How to show hidden autofield in django formset - python

A Django autofield when displayed using a formset is hidden by default. What would be the best way to show it?
At the moment, the model is declared as,
class MyModel:
locid = models.AutoField(primary_key=True)
...
When this is rendered using Django formsets,
class MyModelForm(ModelForm):
class Meta:
model = MyModel
fields = ('locid', 'name')
it shows up on the page as,
<input id="id_form-0-locid" type="hidden" value="707" name="form-0-locid"/>
Thanks.
Edit
I create the formset like this -
LocFormSet = modelformset_factory(MyModel)
pformset = LocFormSet(request.POST, request.FILES, queryset=MyModel.objects.order_by('name'))
Second Edit
Looks like I'm not using the custom form class I defined there, so the question needs slight modification..
How would I create a formset from a custom form (which will show a hidden field), as well as use a custom queryset?
At the moment, I can either inherit from a BaseModelFormSet class and use a custom query set, or I can use the ModelForm class to add a custom field to a form. Is there a way to do both with a formset?
Third Edit
I'm now using,
class MyModelForm(ModelForm):
def __init__(self, *args, **kwargs):
super(MyModelForm, self).__init__(*args, **kwargs)
locid = forms.IntegerField(min_value = 1, required=True)
self.fields['locid'].widget.attrs["type"] = 'visible'
self.queryset = MyModel.objects.order_by('name')
class Meta:
model = MyModel
fields = ('locid', 'name')
LocFormSet = modelformset_factory(MyModel, form = MyModelForm)
pformset = LocFormSet()
But this still doesn't
Show locid
Use the custom query that was specified.

Try changing the default field type:
from django import forms
class MyModelForm(ModelForm):
locid = forms.IntegerField(min_value=1, required=True)
class Meta:
model = MyModel
fields = ('locid', 'name')
EDIT: Tested and works...

As you say, you are not using the custom form you have defined. This is because you aren't passing it in anywhere, so Django can't know about it.
The solution is simple - just pass the custom form class into modelformset_factory:
LocFormSet = modelformset_factory(MyModel, form=MyModelForm)
Edit in response to update 3:
Firstly, you have the redefinition for locid in the wrong place - it needs to be at the class level, not inside the __init__.
Secondly, putting the queryset inside the form won't do anything at all - forms don't know about querysets. You should go back to what you were doing before, passing it in as a parameter when you instantiate the formset. (Alternatively, you could define a custom formset, but that seems like overkill.)
class MyModelForm(ModelForm):
locid = forms.IntegerField(min_value=1, required=True)
def __init__(self, *args, **kwargs):
super(MyModelForm, self).__init__(*args, **kwargs)
self.fields['locid'].widget.attrs["type"] = 'visible'
class Meta:
model = MyModel
fields = ('locid', 'name')
LocFormSet = modelformset_factory(MyModel, form = MyModelForm)
pformset = LocFormSet(request.POST, request.FILES,
queryset=MyModel.objects.order_by('name')))

Okay, none of the approaches above worked for me. I solved this issue from the template side, finally.
There is a ticket filed (http://code.djangoproject.com/ticket/10427), which adds a "value" option to a template variable for a form. For instance, it allows,
{{form.locid.value}}
to be shown. This is available as a patch, which can be installed in the SVN version of django using "patch -p0 file.patch"
Remember, the {{form.locid.value}} variable will be used in conjunction with the invisible form - otherwise, the submit and save operations for the formset will crash.
This is Not the same as {{form.locid.data}} - as is explained in the ticket referred to above.

