We Keep Coding

How to save form data from base.html in django?

In my app, I have created a context_proccessors.py to show the form to base.html file.
I am able to show the form in the base.html file. But the problem I am facing is I have no idea how to save that form data from base.html since there is no view for the base.html. Below is my code:
models.py
class Posts(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='user_posts')
post_pic = models.ImageField(upload_to='post_pic', verbose_name="Image")
post_caption = models.TextField(max_length=264, verbose_name="Caption")
created_date = models.DateTimeField(auto_now_add=True)
edited_date = models.DateTimeField(auto_now=True)
def __str__(self):
return f"{self.user.username}"
forms.py
from django import forms
from post_app.models import Posts
class PostForm(forms.ModelForm):
class Meta:
model = Posts
exclude = ('user',)
context_proccessors.py
from post_app.forms import PostForm
def post_form(request):
form = PostForm
return {
'post_form': form,
}
base.html
<form method="POST" enctype="multipart/form-data">
{{ post_form|crispy }}
{% csrf_token %}
<button type="submit" class="btn btn-primary">Post</button>
</form>
I want the form to be displayed on every page so that the user can submit data from anywhere

def PostView(request):
form = PostForm()
if request.method == 'GET':
return render(request, 'base.html', {form:form})
elif request.method == 'POST':
form.save(request.data)
In the views.py of your app you can define this view, and the you have to provide it an url in the urls.py of the root directory. So evere time there is a request on that url, if the method is GET, the form will be rendered on base.html file, if the method is POST, the post will be saved.

By following the answer by N T I have implemented this. So, I had to make a URL pattern for the view and use that URL pattern in the action in the form of base.html.
view.py
#login_required
def postsaveview(request):
form = PostForm()
if request.method == 'POST':
form = PostForm(request.POST, request.FILES)
if form.is_valid():
user_obj = form.save(commit=False)
user_obj.user = request.user
user_obj.slug = str(request.user) + str(uuid.uuid4())
user_obj.save()
return HttpResponseRedirect(reverse('profile_app:profile'))
urls.py
urlpatterns = [
path('post-save/', views.postsaveview, name='post-save'),
]
base.html
<form action="{% url "post-save" %}" method="POST" enctype="multipart/form-data">
{{ post_form|crispy }}
{% csrf_token %}
<button type="submit" class="btn btn-primary">Post</button>
</form>

Django ModelForm not saving data to database, Form.save is not working?

List item
Hello I am django beginner having tough time could someone please help me I don't know what am I doing wrong ?
I am trying to create a form and saving some data through it by using form.save(). And I am new to here also so don't mind any mistakes.
Here is my model:
from django.db import models
from stores.models import Store
class Category(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=30)
def __str__(self):
return self.name
class Product(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=30)
price = models.DecimalField(max_digits=5, decimal_places=5)
image = models.ImageField(upload_to='upload_to/')
category = models.ForeignKey(Category, default='Default', on_delete=models.CASCADE, blank=False, null=False)
store = models.ForeignKey(Store, on_delete=models.CASCADE, blank=False, null=False)
Here is my view:
from django.shortcuts import render, redirect
from .forms import NewPro
def pro(request):
if request.method == 'POST':
form = NewPro(request.POST)
if form.is_valid():
form.save()
return redirect('stores_list')
else:
form = NewPro()
return render(request, "default/add_product.html", {'form': form})
def product_list(request):
return render(request, 'default/product_list.html')
Here is my form:
from django import forms
from .models import Product
class NewPro(forms.ModelForm):
class Meta:
model = Product
fields = ('name', 'price', 'image','category', 'store',)
default/add_product.html :
{% extends 'default/base.html' %}
<html>
<head><title>E-Commerce App</title></head>
{% block content %}
<h1>Add Product details</h1>
<form method="post">
{% csrf_token %}
{{ form.as_p }}
<button type="submit">Add Product</button>
</form>{% endblock %}
</html>
Settings.py settings
MEDIA_ROOT = '/home/saifi/Saif_project/final_project/MEDIA_ROOT/upload_to'

I can see some indentation issues in the view - but I'll guess that's just formatting when copying into Stackoverflow.
the form.is_valid() check will validate all your form fields and will only write to the database if all the input fields are valid. If it's not saving, the first place I'd check would be for form errors.
In your template you can render the errors with {{form.errors}} and it will list each field and error.

