I am trying to include some dummy fields in my model form and using the same in model formset factory.
Model form:
class DistForm(forms.ModelForm):
dist_from_txt = forms.CharField()
...
class Meta:
model: Distance
fields = ('dist_from', 'dist_to', 'distance')
widgets = {
...
}
However, when rendered the extra field does not show up on the form.
Needlessly to mention here that I have searched (including here) and failed to find a possible solution.
Question is: How to correctly add and render extra field/s in model form?
forms.TextInput is a widget instance, not a form field instance. Use CharField:
class DistForm(forms.ModelForm):
dist_from_txt = forms.CharField()
...
class Meta:
model: Distance
fields = ('dist_from', 'dist_to', 'distance')
widgets = {
...
Related
TL;DR: I need a some kind of formset for formsets.
I have two different models related to one buisness-entity, and I need to make a form to edit both models like a one form. And I need to create a lot of such forms on the one page like Django inline formset does.
Now I have the following thing:
class Parent(models.Model):
name = models.Charfield()
class FirstChild(models.Model):
name = models.Charfield()
e_id = models.IntegerField()
parent = models.ForeignKey(Parent)
class FirstChildForm(django.forms.ModelForm):
class Meta:
model = Child
fields = ('name', 'e_id', 'parent')
widgets = {'parent': forms.TextInput}
And I render a lot of them using inline formsets:
formset_class = inlineformset_factory(Parent, FirstChild,
form=FirstChildForm, extra=1)
But now I have to add second child model and a form for it, and still render it like an one inline form, but make it form actually edit two models. Like this:
class SecondChild(models.Model):
name = models.Charfield()
e_id = models.IntegerField()
parent = models.ForeignKey(Parent)
class SecondChildForm(django.forms.ModelForm):
class Meta:
model = Child
fields = ('name', 'e_id', 'parent')
widgets = {'parent': forms.TextInput}
formset_class = inlineformset_factory(models=[Parent, FirstChild],
forms=[FirstChildForm, SecondChildForm],
extra=1)
As far as I understand, Django formsets cannot work with multiple models right now.
So which way should I choose to implement this behaviour and do not broke all django conceptions?, I cannot use some extra libraries so I have to implement everything by myself and I use django 1.6 if it is important.
So, finally I used this approach as a base: https://micropyramid.com/blog/how-to-use-nested-formsets-in-django/
So to make the above possible I have found out that I have to have ManytoMany Field that is not a problem.
That field is in the form as follows:
class Form(forms.ModelForm):
class Meta:
model = MyModel
fields = ['notes', 'scan']
widgets = {
'scan': forms.CheckboxSelectMultiple(),
}
In the view I have this then:
form = Form(request.POST)
if from.is_valid():
inst = from.save(commit=False)
inst.something = something
inst.save()
Now what do I do, to save the test or scan from the form?
I tried :
inst.test.add(form.cleaned_data['test'])
But that doesn't work for test or scan.
The Model looks like this:
class MyModel(models.Model):
id = models.AutoField(primary_key=True)
notes = models.TextField(default='')
scan = models.ManyToManyField(Scan)
....
Please help I wasn't able find anything in the Internet about this
Thanks!
The documentation of the Form's save method tells it all: If you have a ModelForm that contains the model's ManyToManyField like this:
class MyForm(forms.ModelForm):
class Meta:
model = MyModel
fields = ['__all__'] # or fields = ['scans'] assuming scans is the M2M field in MyModel
Then you have two ways to save the relationships:
Directly, using form.save()
Calling save_m2m() is only required if you use save(commit=False). When you use a simple save() on a form, all data – including many-to-many data – is saved without the need for any additional method calls.
Or indirectly because you want to manipulate the instance before saving:
if form.is_valid():
instance = form.save(commit=False)
instance.some_field = some_value
instance.save()
form.save_m2m() # this saves the relationships
I'd like to add some info to a model field to use at form rendering time. My real model has about 15 values of varying field types (adding and removing as I dev), and it does almost everything I need, so I'd rather not create custom model fields for all of them.
I'd like to do something like this:
from django.db import models
class MyModel(models.Model):
cost = models.DecimalField(max_digits=5,
decimal_places=2,
custom_info= {'glyph': 'glyphicon glyphicon-usd' }
)
And then in my form template use that glyph much like I'd use a verbose_name or help_text.
Something I learned from a post just the other day. Will defining the custom information on the form instead of the model work?
When you define formfield_callback on a forms.ModelForm it will iterate over the form fields and you can manipulate them. This comes in handy when you need to add a css class to widgets and don't want to explicitly override the field. Now you only need to put formfield_callback = modify_form_field on any forms.ModelForm where you want the custom_info to show up.
from django.db import models
def add_glyphicons(model_field):
form_field = model_field.formfield()
if isinstance(model_field, models.IntegerField):
form_field.custom_info = {'glyph': 'glyphicon glyphicon-usd'}
elif isinstance(model_field, models.CharField):
form_field.custom_info = {'glyph': 'glyphicon glyphicon-yen'}
return form_field
class MyModel(models.Model):
formfield_callback = add_glyphicons
class Meta:
model = MyModel
class MyOtherModel(models.Model):
formfield_callback = add_glyphicons
class Meta:
model = MyOtherModel
class Label(models.Model):
name = ...
slug_name = ...
and here is the form I created straightly from model
class LabelForm(models.Model):
class Meta:
model = Label
How can I give url the widget HiddenInput? slug_name also cannot be blank because it is required...
Thanks
For the required field, use the blank field option with a TextField:
slug_name = TextField(blank=False)
For the HiddenInput widget, I assume there exists a url field in your model (which is not the case in the code you posted) check Django's own guide to overriding the default field types or widgets on a ModelForm:
class LabelForm(models.Model):
class Meta:
model = Label
widgets = {'url': HiddenInput()}
When I render my formset, one of the field renders as a select box because it is a foreign field in the model. Is there a way to change this to a text input? I want to populate that field by using Ajax auto complete. Adding a widget to the modelform is not working because the modelformset_factory takes a model and not a model form.
EDIT
My Model Form
class RecipeIngredientForm(ModelForm):
class Meta:
model = RecipeIngredient
widgets = { 'ingredient' : TextInput(), }
I use it in my view
RecipeIngredientFormSet = modelformset_factory(RecipeIngredient, form=RecipeIngredientForm)
objRecipeIngredients = RecipeIngredientFormSet()
EDITED MODEL FORM
class RecipeIngredientForm(ModelForm):
ingredient2 = TextInput()
class Meta:
model = RecipeIngredient
I create the form set like this
RecipeIngredientFormSet = modelformset_factory(RecipeIngredient, form=RecipeIngredientForm)
objRecipeIngredients = RecipeIngredientFormSet()
QUESTION
Do I have to use the formset in html? Can I just hard code the fields that get generated and using javascript I can create new fields and increment the "form-TOTAL-FORMS"? If I can then I do not have to worry about my model form.
Thanks
modelformset_factory does take a form. Here's the function signature from django.forms.models:
def modelformset_factory(
model, form=ModelForm, formfield_callback=lambda f: f.formfield(),
formset=BaseModelFormSet,
extra=1, can_delete=False, can_order=False,
max_num=0, fields=None, exclude=None):
If this isn't working for you, show some code and I'll try and see what is going wrong.
Edit after various comments As you point out, the widget argument is buggy when used in this way. So the solution is not to use it - it's a very recent addition in any case. Instead, define the field directly on the form:
class RecipeIngredientForm(forms.ModelForm):
ingredient = forms.ModelChoiceField(widget=forms.TextInput))
class Meta:
model = RecipeIngredient