The reason that the autofield is hidden, is that both BaseModelFormSet and BaseInlineFormSet override that field in add_field. The way to fix it is to create your own formset and override add_field without calling super. Also you don't have to explicitly define the primary key.
you have to pass the formset to modelformset_factory:
LocFormSet = modelformset_factory(MyModel,
formset=VisiblePrimaryKeyFormSet)
This is in the formset class:
from django.forms.models import BaseInlineFormSet, BaseModelFormSet, IntegerField
from django.forms.formsets import BaseFormSet
class VisiblePrimaryKeyFormset(BaseModelFormSet):
def add_fields(self, form, index):
self._pk_field = pk = self.model._meta.pk
if form.is_bound:
pk_value = form.instance.pk
else:
try:
pk_value = self.get_queryset()[index].pk
except IndexError:
pk_value = None
form.fields[self._pk_field.name] = IntegerField( initial=pk_value,
required=True) #or any other field you would like to display the pk in
BaseFormSet.add_fields(self, form, index) # call baseformset which does not modify your primary key field

Related

Pass a argument to Django meta class

I can alternatively create different types of form, but that's tedious.
So is it possible to pass the type to the form,then show the form accordingly?
This code shows NameError: name 'review_type' is not defined
class Contest1_for_review(ModelForm, review_type):
class Meta:
model = Contest1
decision = review_type + '_decision'
comment = review_type +'comment'
fields = [
decision,
comment,
]
Is it possible to pass a argument to meta class, like this?
Form is a class and when its rendered in the HTML, its rendering an instance of the form class. So when passing a value to that instance, you can use its __init__ method. For example:
class Contest1_for_review(ModelForm):
def __init__(self, *args, **kwargs):
review_type = kwargs.pop('review_type') # <-- getting the value from keyword arguments
super().__init__(*args, **kwargs)
self.fields[f'{review_type}_decision'] = forms.CharField()
self.fields[f'{review_type}_comment'] = forms.CharField()
class Meta:
model = Contest1
fields = "__all__"
Also, you need to send the value of review_type from view to form. Like this in function based view:
form = Contest1_for_review(review_type="my_value")
Or use get_form_kwargs to send the value from a Class based view. FYI: you don't need to change anything in Meta class.
Update:
From discussion in comments, OP should use forms.Form instead of ModelForm as using model form requires fields /exclude value in Meta class.

Pass argument to Django form

There are several questions on stackoverflow related to this topic but none of them explains whats happening neither provides working solution.
I need to pass user's first name as an argument to Django ModelForm when rendering template with this form.
I have some basic form:
class MyForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
self.first_name = ???
super(MyForm, self).__init__(*args, **kwargs)
class Meta:
model = MyModel
fields = [***other fields***, 'first_name']
Here's my sample class-based view:
class SomeClassBasedView(View):
def get(self, request):
user_first_name = request.user.first_name
form = MyForm(user_first_name)
return render(request, 'my_template.html', {'form': form})
What do I pass to MyForm when initialising it and how do I access this value inside the __init__ method?
I want to use 'first_name' value as a value for one of the fields in template by updating self.fields[***].widget.attrs.update({'value': self.first_name}).
The solution is as simple as passing initial data dictionary to MyForm for fields you need to set initial value.
This parameter, if given, should be a dictionary mapping field names to initial values. So if MyModel class has a field my_field and you want to set its initial value to foo_bar then you should create MyForm object with initial parameters as follows:
form = MyForm(initial={'my_field': 'foo_bar'})
A link to Django documentation on this topic.
You can pass the parameter instance to your form like this:
obj = MyModel(...)
form = MyForm(instance=obj)
If obj has a user first name it will be attached to your form.

