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What would be the robust design pattern for multiple class forms render/process (create, update) on single page in django

I am working on a django project in which I have to design a sales order page for order processing,now as per requirement I have to create multiple information forms on single page with different fields (formset does not helpful in this case) which I achieved by creating specific form class and then render them in template through class view function
def get_context_data(self, **kwargs):
context = super(CreateOrderView, self).get_context_data(**kwargs)
context['ord_info_form'] = OrderInformationForm()
context['billing_form'] = BillingInformationForm(prefix='billing')
context['shipping_form'] = ShipingInformationForm(prefix='shipping')
context['payment_form'] = PaymentInfoForm()
context['summary_form'] = OrderProductsForm()
return context
My template look like this:
Now, I have to save data for all forms which belongs to multiple models like billing, contact and shipping information. So for save information I created another form class for create order and set it in form_class variable in Createview class, and I override the save function in form class in which I manipulate the data for all model forms. For update order (put operation) I did the same I create another form class for update like create one in form class.
As per my understanding I implement this strategy, but this seems to me repetitive design. How can I design it better for multiple forms with different fields. Please let me know what kind of design pattern I can apply for this.
P.S: If require I'll post more code and assets for clear understanding.
Thanks in advance.

I've been working on exactly the same requirement. What we have done is the following:
Create the classbasedview to render the forms and add them to the context just as you did on the top.
I see you've created the template based on the forms you added to the context, but as someone else pointed out you need to use an alternative rather than just django to make the post of each item. In my case I used plain old ajax to perform my post. So it looks something like this:
function firePost() {
var endpoint = '/sales_center/orders/add';
var channel = getChannelForm();
var order = getOrderForm()
var ip = getIPForms();
var collections = getCollections();
var subscriptions = getBookSubs();
var perpetual = getBookPerp();
$.ajax({
type: 'POST',
url: endpoint,
data: {
'order':JSON.stringify(order),
'channel': JSON.stringify(channel),
'ips':JSON.stringify(ip),
'collections': JSON.stringify(collections),
'subscriptions': JSON.stringify(subscriptions),
'perpetual': JSON.stringify(perpetual)
},
success: function(data) {
console.log(data);
},
error: function(data) {
console.log(data);
}
});
}
In the data section I divided the information by name. My order is composed of a channel, multiple collections, multiple ips, multiple subscriptions and multiple perpetual purchases. The endpoint is the same as the one in the page so in the backend in order to process that we do it like this:
#transaction.atomic
def post(self, request):
channel = self._create_channel(json.loads(request.POST.get('channel')))
order = self._create_order(json.loads(request.POST.get('order')), canal)
ips = self._assign_ips(json.loads(request.POST.get('ips')), canal)
return JsonResponse({}, status=200)
The code above is just an example the _create_channel is a function in our classbasedview. Each function performs the posts in our backend for each model. That way it looks clean and easy to read each step of the order.
I added the #transaction.atomic to perform
any rollbacks in case in one critical step of the order we have
inconsistent data and we need to rollback or delete the data saved before in the database.
In each function we then use the forms we sent on the template in the context data, we use them to save or get errors in case something isn't filled out correctly. We do something like this:
def _create_channel(self, data):
form = ChannelForm(data)
if form.is_valid():
new_channel = form.save(commit=False)
new_channel.save()
return new_channel
else:
return JsonResponse(form.errors, status=400)
I hope the example above helps you.

My suggestion would be:
Build a stable(take your time here, you don't want any redundant data) database using django models.
Use Django Rest Framework to build proper serializers and validators against every model.
Route them to urls and perform other linking operations on your views.
Choose a javascript framework, you are comfortable with(Vue.js,Angular2,React which ever you want) and write the frontend separately(this part will have your multiple forms and stuff).
Use axios to post your data in django.
This is my personal suggestion, you can also go for django formsets but that will not be suitable in your example.

django - how to implement a 2-step publish mechanism

I'm new to both web development and django so maybe that's a noob question.
I want to do the following:
Ask user to fill some form and submit it.
Then, parse and format the content and display it back to the user to let him verify it.
User can accept the result or go back to the previous view, update data and resend.
This is as far as I can think:
views.py
def add_content(request):
if request.method == 'POST':
form = AddContentForm(request.POST)
if form.is_valid():
content = form.save(commit=False)
return verify_content(request, content)
else:
form = AddContentForm()
return render(request, 'myapp/add_content.html', {'form' : form})
def verify_content(request, content):
return render(request, 'myapp/verify_content.html', {'content' : content})
The verify_content template will obviously contain two buttons ('back', 'ok'), but I don't know how to pass the content object to a view for saving it in the db, or send it back to the previous view from there. Should I use js? Can i do it with just server side code?
Maybe my whole logic is wrong. Should I save the object in the db before verification and then delete it if needed (sounds ugly)? What is a good way to implement this?
Thanks in advance for your time.