You forgot request.FILES in your pro view function, you have an image file after all.
def pro(request):
if request.method == 'POST':
form = NewPro(request.POST, request.FILES)
if form.is_valid():
form.save()
return redirect('stores_list')
else:
form = NewPro()
return render(request, "default/add_product.html", {'form': form})
Try using the form this way:
<form action="YOUR_URL_HERE" method="post" enctype="multipart/form-data">
{% csrf_token %}
{{ form.as_p }}
<button type="submit">Add Product</button>
</form>
I hope this will help. Welcome aboard ;)

Your indentation is wrong, the else should be for first 'if'
def pro(request):
form = NewPro()
if request.method == 'POST':
form = NewPro(request.POST)
if form.is_valid():
form.save()
return redirect('stores_list')
else:
form = NewPro()
return render(request, "default/add_product.html", {'form': form})

Django: ModelForm and show data on the same url

I'm new to django and trying to create my first app and I think I might need some little help :)
I have a ModelForm on a site to submit and want to show the data on the same page. I'm having trouble to set up two functions on the same page, I think i might have to use a class and set it in urls.py but I'm not able to make it work :( the code looks like this:
forms.py:
from django import forms
from .models import Eintrag
class NameForm(forms.ModelForm):
class Meta:
model = Eintrag
fields = ['Anmeldung', 'Essen']
urls.py
from django.urls import path
from . import views
app_name = 'form'
urlpatterns = [
path('', views.get_name, name='form'),
]
views.py
from django.shortcuts import render
from django.utils import timezone
from django.contrib.auth.decorators import login_required
from .forms import NameForm
from .models import Eintrag
#login_required()
def get_name(request):
# if this is a POST request we need to process the form data
if request.method == 'POST':
# create a form instance and populate it with data from the request:
form = NameForm(request.POST)
# check whether it's valid:
if form.is_valid():
eintrag = form.save(commit=False)
# process the data in form.cleaned_data as required
# ...
# redirect to a new URL:
eintrag.Name = request.user # Set the user object here
eintrag.pub_date = timezone.now() # Set the user object here
eintrag.save()
return render(request, 'form/name.html', {'form': form})
# if a GET (or any other method) we'll create a blank form
else:
form = NameForm()
return render(request, 'form/name.html', {'form': form})
def post_list(request):
posts = Eintrag.objects.all()
return render('form/post_list.html', {'posts': posts})
name.html
...
{% include "form/post_list.html" %}
<form action="/form/" method="post">
{% csrf_token %}
{{ form }}
<input type="submit" value="Submit" />
</form>
...
post_list.html
{% for post in posts %}
{{ post }}
{% endfor %}
So the problem is in urls.py only get_name is handled and I'm clueless how I should include post_list. I rather not want to use different url's, do I have to?
Thanks for any help and advice!

You don't need a separate URL or view for the list. Just include the queryset in the context of your get_name view.
posts = Eintrag.objects.all()
return render(request, 'form/name.html', {'form': form, 'posts': posts})

with [Class Based View] it would be better.
But with your view, you can send multiple data via context.
#login_required()
def get_name(request):
# if this is a POST request we need to process the form data
if request.method == 'POST':
# create a form instance and populate it with data from the request:
''' codes '''
eintrag.save()
return HttpResponseRedirect(request.path) # generate an empty form
# if a GET (or any other method) we'll create a blank form
else:
form = NameForm()
posts = Eintrag.objects.all() # the queryset is here, and sent via context
return render(request, 'form/name.html', {'form': form,'posts':posts})
I your html remain the same, but keep your form action='' empty
{% include "form/post_list.html" %}
<form action="" method="post">
{% csrf_token %}
{{ form }}
<input type="submit" value="Submit" />
</form>

How can I call multiple views in one url address in Django?

I'm trying to show forms defined by new_measurement on index.html, but I only manage to get IndexView() to work. I tried various combinations between IndexView() and new_measurement(), but those didn't work out at all. I know that IndexView() doesn't pass anything related to new_measurement(), and new_measurement() isn't called, which is the core of my problem. I'd really appreciate if someone more experienced with Django could tell me what I could, or should do. Thank you.
Here's my views.py:
from django.shortcuts import render
from django.utils import timezone
from .models import Measurement
from .forms import MeasurementForm
from django.views import generic
class IndexView(generic.ListView):
model = Measurement
context_object_name = 'measurement_list'
template_name = 'index.html'
queryset = Measurement.objects.all()
def new_measurement(request):
if request.method == "POST":
form = MeasurementForm(request.POST)
if form.is_valid():
measurement = form.save(commit=False)
measurement.measurement_date = timezone.now()
measurement.save()
else:
form = MeasurementForm()
return render(request, 'index.html', {'form': form})
urls.py:
from django.urls import path
from . import views
urlpatterns = [
path('', views.IndexView.as_view(), name='index'),
]
forms.py:
class MeasurementForm(forms.ModelForm):
class Meta:
model = Measurement
fields = ('measurement_value', 'measurement_unit')
index.html:
{% extends "base.html" %}
{% block content %}
<h1>Climate Measurement Tool</h1>
<h2>Add a new measurement</h2>
<form method="POST" class="post-form">
{% csrf_token %}
{{ form.as_p }}
<button type="submit" class="save">Add</button>
</form>
<h2>Measurements</h2>
{% if measurement_list %}
<ul>
{% for measurement in measurement_list %}
<li>
<p>{{ measurement }}</p>
</li>
{% endfor %}
</ul>
{% else %}
<p>No measurements yet</p>
{% endif %}
{% endblock %}