Change a Django form field to a hidden field

I have a Django form with a RegexField, which is very similar to a normal text input field.
In my view, under certain conditions I want to hide it from the user, and trying to keep the form as similar as possible. What's the best way to turn this field into a HiddenInput field?
I know I can set attributes on the field with:
form['fieldname'].field.widget.attr['readonly'] = 'readonly'
And I can set the desired initial value with:
form.initial['fieldname'] = 'mydesiredvalue'
However, that won't change the form of the widget.
What's the best / most "django-y" / least "hacky" way to make this field a <input type="hidden"> field?
This may also be useful: {{ form.field.as_hidden }}
If you have a custom template and view you may exclude the field and use {{ modelform.instance.field }} to get the value.
also you may prefer to use in the view:
field = form.fields['field_name']
field.widget = field.hidden_widget()
but I'm not sure it will protect save method on post.
edit: field with multiple values don't supports HiddenInput as input type, so use default hidden input widget for this field instead.
an option that worked for me, define the field in the original form as:
forms.CharField(widget = forms.HiddenInput(), required = False)
then when you override it in the new Class it will keep it's place.
Firstly, if you don't want the user to modify the data, then it seems cleaner to simply exclude the field. Including it as a hidden field just adds more data to send over the wire and invites a malicious user to modify it when you don't want them to. If you do have a good reason to include the field but hide it, you can pass a keyword arg to the modelform's constructor. Something like this perhaps:
class MyModelForm(forms.ModelForm):
class Meta:
model = MyModel
def __init__(self, *args, **kwargs):
from django.forms.widgets import HiddenInput
hide_condition = kwargs.pop('hide_condition',None)
super(MyModelForm, self).__init__(*args, **kwargs)
if hide_condition:
self.fields['fieldname'].widget = HiddenInput()
# or alternately: del self.fields['fieldname'] to remove it from the form altogether.
Then in your view:
form = MyModelForm(hide_condition=True)
I prefer this approach to modifying the modelform's internals in the view, but it's a matter of taste.
For normal form you can do
class MyModelForm(forms.ModelForm):
slug = forms.CharField(widget=forms.HiddenInput())
If you have model form you can do the following
class MyModelForm(forms.ModelForm):
class Meta:
model = TagStatus
fields = ('slug', 'ext')
widgets = {'slug': forms.HiddenInput()}
You can also override __init__ method
class Myform(forms.Form):
def __init__(self, *args, **kwargs):
super(Myform, self).__init__(*args, **kwargs)
self.fields['slug'].widget = forms.HiddenInput()
If you want the field to always be hidden, use the following:
class MyForm(forms.Form):
hidden_input = forms.CharField(widget=forms.HiddenInput(), initial="value")
If you want the field to be conditionally hidden, you can do the following:
form = MyForm()
if condition:
form.fields["field_name"].widget = forms.HiddenInput()
form.fields["field_name"].initial = "value"
Example of a model:
models.py
from django.db import models
class YourModel(models.Model):
fieldname = models.CharField(max_length=255, default="default")
In your form, you can add widgets with ModelForm. To make it hidden add 'type': 'hidden' as shown below👇
forms.py
from .models import YourModel
from django import forms
class YourForm(forms.ModelForm):
class Meta:
model = YourModel
fields = ('fieldname',)
widgets = {
'fieldname': forms.TextInput(attrs={'type': 'hidden'}),
}
If you don't know how to add it to your views.py file, here is some videos about it.
If you use Function Based View:
https://www.youtube.com/watch?v=6oOHlcHkX2U
If you use Class Based View:
https://www.youtube.com/watch?v=KB_wDXBwhUA
{{ form.field}}
{{ form.field.as_hidden }}
with this jinja format we can have both visible form fields and hidden ones too.
if you want to hide and disable the field to protect the data inside. as others mentioned use the hiddenInput widget and make it disable
in your form init
example:
if not current_user.is_staff:
self.base_fields['pictureValid'].disabled = True
self.base_fields['pictureValid'].widget = forms.HiddenInput()
With render_field is easy
{% render_field form.field hidden=True %}
You can just use css :
#id_fieldname, label[for="id_fieldname"] {
position: absolute;
display: none
}
This will make the field and its label invisible.