You could use the users session for this:
request.session['content'] = content
and in the view where the user should verify his input do:
content = request.session['content']
and voilá you got the content between 2 views.
Django also secures that users can't tinker with its data by either saving it server side, or in a signed cookie.

I would save the form with commit=True in the add_content view, and would add a verified field or something to the model. Then you can append the pk as GET parameter to the link which will get you back to add_content view from verify. You can extract the parameter from request.GET dict.

Initialising the flask-wtf form when the page loads

Whenever the flask server is started, all the forms are initialised at that time itself. The data is interconnected between the pages, so for one form, the choices come from the database and those choices can be edited using another form on another page but after the choices are updated on that page, they remain the same for the first form. To get the new values I need to restart the server. Is there any way to refresh the values without restarting the server?
This is how the form looks like
class AddExpenses(Form):
reason = wtforms.StringField('reason', [validators.Required()])
amount = wtforms.IntegerField('amount', [validators.Required()])
allnames = []
allnames = getSalesman()
salesperson = wtforms.SelectField('salesperson', choices=[names for names in allnames])
submitfield = wtforms.SubmitField('Submit')
getSalesman() function is used to query the database and get the choices.

Why not get the choices in the view function that requires the form? For example
#app.route('/index')
def index():
form = AddExpenses()
allnames = getSalesman()
form.salesperson.choices = [names for names in allnames]
...
You may also want to use some caching on getSalesman() if the database will not change often.

WTForms: passing form values back to the user on post

When both the GET and POST methods are in the same handler class, and I want to populate form fields with user input after failed form validation, I do this...
Class CommentHandler(BaseHandler):
def get(self, form=None):
if form is None: # create new form unless populated form is passed in
form = CommentForm()
# query DB and create template context
self.render('page.html', **context)
def post(self):
form = CommentForm(self.request.POST)
if form.validate():
# populate entity with form data and save to DB
return self.redirect_to('page')
self.get(form=form) # pass populated form back to user for editing
I don't know if this is the best way to get form data back to the user, but it seems to work. My question is: how do I pass that data back into the form if the GET and POST methods are in different handler classes?
class PageHandler(BaseHandler):
def get(self):
# displays form to user
class CommentHandler(BaseHandler):
def post(self):
# processes POSTed form data...
# but if form.validate() fails,
# how can I pass the form data back to the user
# so they can edit their form input?

When you post and call validate() WTForms binds the form encoded data to the Form instance. An http POST can return a response just like a GET this is why you sometimes get those funny messages in your browser when a server application has failed validation and you try to refresh. Its because the refresh action is going to invoke the GET processing pipeline and you will lose your POST data.
What you need to do instead of delegating back to the get implementation you just need to render a response from your post implementation that passes Form instance with the data bound to it back in the response. If you have set your template up in the recommended way the data will automatically show up in the appropriate fields. Below is a snippet of what your Handler might look like.
def post(self):
form = CommentForm(self.request.POST)
if form.validate():
# populate entity with form data and save to DB
return self.redirect_to('page')
# If we are here it means we failed validation
# We need to send back the data the use supplied
# with error messages so we can re-render the form
# with their data and error messages indicating why
# it was rejected.
self.render('page.html', form=form) # pass populated form back to user for editing
This of course assumes that your page.html knows what to do with the Form instance.

Here's one workflow that may help to solve your problem. I chose to provide a workflow instead of a specific code snippet in order to avoid being too prescriptive in the technology you use to solve the problem.
GET
if session contains form data:
add form data from session to template context
display form
POST
if form validates:
clear this form data from session
save to db
else:
save form data to session
redirect to GET handler

How to redirect with post data (Django)

When processing a POST request in the Django views.py file, I sometimes need to redirect it to another url. This url I'm redirecting to is handled by another function in the same Django views.py file. Is there a way of doing this and maintaining the original POST data?
UPDATE: More explanation of why I want to do this.
I have two web apps (let's call them AppA and AppB) which accept data entered into a text field by the user. When the the user clicks submit, the data is processed and detailed results are displayed. AppA and AppB expect different types of data. Sometimes a user mistakenly posts AppB type data to AppA. When this happens I want to redirect them to AppB and show the AppB results or at least have it populated with the data they entered into AppA.
Also:
The client wants two separate apps rather than combining them into just one.
I can't show the code as it belongs to a client.
UPDATE 2:
I've decided that KISS is the best principle here. I have combined the two apps into one which makes things simpler and more robust; I should be able to convince the client it's the best way to go too. Thanks for all the great feedback. If I was going to maintain two apps as described then I think sessions would be the way to do this - thanks to Matthew J Morrison for suggesting that. Thanks to Dzida as his comments got me thinking about the design and simplification.

