my mission is to get a list of articles. This article come form a simple FeinCMS ContentType.
class Article(models.Model):
image = models.ForeignKey(MediaFile, blank=True, null=True, help_text=_('Image'), related_name='+',)
content = models.TextField(blank=True, help_text=_('HTML Content'))
style = models.CharField(
_('template'),max_length=10, choices=(
('default', _('col-sm-7 Image left and col-sm-5 Content ')),
('fiftyfifty', _('50 Image left and 50 Content ')),
('around', _('small Image left and Content around')),
),
default='default')
class Meta:
abstract = True
verbose_name = u'Article'
verbose_name_plural = u'Articles'
def render(self, **kwargs):
return render_to_string('content/articles/%s.html' % self.style,{'content': self,})
I would like to use that in different subpages.
Now it would be great to get a list of all articels on the main page (my projects -> list of project1, project2, project3, ).
Something like: Article.objects.all()
Template:
{% for entry in article %}
{% if content.parent_id == entry.parent_id %} #only projects
<p>{{ entry.content|truncatechars:180 }}</p>
{% endif %}
{% endfor %}
but i get a error "type object 'Articels' has no attribute 'objects'...
Do you have a smart idea? It would be grade to use Feincms ContentType.
FeinCMS content types are abstract, that means there is no data and no database table associated with them. Therefore, there's no objects manager and no way to query.
When doing Page.create_content_type(), FeinCMS takes the content type and the corresponding Page class and creates a (non-abstract) model which contains the actual data. In order to access that new, concrete model, you need to use content_type_for. In other words, you're looking for:
from feincms.module.page.models import Page
PageArticle = Page.content_type_for(Article)
articles = PageArticle.objects.all()
Related
I'm struggling getting the right query for my project. Here is an example or my model :
from django.db import models
class Pictures(models.Model):
name = models.CharField(max_length=100)
bild = models.FileField(upload_to='article_pictures/')
articel = models.ForeignKey('articles', on_delete=models.CASCADE)
def __str__(self):
return self.name
class Articles(models.Model):
name = models.CharField(max_length=100)
text = models.TextField(max_length=2000)
published = models.BooleanField(default=False)
def __str__(self):
return self.name
how do I get the published artikles from the artikles class including the pictures (if there is one, or more)?
Thank you for your help
I don't think there is any exact query for this, but you can use prefetch_related to pre-load data from database. For example:
articles = Artikles.objects.filter(published=True).prefetch_related('pictures_set')
for article in articles:
article.pictures_set.all() # will not hit database
All published articles:
Articles.objects.filter(published=True)
A single published Article(Example):
article = Articles.objects.filter(published=True).first()
# and it's pictures
for picture in article.pictures_set.all():
print(picture)
Note: models have singular names, so you should rename Articles to Article and Pictures to Picture.
The related Pictures of an article article can be obtained with:
my_article.picture_set.all()
this is a queryset that contains all the related pictures.
We can obtain the Articles that are publised, and then fetch the related Pictures in two extra queries with:
articles = Article.objects.filter(published=True).prefetch_related('picture_set')
So in a template you can then render it like:
{% for article in articles %}
{{ article.name }}
{% for picture in article.picture_set.all %}
{{ picture.name }}
{% endfor %}
{% endfor %}
I am compiling a database of articles and have my model set up like this:
class articles(models.Model):
ArticleID = models.IntegerField(primary_key=True)
Title = models.CharField(max_length=500)
Author = models.CharField(max_length=200, null=True)
Journal = models.CharField(max_length=500, null=True)
Date = models.IntegerField(null=True)
Issue = models.IntegerField(null=True)
Link = models.URLField(max_length=800, null=True)
Content = models.TextField()
class Meta:
db_table = 'TEST'
def __str__(self):
return f'{self.Title}, {self.Author}, {self.Journal},{self.Date}, {self.Issue}, {self.Content}'
def get_absolute_url(self):
return reverse('article-detail', args=[str(self.ArticleID)])
The idea is pretty simple. Each meta data type (i.e. title, author) is it's own field, and the actual content of the article is in the field Content.
My view for this model:
def article_detail(request, ArticleID):
ArticleID = get_object_or_404(articles, ArticleID=ArticleID)
context = {'ArticleID': ArticleID}
return render(request, 'article_detail.html', context)
The HTML template for the view:
{% extends 'base.html' %}
{% block content %}
<div class="container">
{{ ArticleID }}
</div>
{% endblock %}
The data displayed in on the HTML page is one big block of text in one single HTML element. How can I make it so that I can use CSS to target each individual field from the model? Must I make separate models for each field (and bound them with foreign keys)?
No of course not. You can access fields with normal dot notation: ArticleID.Title, ArticleID.Author, etc.
(But you shouldn't call your context variable ArticleID; it's the whole article, not the ID. Also, Python style is to use lower_case_with_underscore for variables and attribute names.)
There is already a primary key in for every model which is called id, you don't have to explicitly declare that.
Secondly you are getting an article object with get_object_or_404 so if you use . (dot) notation you will get your desired value in your template.
Something like-
<h2>{{article.Title}}</h2>
<p>{{article.Content}}</p>
though you have to send article names instead of ArticleID in context variable.
In addition to Mr. Daniel Roseman's comment you should use class name Article instead of articles which is not pythonic.