Creating modelformset from a modelform

I have a model MyModel which contains a PK - locid, that is an AutoField.
I want to construct a model formset from this, with some caveats:
The queryset for the formset should be a custom one (say, order_by('field')) rather than all()
Since locid for MyModel is an AutoField and thus hidden by default, I want to be able to show it to the user.
I'm not sure how to do this. I've tried multiple approaches,
MyModelFormSet = modelformset_factory(MyModel, fields=('locid', 'name', 'dupof'))
The above gives me the 3 fields, but locid is hidden.
class MyModelForm(ModelForm):
def __init__(self, *args, **kwargs):
super(MyModelForm, self).__init__(*args, **kwargs)
self.fields['locid'].widget.attrs["type"] = 'visible'
locid = forms.IntegerField(min_value = 1, required=True)
class Meta:
model = MyModel
fields = ('locid', 'name', 'dupof')
The above gives me a ManyToMany error.
Has anyone done something like this before?
Edit 2
I can now use a custom query when I instantiate the formset - but I still need to show the locid field to the user, because the id is important for the application's use. How would I do this? Is there a way to override the default behavior of hiding a PK if its an autofield?
It makes no sense to show an autofield to the user, as it's an autoincremented primary key -- the user can not change it and it will not be available before saving the record to the database (where the DBMS selectes the next available id).
This is how you set a custom queryset for a formset:
from django.forms.models import BaseModelFormSet
class OrderedFormSet(BaseModelFormSet):
def __init__(self, *args, **kwargs):
self.queryset = MyModel.objects.order_by("field")
super(OrderedFormSet, self).__init__(*args, **kwargs)
and then you use that formset in the factory function:
MyModelFormSet = modelformset_factory(MyModel, formset=OrderedFormSet)
I ended up using a template side variable to do this, as I mentioned here:
How to show hidden autofield in django formset
If you like cheap workarounds, why not mangle the locid into the __unicode__ method? The user is guaranteed to see it, and no special knowledge of django-admin is required.
But, to be fair, all my answers to django-admin related questions tend along the lines of "don't strain to hard to make django-admin into an all-purpose CRUD interface".

Dynamically update ModelForm's Meta class

I am hoping to dynamically update a ModelForm's inline Meta class from my view. Although this code seems to update the exclude list in the Meta class, the output from as_p(), as_ul(), etc does not reflect the updated Meta exclude.
I assume then that the html is generated when the ModelForm is created not when the as_*() is called. Is there a way to force the update of the HTML?
Is this even the best way to do it? I just assumed this should work.
Thoughts?
from django.forms import ModelForm
from testprogram.online_bookings.models import Passenger
class PassengerInfoForm(ModelForm):
def set_form_excludes(self, exclude_list):
self.Meta.exclude = excludes_list
class Meta:
model = Passenger
exclude = []
The Meta class is used to dynamically construct the form definition - so by the time you've created the ModelForm instance, the fields not in the exclude have already been added as the new object's attributes.
The normal way to do it would be to just have multiple class definitions for each possible exclude list. But if you want the form itself to be dynamic, you'll have to create a class definition on the fly. Something like:
def get_form(exclude_list):
class MyForm(ModelForm):
class Meta:
model = Passenger
exclude = exclude_list
return MyForm
form_class = get_form(('field1', 'field2'))
form = form_class()
UPDATE: I just revisited this post and thought I'd post a little more idiomatic way to handle a dynamic class:
def PassengerForm(exclude_list, *args, **kwargs):
class MyPassengerForm(ModelForm):
class Meta:
model = Passenger
exclude = exclude_list
def __init__(self):
super(MyPassengerForm, self).__init__(*args, **kwargs)
return MyPassengerForm()
form = PassengerForm(('field1', 'field2'))
Another way:
class PassengerInfoForm(ModelForm):
def __init__(self, *args, **kwargs):
exclude_list=kwargs.pop('exclude_list', '')
super(PassengerInfoForm, self).__init__(*args, **kwargs)
for field in exclude_list:
del self.fields[field]
class Meta:
model = Passenger
form = PassengerInfoForm(exclude_list=['field1', 'field2'])
Similar approach, somewhat different goal (generic ModelForm for arbitrary models):
from django.contrib.admin.widgets import AdminDateWidget
from django.forms import ModelForm
from django.db import models
def ModelFormFactory(some_model, *args, **kwargs):
"""
Create a ModelForm for some_model
"""
widdict = {}
# set some widgets for special fields
for field in some_model._meta.local_fields:
if type(field) is models.DateField:
widdict[field.name] = AdminDateWidget()
class MyModelForm(ModelForm): # I use my personal BaseModelForm as parent
class Meta:
model = some_model
widgets = widdict
return MyModelForm(*args, **kwargs)
Use modelform_factory (doc):
from django.forms.models import modelform_factory
from testprogram.online_bookings.models import Passenger
exclude = ('field1', 'field2')
CustomForm = modelform_factory(model=Passenger, exclude=exclude) # generates ModelForm dynamically
custom_form = CustomForm(data=request.POST, ...) # form instance

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