If you faced such problem there's a slight chance that you might need to revise your designs.
This is a restriction of HTTP that POST data cannot go with redirects.
Can you describe what are you trying to accomplish and maybe then we can think about some neat solution.
If you do not want use sessions as Matthew suggested you can pass POST params in GET to the new page (consider some limitations such as security and max length of GET params in query string).
UPDATE to your update:)
It sounds strange to me that you have 2 web apps and those apps use one views.py (am I right?). Anyway consider passing your data from POST in GET to the proper view (in case data is not sensitive of course).

I think how I would probably handle this situation would be to save the post data in session, then remove it when I no longer need it. That way I can access the original post data after a redirect even though that post is gone.
Will that work for what you're trying to do?
Here is a code sample of what I'm suggesting: (keep in mind this is untested code)
def some_view(request):
#do some stuff
request.session['_old_post'] = request.POST
return HttpResponseRedirect('next_view')
def next_view(request):
old_post = request.session.get('_old_post')
#do some stuff using old_post
One other thing to keep in mind... if you're doing this and also uploading files, i would not do it this way.

You need to use a HTTP 1.1 Temporary Redirect (307).
Unfortunately, Django redirect() and HTTPResponseRedirect
(permanent) return only a 301 or 302. You have to implement it yourself:
from django.http import HttpResponse, iri_to_uri
class HttpResponseTemporaryRedirect(HttpResponse):
status_code = 307
def __init__(self, redirect_to):
HttpResponse.__init__(self)
self['Location'] = iri_to_uri(redirect_to)
See also the django.http module.
Edit:
on recent Django versions, change iri_to_uri import to:
from django.utils.encoding import iri_to_uri

use requests package.Its very easy to implement
pip install requests
then you can call any urls with any method and transfer data
in your views import requests
import requests
to post data, follow the format
r = requests.post('http://yourdomain/path/', data = {'key':'value'})
to get the absolute url in django view, use
request.build_absolute_uri(reverse('view_name'))
Thus the django view code looks like
r = requests.post(
request.build_absolute_uri(reverse('view_name')),
data = {'key':'value'}
)
where r is the response object with status_code and content attribute.
r.status_code gives the status code(on success it will be 200) and r.content gives the body of response. There is a json method(r.json()) that will convert response to json format
requests
requests.post

Just call your new view from your old view using the same request object.
Of course it won't result in a redirect as such, but if all you care about is 'transferring' data from one view to the other, then it should work.
I tested the following snippet and it works.
from django.views.generic import View
class MyOldView(View):
def post(self, request):
return MyNewView().post(request)
class MyNewView(View):
def post(self, request):
my_data = request.body
print "look Ma; my data made it over here:", my_data

You can use render and context with with it:
Render(request,"your template path", {'vad name' : var value}
You can recive vars in template :
{% If var name %}
{{ var name }}
{% endif %}

I faced a similar issue recently.
Basically I had a form A, upon submitting it another form B would show up, which contains some results + a form. Upon submitting B, i wanted to display some alert to user and keep user on B only.
The way I solved this, is by displaying the results in a <output> field, in B.
<output name="xyz" value="xyz">{{xyz}}</output>
And I used the same view for A->B and B->B. Now I just had to distinguish if the request is coming from A or B and render accordingly.
def view1(request):
if "xyz" in request.POST:
# request from B
# do some processing
return render(request, 'page.html', {"xyz":request.POST["xyz"]})
else:
# request from A
res = foo() # some random function
return render(request, 'page.html', {"xyz":res})
But this only works if form B is small and not that dynamic.

If you are using a redirect after processing the POST to AppB, you can actually get away with calling the AppB method from the AppA method.
An Example:
def is_appa_request(request):
## do some magic.
return False or True
is_appb_request = is_appa_request
def AppA(request):
if is_appb_request(request):
return AppB(request)
## Process AppA.
return HttpResponseRedirect('/appa/thank_you/')
def AppB(request):
if is_appa_request(request):
return AppA(request)
## Process AppB.
return HttpResponseRedirect('/appb/thank_you/')
This should yield a transparent experience for the end-user, and the client who hired you will likely never know the difference.
If you're not redirecting after the POST, aren't you worried about duplicate data due to the user refreshing the page?

You can redirect with session using request.session["key"] as shown below:
# "views.py"
from django.shortcuts import redirect
def my_view(request):
# Here
request.session["message"] = "success"
return redirect("https://example.com")
# "index.html"
{{ request.session.message }} {# success #}