Develop Reference

How to save multiple objects in Django models at once

What I'm trying to do is saving multiple scraped data(actor names) in my Actor table inside Django models.
So far I've written the for loop below to achieve this goal but it only saves the last object.
class Actor(models.Model):
name = models.CharField(max_length=50)
url = models.URLField()
def __str__(self):
return self.name
def save(self, *args, **kwargs):
i = 0
for num in range(5):
title, year, actor_list = scraper(self.url)
self.name = actor_list[i]
super().save(*args, **kwargs)
i += 1
I get the URL from users within a form, then save the form and send it to my models and the scraping begins.
I scrape 5 actor names inside the scraper() function and append them to actor_list.
But when I try to save the actor_list under the save function, it only saves the 5th actor which I think overwrites the previously saved objects.
Is there something wrong with my for loop? or should I completely change my approach for this purpose?
I'd also prefer to save the objects using Actor.objects.get_or_create() to skip the already existing objects, but I don't know how and where.
I would appreciate it if someone could help me with this.

Yes, you should use a completely different approach. A Model is just a code representation of your database table.
You are also executing the same scraping 5 times on the same url, but that is another point.
Doing scraping and creating multiple actors should not be done within the model. Bear in mind that the model gets instantiated as an actor object (a single record in the actor table), so every time you do a super().save() it overwrites the same record.
The standard way is to do it in the view you use to print the form and have that form submit to its own URL (i.e. the form action should point to the same view you use to print it).
Something like this:
forms.py
from django import forms
class MyForm(forms.Form):
url = forms.CharField(label='Your name', max_length=255)
views.py
from my_app.forms import MyForm
from my_app.models import Actor
from django.shortcuts import render
def my_view(request):
if request.method == "POST":
form = MyForm(request.POST)
if form.is_valid():
title, year, actor_list = scraper(form.cleaned_data['url'])
for actor in actor_list:
# Here you are creating new instances of Actor (new records)
# at each iteration, unless they already exist.
Actor.objects.get_or_create(name=actor)
else:
form = MyForm()
return render(request, 'your_template.html', {'form': form})

The save() method acts on a single Actor object. It's wrong to try to save multiple actors there.
Do this in a view:
title, year, actor_list = scraper(self.url)
for i in range(5):
actor = Actor()
actor.name = actor_list[i]
actor.save()
And delete the save() method in the model

How to access object count in template using model manager?

I have a model which creates Memo objects. I would like to use a custom Model Manager's posted method to return the total number of Memo objects - then use this number within a template. I am trying to keep as much of my code as possible within my Models and Model Managers and less within my Views as I read that this was a best practice in 'Two Scoops of Django'.
In the shell I can get the number of memos as such:
>>> from memos.models import Memo
>>> Memo.objects.all()
<QuerySet [<Memo: Test Memo 2>, <Memo: Test Memo 1>]>
>>> Memo.objects.all().count()
2
This is what my Model and Model Manager look like:
class MemoManager(models.Manager):
use_for_related_fields = True
def posted(self):
return self.count()
class Memo(models.Model):
title = models.CharField(max_length=100)
content = models.TextField()
date_time = models.DateTimeField(default=timezone.now)
author = models.ForeignKey(User, on_delete=models.CASCADE)
objects = MemoManager()
def __str__(self):
return self.title
def get_absolute_url(self):
return reverse('memos-detail', kwargs={'pk': self.pk})
I know this is clearly the wrong way to do it but I have confused myself here. So how do I use my Model Manager to get the count of objects and use it in a template like: {{ objects.all.count }}?
P.S. I see other posts that show how to do this within the view but as stated I am trying not to use the view. Is using the view required? I also understand my posted method is written incorrectly.

I'm sorry but you have misinterpreted what was written in TSD. The Lean View Fat Model is meant to keep code which pertains to 'business logic' out of the views, and certain model specific things. A request should be handled by a view. So when you want to load a template, you must first have a GET request to your app.
A view function should be written such that Validation of POST data or the Creation of a new object in DB or Querying/Filtering for GET requests should be handled in the corresponding serializer/model/model manager.
What should be happening while you want to load your template.
Have a url for the template that you have created and a view function mapped for it
In the view function you should render said template and pass the necessary data inside the context.
To keep in line with the Lean View Fat Model style, if you want to get a Queryset of of Memo's but only those which have their is_deleted fields set to False, you can overwrite the model manager get_queryset() method for Memo model.
If you want to create a new Memo with a POST request, you can handle
the creation using a ModelForm!
Hope this clears things up!
EDIT:
How to pass a context to a template, in your case the memo count.
def random_memo_view(request):
context = {'memo_count': Memo.posted()}
return render(request, 'template.html', context=context)
RE-EDIT
I just checked that you were using DetailView. In this case follow this from the django docs.
Class Based Views: Adding Extra Context

