I have 2 forms in my Django view. How can I do a check to see which one has been submitted?
Thanks
Put a different name attribute on the submit buttons for each form, then check for that key in request.POST in your view.
Also don't forget to give each form a separate prefix attribute when you instantiate them, to avoid any possible field name collisions.
Here are some ideas:
Use different action URLs for the forms, associated with different views.
Use different action URLs for the forms, associated with the same view but with using different parameters to the view (using the URLconf)
Use an <input type="hidden" /> to differentiate between the forms.
Philip
Related
i am new to django and to this site, so apologies if this has been solved before but i haven't found it
So i have 2 django models
ModelA(Model):
ModelB(Model):
modelA = ForeignKey(ModelA, on_delete=models.CASCADE)
A form for the ModelB
ModelBForm(ModelForm):
class Meta:
model=ModelB
exclude=()
and a view
createModelBView(CreateView):
model = ModelB
form_class = ModelBForm
the template only does
{{form}}
When rendered, there is a dropdown list for the ModelA field so I can choose from existing instances of the ModelA, but what if a new one needs to be created? In the admin there is an option next to edit or create a new ModelA in a popup. Is there an option to do this with CreateView?
Thanks
There is no built-in functionality like that.
However you can build it easily yourself.
You will have to add a link (or a HTML form) in your template which points to the URL corresponding to the view you implemented to create the given model.
Following is a very abstract example.
In your template:
<form>
{{csrf_token}}
{{ form }}
Create model A if you want
<input type="submit" value="Submit">
<\form>
In your urls.py
url(r'^models/createA/$', views.CreateModelAView.as_view(), name="optional")
In your views.py
createModelAView(CreateView):
model = ModelA
form_class = ModelAForm
Then you'll need to create a form called ModelAForm.
On a different note, I'd suggest to start off with functional views if you're new to Django. It is more coding but you get a better feel of what's going on
In the admin there is an option next to edit or create a new ModelA in a popup. Is there an option to do this with CreateView?
No, not built in. That functionality in the admin involves a lot of front-end work involving templates and routing that would have to come from somewhere; since a Form/ModelForm instance can't assume it has access to the admin (which is a contrib module, may not be enabled, and is permission-sensitive), the infrastructure required for that can't be assumed to be available in the general case.
Keep in mind that {{ form }} doesn't even render <form> tags or any kind of submit element. It's intended to be a very, very basic way to render a very, very basic set of fields, while the admin is built specifically to be a (reasonably) powerful, flexible way to put a UI in front of your models.
You could certainly build that functionality yourself, or find a reusable app that does the same thing, but there is no facility distributed with Django to generate it automatically.
I have a django project with a single html template. i am want to know if it is possible to pass multiple forms into a single template. I want to process the form differently for each of the forms that are passed. Is that possible to do and how can I differentiate the different forms for processing... if anyone can help, I would much appreciate it.
You can. all the forms can be rendered inside single . If you want to submit them in different events, you can add multiple submit buttons and give name attribute for each of them. An in your view you can check:
if 'submit_button_name' in request.POST:
If you just want to submit them altogether, use just one submit button.
To render mulitple model forms in same template you can do like this:
<form>{{ form1.as_p }} {{ form2.as_p }}</form>
I am using materializecss to give my django site some material elements. I have put together a form (the 'old' way using html) but now realised I need to use a django form instead. The problem is, these forms don't play well with materialises built in column system (they use classes to determine rows and column spacing). Here is an example of the layout I set up so far. However when defining the form through form.py, it spits out one input per layer.
My question is: what can I do to either a) get django to work with the html-defined form or b) make a 'form template' to give the input fields the appropriate classes?
If you want to see the code I can post some but I'm quite a new coder so it's messy.
Thanks!
There are three ways I can think of off the top of my head.
If you want full control over the HTML form, in a Django template or HTML form, simply map the names of your fields to match the underlying field names in the Django form. This way, when POSTed back to your view, Django will automatically link up the POSTed fields with the Django form fields.
For example, if you have a field username in your Django form (or Django model if using ModelForm), you could have an element <input type="text" name="username" maxlength="40"> (that you can style any way you need) on your HTML form that Django will happily parse into your Django form field, assuming your view is plumbed correctly. There is an example of this method in the Django documentation.
Another way is to customize the Django form field widgets in your Django form definition. The Django documentation talks a little bit about how to do this. This is great for one offs, but is probably not the best approach if you expect to reuse widgets.
The final approach would be to subclass Django form field widgets to automatically provide whatever attributes you need. For example, we use Bootstrap and have subclassed nearly all of the widgets we use to take advantage of Bootstrap classes.
class BootstrapTextInput(forms.TextInput):
def __init__(self, attrs=None):
final_attrs = {'class': 'form-control'}
if attrs is not None:
final_attrs.update(attrs)
super().__init__(attrs=final_attrs)
Then it's simply a matter of letting the Django form know which widget to use for your form field.
class UsernameForm(forms.ModelForm):
class Meta:
model = auth.get_user_model()
fields = ['username']
widgets = {'username': BootstrapTextInput()}
Hope this helps. Cheers!
I need to create a form to admin-side with two fields
Number of code: integer
Value of code: float
How can I do that. This form is not related to any model.
You can implement your modelless form as explained in #levi's answer.
Then, you can place it in the admin site in a number of different ways, depending your needs:
Make instances of the form available to all templates via a context processor, and override the admin templates to have it rendered wherever you want. You can create a view for only processing the form.
Create a view for both rendering and processing the form in a unique place, and hook it up to the admin as explained in the old Django Book, you'll need to make sure the template for that view extends one of the admin's templates (admin/change_form.html may be a good choice).
from django import forms
class Your_Form(forms.Form):
number_code = forms.IntegerField()
value_code = forms.FloatField()
I want to change the shape of some fields showed in the admin site.
I found that the template that manage everything is change_form.html with fieldset.html but I cannot find where the fields are actually transformed in html.
Basically I want to change the field of the foreign key adding a link to another page.
Do you have any idea?
Thanks,
Giovanni
The HTML for a given field is handled by its widget in the render function. If you want to customize the look of a field you could create a custom widget which has the additional HTML you need in the render.
You can check out the render some of the built in widgets in django/forms/widgets.py (links to the Django trunk).
In fieldset.html, the code {{ field.field }} renders the field's HTML representation. to achieve what you want, you'll probably need to define your own widget. you can take a look at admin's widgets.py