
if fields don't validate, return the values along with error messages to HTML,
if all fields validate, provide form.cleaned_data dictionary as a convenient way to access these values in view.
With these values, I could then manually create a new instance of a MyModel and save it. Of course, I would have to define a Field in the Form for every Field in MyModel model.
This means that, basically, I could do something like this:
(forgive me for not testing this code, so I can't vouch that it's 100% correct)
models.py:
class MyModel(models.Model):
field1 = models.CharField(max_length=40, blank=False, null=False)
field2 = models.CharField(max_length=60, blank=True, null=True)
forms.py:
class FormForMyModel(forms.Form):
form_field1 = forms.CharField(max_length=40, required=True)
form_field2 = forms.CharField(max_length=60, required=False)
views.py:
def create_a_my_model(request):
if request.method == 'POST':
form = FormForMyModel(request.POST)
if form.is_valid():
my_model = MyModel()
my_model.field1 = form.cleaned_data.get('form_field1', 'default1')
my_model.field2 = form.cleaned_data.get('form_field2', 'default2')
my_model.save()
else:
form = FormForMyModel()
context_data = {'form': form}
return HttpResponse('templtate.html', context_data)
(this could be written with a few lines of code less, but it's meant to be as clear as possible)
Notice there are no relation between model Fields and form Fields! We have to manually assign values to MyModel instance when creating it.
The above example outlines generic form workflow. It is often needed in complex situations, but not in such a simple one as is this example.
For this example (and a LOT of real-world examples), Django can do better than that...
You can notice two annoying issues in the above example:
I have to define Fields on MyModel and Fields on FormForMyModel separately. However, there is a lot of similarity between those two groups (types) of Fields, so that's kind of duplicate work. The similarity grows when adding labels, validators, etc.
creating of MyModel instance is a bit silly, having to assign all those values manually.
This is where a ModelForm comes in.
These act basically just like a regular form (actually, they are extended from regular forms), but they can save me some of the work (the two issues I just outlined, of course :) ).
So back to the two issues:
Instead of defining a form Field for each model Field, I simply define model = MyModel in the the Meta class. This instructs the Form to automatically generate form Fields from model Fields.
Model forms have save method available. This can be used to create instance of model in one line in the view, instead of manually assigning field-by-field.
So, lets make the example above with a ModelForm:
models.py:
class MyModel(models.Model):
field1 = models.CharField(max_length=40, blank=False, null=False)
field2 = models.CharField(max_length=60, blank=True, null=True)
forms.py:
class MyModelForm(forms.ModelForm): # extending ModelForm, not Form as before
class Meta:
model = MyModel
views.py:
def create_a_my_model(request):
if request.method == 'POST':
form = MyModelForm(request.POST)
if form.is_valid():
# save the model to database, directly from the form:
my_model = form.save() # reference to my_model is often not needed at all, a simple form.save() is ok
# alternatively:
# my_model = form.save(commit=False) # create model, but don't save to database
# my.model.something = whatever # if I need to do something before saving it
# my.model.save()
else:
form = MyModelForm()
context_data = {'form': form}
return HttpResponse('templtate.html', context_data)
Hope this clears up the usage of Django forms a bit.
Just one more note - it is perfectly ok to define form Fields on a ModelForm. These will not be used in form.save() but can still be access with form.cleaned_data just as in a regular Form.
Have you tried working with ModelForms before? As I understand, you're looking to create a form based on the model you created right?
Lets say your model is called Temp. You can create a form that correlates with this model (and your question) like this:
forms.py
from django.forms import ModelForm
class TempForm(ModelForm):
class Meta:
model = Temp
The ModelForm will automatically map the selections/choices from your model to a form version.
If you plan on using this in a template later, doing something like this will automatically create a drop-down menu with choices:
<form>
<label for="id_size">Size</label>
{{ form.size }}
</form>
Hope that answers your question!
Simply use CharField in your modelform as below:
SIZES_CHOICES = (
('size1', 'M'),
('size2', 'L'),
)
size = models.CharField(max_length=100, choices=SIZES_CHOICES, default=size1)
in the above code, size1 is the value which will be going to store in your database as name 'size1' and in the drop-down menu, there will be an option is 'M' of right side.you can mentioned any name to these options.