You can't map multiple views in one url but you can do mutiple works in one view.
update your views.py as you can see that I am sending (querylist and form) both in that view
views.py
def new_measurement(request):
if request.method == "POST":
form = MeasurementForm(request.POST)
if form.is_valid():
measurement = form.save(commit=False)
measurement.measurement_date = timezone.now()
measurement.save()
else:
form = MeasurementForm()
qs = Measurement.objects.all()
context = {'form': form, 'measurement_list': qs}
return render(request, 'index.html', context)
update urls.py
from django.urls import path
from . import views
urlpatterns = [
path('', views.new_measurement, name='index'),
]

You can't call 2 views for one url. basically each url has to be linked to one view and that's something you can't really change.
But if you want your code to be cleaner and have multiple functions, you can call them in your view, basically what you can do is to make a view and call it when a url or even more than one url has been used and in that view decide which function to use
Example:
def god_view(request):
if request.method == "POST"
return post_func(request)
return get_func(request)
This is a very simple example but you can do so many other things.

It is not possible to have more views in one url, but you can simulate it. I did it like a view and in the template of this view was javascript which loaded the second view with the response of AJAX and filled the belonging element with the second view's content. The second view was not whole template but it started with some div tags which were placed into the first template. I'll try to give you an example
views
def first_view(request):
return render(
request,
'first_template.html',
{
'first_content': 'Some heavy content'
})
def second_view(request):
return render(
request,
'second_template.html',
{
'second_content': 'Some heavier content!'
})
first_template.html
...
<body>
<div id="1">
{{ first_content }}
</div>
<div>
... loading ...
</div>
<script>
window.onload = function() {
$.ajax({
url: {% url 'to_second_view' %},
method: 'GET',
success: function(response) {
$('#2').html(response);
}
})
}
</script>
</body>
...
second_template.html
<div>
{{ second_content }}
</div>

If you're using cbv you can override the get_template_names method for any view that inherits TemplateResponseMixin and return a list of string which are searched in order until one matches or ImporperlyConfigured is raised. For example:
class SomeView(TemplateResponseMixin):
...
def get_template_names(self):
if self.request.method == "POST":
return ['post_template.html']
else:
return ['template.html']

Instead of generic.ListView you can try with rest_framework.views.APIView
from rest_framework.views import APIView
class IndexView(APIView):
def post(self, request: Request):
form = MeasurementForm(request.POST)
if form.is_valid():
measurement = form.save(commit=False)
measurement.measurement_date = timezone.now()
measurement.save()
return render(request, 'index.html', {'form': form})
def get(self, request: Request):
form = MeasurementForm()
return render(request, 'index.html', {'form': form})
This gives you more control on the APIs you call. Also you can raise/return error when you call your API using incorrect methods (PUT, PATCH)

Include template form to django admin

Is it possible to include model form template in django admin as follows?
models.py
class Customer(models.Model):
name = models.CharField(max_length=20)
designation = models.CharField(max_length=20)
gender = models.BooleanField()
forms.py
class CustomerForm(forms.ModelForm)
gender = forms.TypedChoiceField(
choices=GENDER_CHOICES, widget=forms.RadioSelect(renderer=HorizontalRadioRenderer), coerce=int, )
class Meta:
model = Customer
template.html
<form action="{% url 'verinc.views.customerView' %}" method="POST">{% csrf_token %}
{{ form.as_p }}
<input id="submit" type="button" value="Click" /></form>
views.py
def customerView(request):
if request.method == 'POST':
form = CustomerForm(request.POST)
else:
form = CustomerForm()
return render_to_response('myapp/template.html', {'form' : form,})
admin.py
class CustomerInline(admin.StackedInline)
model= Customer
form = CustomerForm
template = 'myapp/template.html
When I view the form in url (localhost/myapp/customer) it displays all the fields correctly. But when I view it in admin page it displays only the submit button in the block. My requirement is to view the form using templates in admin page, so that i could use some AJAX script for further process. Any help is most appreciated.

Well , Its not possible . But you can implement like this :
You should create file called admin_views.py in your app customer.
Then add url in your urls.py like
(r'^admin/customer/customerView/$', 'myapp.customer.admin_views.customerView'),
Write your view implementation inside admin_views.py like :
from myapp.customer.models import Customer
from django.template import RequestContext
from django.shortcuts import render_to_response
from django.contrib.admin.views.decorators import staff_member_required
def customerView(request):
return render_to_response(
"admin/customer/template.html",
{'custom_context' : {}},
RequestContext(request, {}),
)
customerView = staff_member_required(customerView)
And inside your admin template extend base_site.html like this one :
{% extends "admin/base_site.html" %}
{% block title %}Custmer admin view{% endblock %}
{% block content %}
<div id="content-main">
your content from context or static / dynamic...
</div>
{% endblock %}
Thats it . hit that new url after admin login . Note Not Tested :) .