Django REST: serialize url to list of objects of a category

Hard facts:
I am using Django 2.0 with python 3.6, if it makes any difference.
What I am trying to achieve is a link to a list of objects that belong to a summary.
I have a ManyToOne relationship in my models.py.
class Summary(models.model):
type=models.CharField
class Object(models.Model):
summary= models.ForeignKey(Summary, on_delete=models.CASCADE)
in urls.py
object_list= views.ObjectListViewSet.as_view({
'get': 'list'
})
urlpatterns = format_suffix_patterns([
url(r'^summary/(?P<pk>[^/.]+)/objects/$', object_list, name='summary-objects')
])
and now the idea was to give a user the possibility to click the an url in the browsable API and getting all objects.
So, I tried to write a MethodField in serializers.py. I am not able to get any reasonable URL here, the only solution would be to hardcode it.
class SummarySerializer(serializers.HyperlinkedModelSerializer):
url = serializers.HyperlinkedIdentityField(
view_name="app:summary-detail")
objects= serializers.SerializerMethodField('get_obj_url')
def get_obj_url(self, obj):
pass
class Meta:
model = Summary
Is this possible?
Is it necessary to write a MethodField?
If yes, how do I get the url I need?

Actually, reverse, as suggested in the comments, does the trick.
The solution is:
def get_obj_url(self, obj):
request = self.context.get('request')
return request.build_absolute_uri(reverse('api-root')) + 'summary/{id}/objects'.format(
id=obj.id)
EDIT:Typo

Passing form data from my view to thank you page

Is there a way I can pass data from a form submission over to the 'thank you' page. The reason i'd like to do this is because I have a form on the website, where the user will select multiple fields which all contains different PDF's.
So once the user has submitted the form the idea is to re-direct them to a thankyou page, where they can view the list of pdf/files they have selected on the form.
I hope this is enough info to go on. Here are my views / models.
def document_request(request, *args):
# template = kwargs['name'] + ".html"
if request.method == 'POST':
form = forms.ReportEnquiryForm(request.POST)
print(request.POST)
if form.is_valid():
docrequest = form.save()
return HttpResponseRedirect(reverse('thank_you', kwargs={'id': docrequest.id}))
else:
form = forms.ReportEnquiryForm()
return render_to_response('test.html',{'form':form})
def thank_you(request):
docrequest = DocumentRequest.objects.get(pk=id)
return render_to_response('thankyou.html',
{'docrequest' : docrequest },
context_instance=RequestContext(request))
My initial idea was to pass the data to a new view called thank_you. But not this is possible.
class DocumentUpload(models.Model):
name = models.CharField(max_length="200")
document_upload = models.FileField(upload_to="uploads/documents")
def __unicode__(self):
return "%s" % self.name
class DocumentRequest(models.Model):
name = models.CharField(max_length="200")
company = models.CharField(max_length="200")
job_title = models.CharField(max_length="200")
email = models.EmailField(max_length="200")
report = models.ManyToManyField(DocumentUpload)
def __unicode__(self):
return "%s" % self.name
form.py
class ReportEnquiryForm(forms.ModelForm):
class Meta:
model = models.DocumentRequest
fields = ('name', 'company', 'job_title', 'email', 'report')
If you need anymore info, please ask :)

You've saved the user's submission in a DocumentRequest object. So you can pass the ID of that object in the URL when you redirect, and in the thank_you view you can get the DocumentRequest and render the list.
Edit The idea is to make the thank_you page like any other view that accepts a parameter from the URL:
url(r'thanks/(?P<id>\d+)/$, 'thank_you', name='thank_you')
and so the POST part of the form view becomes:
if form.is_valid():
docrequest = form.save()
return HttpResponseRedirect(reverse('thank_you', kwargs={'id': docrequest.id}))
and thank_you is:
def thank_you(request, id):
docrequest = DocumentRequest.objects.get(pk=id)
return render_to_response('thankyou.html',
{'docrequest' : docrequest },
context_instance=RequestContext(request))
Second edit
As others have suggested, this makes it possible for anyone to see the request. So a better solution is to put it in the session:
docrequest = form.save()
request.session['docrequest_id'] = docrequest.id
and in thank_you:
def thank_you(request):
if not 'docrequest_id' in request.session:
return HttpResponseForbidden
docrequest = DocumentRequest.objects.get(request.session['docrequest_id'])

