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Control requests to view and template output in django

This is a view for get all the records in the EducationalRecord model:
def all_education_resume(request):
RESUME_INFO['view'] = 'education'
educations_resume = EducationalRecord.objects.all().order_by('-created_date')
template = 'resumes/all_resume.html'
context = {'educations_resume': educations_resume, 'resume_info': RESUME_INFO}
return render(request, template, context)
Now, if I want to write exactly this view for other models (like job resumes, research resumes , etc.),
I must another view one separately.
My question is:
How can I get a view for all these requests, so first check the URL of
the request and then do the relevant query? How can I control URL
requests in my views?
My other question is exactly the same as my first question,with this difference:
control view that must render in specific template.In other words,in
second question the ratio between the template and the view is instead
of the ratio of the view to the url or how to create a template for
multiple views (for example, for a variety of database resume
resumes, I have a template) and then, depending on which view render,
the template output is different.
I have implemented these two issues as follows:
I wrote a view for each of request!
In each view, I set the value of RESUME_INFO['view'], and then I've checked it in a template page and specified the corresponding template.
What is the best solution to these two questions?

How can I get a view for all these requests, so first check the URL of the request and then do the relevant query? How can I control URL requests in my views?
You can access request.path, or you can let the url(..)s pass a parameter with kwargs that holds a reference to the model for example, but this is usually bad design. Typically if you use different models, you will likely have to order these different as well, filter these differently, render these differently, etc. If not, then this typically indicates that something is wrong with the modeling.
You can however make use of class-based views [Django-doc], to remove as much boilerplate as posssible. Your view looks like a ListView [Django-doc], by using such view, and patching where necessary, we can omit most of the "boilerplate" code:
# app/views.py
from django.views.generic.list import ListView
class MyBaseListView(ListView):
resume_info = None
template = 'resumes/all_resume.html'
def get_context_data(self, *args, **kwargs):
context = super().get_context_data(*args, **kwargs)
context['resume_info'] = {'view': self.resume_info}
return context
In the individual listviews, you then only need to specify the resume_info and the model or queryset to render it with the 'all_resume.html' template, for example:
# app/views.py
# ...
class EducationalResumeView(MyBaseListView):
queryset = EducationalRecord.objects.order_by('-created_date')
resume_info = 'education'
class OtherModelView(MyBaseListView):
model = OtherModel
resume_info = 'other_info'
So we can here use inheritance to define common things only once, and use it in multiple views. In case we need to change something in a specific view, we can override it at that level.
In the urls.py, you define such view with the .as_view() method [Django-doc]. For example:
# app/urls.py
from django.urls import path
from app.views import EducationalResumeView, OtherModelView
urlpatterns = [
path('education/', EducationalResumeView.as_view()),
path('other/', OtherModelView.as_view()),
]

Is it required to add custom views in admin page in ModelAdmin class when we can do it normally by adding views in views.py and urls in urls.py?

According to django docs:
class MyModelAdmin(admin.ModelAdmin):
def get_urls(self):
urls = super(MyModelAdmin, self).get_urls()
my_urls = [
url(r'^my_view/$', self.my_view),
]
return my_urls + urls
def my_view(self, request):
# ...
context = dict(
# Include common variables for rendering the admin template.
self.admin_site.each_context(request),
# Anything else you want in the context...
key=value,
)
return TemplateResponse(request, "sometemplate.html", context)
If I am not wrong, we can do the same thing by adding url in urls.py and the views in views.py as it is normally done then, what is the use of introducing this way? I am a newbie to django and I may be missing something here.
Can you please provide an example where we cannot do it in views.py and we must use the above method?
Any guidance/help would be appreciated.

I think I figured out, both of them can be used to do the same thing but the key difference is that the views which you write using above method will belong to admin app while the general views in views.py belongs to the particular app in you have written.
Hence, the url in ModelAdmin need to be called using name admin:url_name since the url goes as admin/my_views/ in given example.

How do I test a Django CreateView?

I want to practice testing on Django, and I have a CreateView I want to test. The view allows me to create a new post and I want to check if it can find posts without a publication date, but first I'm testing posts with published date just to get used to syntax. This is what I have:
import datetime
from django.test import TestCase
from django.utils import timezone
from django.urls import reverse
from .models import Post, Comment
# Create your tests here.
class PostListViewTest(TestCase):
def test_published_post(self):
post = self.client.post('/post/compose/', {'author':"manualvarado22", 'title': "Super Important Test", 'content':"This is really important.", 'published_date':timezone.now()})
response = self.client.get(reverse('blog:post_detail'))
self.assertContains(response, "really important")
But I get this:
django.urls.exceptions.NoReverseMatch: Reverse for 'post_detail' with no
arguments not found. 1 pattern(s) tried: ['post/(?P<pk>\\d+)/$']
How do I get the pk for that newly created post?
Thank you!

You can get it directly from the database.
Note, you shouldn't call two views in your test. Each test should only call the code it is actually testing, so this should be two separate views: one to call the create view and assert that the entry is in the db, and one that creates an entry directly and then calls the detail view to check that it displays. So:
def test_published_post(self):
self.client.post('/post/compose/', {'author':"manualvarado22", 'title': "Super Important Test", 'content':"This is really important.", 'published_date':timezone.now()})
self.assertEqual(Post.objects.last().title, "Super Important Test")
def test_display_post(self):
post = Post.objects.create(...whatever...)
response = self.client.get(reverse('blog:post_detail', pk=post.pk))
self.assertContains(response, "really important")

What you wanna do with testing is using the relyable Django Databse API for recevieng the created data and see if your view represents this data.
As you only create 1 model instance and save it. You may obtain its pk via
model_pk = Post.objects.get(author="manualvarado22").pk
This pk then should be inserted into your url as the Exception states.
But i also recommend abseconded test, where you directly check if the newly created "Post" exists in the DB via django model API.
Edit:
When testing, django or the test Module you are using, creates a clean database only for testing and destroys it after the test-run. So if you want to acces a User while testing you must have created the user in your Test setup or in the Test method itself. Otherwise the Usertable would be empty.

I was finally able to solve the issue thanks to your great answers as well as some extra SO research. This is how the test looks like:
def test_display_no_published_post(self):
test_user = User.objects.create(username="newuser", password="securetestpassword")
post = Post.objects.create(author=test_user, title="Super Important Test", content="This is really important.")
response = self.client.get(reverse('blog:post_detail', kwargs={'pk':post.pk}))
self.assertEqual(response.status_code, 404)
And this are the create and detail views:
class PostDetailView(DetailView):
model = Post
def get_queryset(self):
return Post.objects.filter(published_date__lte=timezone.now())
class PostCreateView(LoginRequiredMixin, CreateView):
login_url = '/login/'
redirect_field_name = 'blog/post_detail.html'
form_class = PostForm
model = Post

Saving a Form to a Database (Django Python)

In my Django app I have a form with an email and a text area that needs to go into the database but am struggeling to do so. At the moment I have 2 different solutions in my code:

OK, let me write this from the beginning. Maybe it is better if you do the below steps with me now for an exercise.
Now, the correct way to do the above task (I try to clear this subject a little more because I know that many other people will read this problem and they can learn from this too. And it is important to understand it for your future tasks.
If you want to create a Form and you know that you will want to save the submitted data from that Form to the database, then of course you should start the whole task with creating a Model, thus a table for that in the database.
So, first you create a Model (which you will call ie. “Questions” in this case, since you want to call your Form ie. QuestionForm, so it is better if your Model and table will be related to that with their names too).
So your Model will be in your models.py file:
from django.db import models
# Create your models here.
class Questions(models.Model):
contact_email = models.EmailField(max_length=60)
question = models.CharField(max_length=600)
Then you will create a ModelForm from this in your forms.py file the following way:
from django import forms
from django.forms import ModelForm, Textarea
from . import models
class QuestionForm(ModelForm):
class Meta:
model = models.Questions
fields = ['contact_email', 'question']
widgets = {
'question': Textarea(attrs={'cols': 40, 'rows': 20}),
}
Then you create your view function in your views.py file:
from django.shortcuts import render, redirect
from . import forms
from . import models
from django.http import HttpResponse
def get_question(request):
form = forms.QuestionForm()
if request.method == 'POST':
form = forms.QuestionForm(request.POST)
if form.is_valid():
form.save()
return redirect(‘success.html’) # or you redirect anywhere you want to
else:
form = forms.QuestionForm()
return render(request, 'contact.html', {'form':form})
And at this point you will create your urlpattern in your urls.py to call the get_question view function. It will look like the following:
from django.conf.urls import url
from basic_app import views
# app_name = 'basic_app' # Important for referencing urls in HTML pages(use your own app_name here). But for this example task this is not important.
urlpatterns = [
url(r'^$', views.home, name='home'),
url(r'^questions/', views.get_question, name="questions_page"),
]
I hope I did not confuse you more. If you do the above steps, it should work for you. And you can create as many Forms as you want with the above steps and saving them easily in the Database.
The only other thing you have to have to run the above, is your 'contact.html' page which you already have and you already created.
(do not forget to run: python manage.py migrate)
So, I hope that you see, in the above example you do not mismatch fields and field names, and you do not get confused about what to save where. Since the Model and the Form is working together and created together with the same field names.

Dynamically filter Haystack SearchQuerySet based on the user

I need to filter Haystack's SearchQuerySet based on the current user. I know I can get the current user via request.user in other views, so I imagine I need to override Haystack's default SearchView, but I'm not sure how. Below is a sample of my urls.py and views.py.
urls.py:
urlpatterns = patterns('mysite.search.views',
url(r'^mymodel/$', search_view_factory(
view_class=MySearchView,
template='mymodel_search.html',
form_class=MyModelSearchForm,
searchqueryset = SearchQuerySet().models(MyModel),
), name='mymodel_search'),
...
views.py:
class MySearchView(SearchView):
def build_form(self, form_kwargs=None):
prop = UserProfile.objects.get(user = self.request.user).property
self.searchqueryset = self.searchqueryset.filter(property = prop)
return super(MySearchView, self).build_form(form_kwargs)
This doesn't seem to be working, though. Results are limited to MyModel (as filtered in urls.py), but the filtering in MySearchView doesn't seem to do anything. Is there a different way or place I should be doing this?

It was my own mistake. In my custom MyModelSearchForm's search() method, I was resetting the search query set to SearchQuerySet().models(MyModel), so it replaced all previous filtering. The method I was attempting with MySearchView works.