You can do as Daniel Roseman said but in this case the thank you pages can be accessed by anyone with the Ids.
Some ways to pass data between views are the following(the list is not mine.):
GET request - First request hits view1->send data to browser -> browser redirects to view2
POST request - (as you suggested) Same flow as above but is suitable when more data is involved
Using django session variables - This is the simplest to implement
Using client-side cookies - Can be used but there is limitations of how much data can be stored.
Maybe using some shared memory at web server level- Tricky but can be done.
Write data into a file & then the next view can read from that file.
If you can have a stand-alone server, then that server can REST API's to invoke views.
Again if a stand-alone server is possible maybe even message queues would work.
Maybe a cache like memcached can act as mediator. But then if one is going this route, its better to use Django sessions as it hides a whole lot of implementation details.
Lastly, as an extension to point 6, instead of files store data in some persistent storage mechanism like mysql.
The simplest way is to use sessions. Just add the id to the session and redirect to the thank you view, you read the id value and query the db with that id.

Inline-like solution for Django Admin where Admin contains ForeignKey to other model

I have several Customers who book Appointments. Each Appointment has exactly one customer, though a customer can be booked for multiple appointments occurring at different times.
class Customer(model.Model):
def __unicode__(self):
return u'%s' % (self.name,)
name = models.CharField(max_length=30)
# and about ten other fields I'd like to see from the admin view.
class Appointment(models.Model):
datetime = models.DateTimeField()
customer = models.ForeignKey("Customer")
class Meta:
ordering = ('datetime',)
Now when an admin goes to browse through the schedule by looking at the Appointments (ordered by time) in the admin, sometimes they want to see information about the customer who has a certain appointment. Right now, they'd have to remember the customer's name, navigate from the Appointment to the Customer admin page, find the remembered Customer, and only then could browse their information.
Ideally something like an admin inline would be great. However, I can only seem to make a CustomerInline on the Appointment admin page if Customer had a ForeignKey("Appointment"). (Django specifically gives me an error saying Customer has no ForeignKey to Appointment). Does anyone know of a similar functionality, but when Appointment has a ForeignKey('Customer')?
Note: I simplified the models; the actual Customer field currently has about ~10 fields besides the name (some free text), so it would be impractical to put all the information in the __unicode__.

There is no easy way to do this with django. The inlines are designed to follow relationships backwards.
Potentially the best substitute would be to provide a link to the user object. In the list view this is pretty trivial:
Add a method to your appointment model like:
def customer_admin_link(self):
return 'Customer' % reverse('admin:app_label_customer_change %s') % self.id
customer_admin_link.allow_tags = True
customer_admin_link.short_description = 'Customer'
Then in your ModelAdmin add:
list_display = (..., 'customer_admin_link', ...)
Another solution to get exactly what you're looking for at the cost of being a bit more complex would be to define a custom admin template. If you do that you can basically do anything. Here is a guide I've used before to explain:
http://www.unessa.net/en/hoyci/2006/12/custom-admin-templates/
Basically copy the change form from the django source and add code to display the customer information.

Completing #John's answer from above - define what you would like to see on the your changelist:
return '%s' % (
reverse('admin:applabel_customer_change', (self.customer.id,)),
self.customer.name # add more stuff here
)
And to add this to the change form, see: Add custom html between two model fields in Django admin's change_form

In the ModelAdmin class for your Appointments, you should declare the following method:
class MySuperModelAdmin(admin.ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
if obj:
# create your own model admin instance here, because you will have the Customer's
# id so you know which instance to fetch
# something like the following
inline_instance = MyModelAdminInline(self.model, self.admin_site)
self.inline_instances = [inline_instance]
return super(MySuperModelAdmin, self).get_form(request, obj, **kwargs)
For more information, browser the source for that function to give you an idea of what you will have access to.
https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/options.py#L423

There is a library you can use it.
https://github.com/daniyalzade/django_reverse_admin
But if you want to use link to object in showing table you can like this code:
def customer_link(self, obj):
if obj.customer:
reverse_link = 'admin:%s_%s_change' % (
obj.customer._meta.app_label, obj.customer._meta.model_name)
link = reverse(reverse_link, args=[obj.customer.id])
return format_html('More detail' % link)
return format_html('<span >-</span>')
customer_link.allow_tags = True
customer_link.short_description = 'Customer Info'
And in list_display:
list_display = (...,customer_link,...)