I have three different models for my app. All are working as I expected.
class Tender(models.Model):
title = models.CharField(max_length=256)
description = models.TextField()
department = models.CharField(max_length=50)
address = models.CharField(max_length=50)
nature_of_work = models.CharField(choices=WORK_NATURE, max_length=1)
period_of_completion = models.DateField()
pubdat = models.DateTimeField(default=timezone.now)
class Job(models.Model):
user = models.ForeignKey(settings.AUTH_USER_MODEL)
title = models.CharField(max_length=256)
qualification = models.CharField(max_length=256)
interview_type = models.CharField(max_length=2, choices=INTERVIEW_TYPE)
type_of_job = models.CharField(max_length=1, choices=JOB_TYPE)
number_of_vacancies = models.IntegerField()
employer = models.CharField(max_length=50)
salary = models.IntegerField()
pubdat = models.DateTimeField(default=timezone.now)
class News(models.Model):
user = models.ForeignKey(settings.AUTH_USER_MODEL)
title = models.CharField(max_length=150)
body = models.TextField()
pubdat = models.DateTimeField(default=timezone.now)
Now I am displaying each of them at separate page for each of the model (e.g. in the jobs page, I am displaying only the jobs.). But now at the home page, I want to display these according to their published date at the same page. How can I display different objects from different models at the same page? Do I make a separate model e.g. class Post and then use signal to create a new post whenever a new object is created from Tender, or Job, or News? I really hope there is a better way to achieve this. Or do I use multi-table inheritance? Please help me. Thank you.
Update:
I don't want to show each of the model objects separately at the same page. But like feeds of facebook or any other social media. Suppose in fb, any post (be it an image, status, share) are all displayed together within the home page. Likewise in my case, suppose a new Job object was created, and after that a new News object is created. Then, I want to show the News object first, and then the Job object, and so on.
A working solution
There are two working solutions two other answers. Both those involve three queries. And you are querying the entire table with .all(). The results of these queries combined together into a single list. If each of your tables has about 10k records, this is going to put enormous strain on both your wsgi server and your database. Even if each table has only 100 records each, you are needlessly looping 300 times in your view. In short slow response.
An efficient working solution.
Multi table inheritance is definitely the right way to go if you want a solution that is efficient. Your models might look like this:
class Post(models.Model):
title = models.CharField(max_length=256)
description = models.TextField()
pubdat = models.DateTimeField(default=timezone.now, db_index = True)
class Tender(Post):
department = models.CharField(max_length=50)
address = models.CharField(max_length=50)
nature_of_work = models.CharField(choices=WORK_NATURE, max_length=1)
period_of_completion = models.DateField()
class Job(Post):
user = models.ForeignKey(settings.AUTH_USER_MODEL)
qualification = models.CharField(max_length=256)
interview_type = models.CharField(max_length=2, choices=INTERVIEW_TYPE)
type_of_job = models.CharField(max_length=1, choices=JOB_TYPE)
number_of_vacancies = models.IntegerField()
employer = models.CharField(max_length=50)
salary = models.IntegerField()
class News(models.Model):
user = models.ForeignKey(settings.AUTH_USER_MODEL)
def _get_body(self):
return self.description
body = property(_get_body)
now your query is simply
Post.objects.select_related(
'job','tender','news').all().order_by('-pubdat') # you really should slice
The pubdat field is now indexed (refer the new Post model I posted). That makes the query really fast. There is no iteration through all the records in python.
How do you find out which is which in the template? With something like this.
{% if post.tender %}
{% else %}
{% if post.news %}
{% else %}
{% else %}
Further Optimization
There is some room in your design to normalize the database. For example it's likely that the same company may post multiple jobs or tenders. As such a company model might come in usefull.
An even more efficient solution.
How about one without multi table inheritance or multiple database queries? How about a solution where you could even eliminate the overhead of rendering each individual item?
That comes with the courtesy of redis sorted sets. Each time you save a Post, Job or News, object you add it to a redis sorted set.
from django.db.models.signals import pre_delete, post_save
from django.forms.models import model_to_dict
#receiver(post_save, sender=News)
#receiver(post_save, sender=Post)
#receiver(post_save, sender=Job)
def add_to_redis(sender, instance, **kwargs):
rdb = redis.Redis()
#instead of adding the instance, you can consider adding the
#rendered HTML, that ought to save you a few more CPU cycles.
rdb.zadd(key, instance.pubdat, model_to_dict(instance)
if (rdb.zcard > 100) : # choose a suitable number
rdb.zremrangebyrank(key, 0, 100)
Similarly, you need to add a pre_delete to remove them from redis
The clear advantage of this method is that you don't need any database queries at all and your models continue to be really simple + you get catching thrown in the mix. If you are on twitter your timeline is probably generated through a mechanism similar to this.
The following should do want you need. But to improve performance you can create an extra type field in each of your models so the annotation can be avoided.
Your view will look something like:
from django.db.models import CharField
def home(request):
# annotate a type for each model to be used in the template
tenders = Tender.object.all().annotate(type=Value('tender', CharField()))
jobs = Job.object.all().annotate(type=Value('job', CharField()))
news = News.object.all().annotate(type=Value('news', CharField()))
all_items = list(tenders) + list(jobs) + list(news)
# all items sorted by publication date. Most recent first
all_items_feed = sorted(all_items, key=lambda obj: obj.pubdat)
return render(request, 'home.html', {'all_items_feed': all_items_feed})
In your template, items come in the order they were sorted (by recency), and you can apply the appropriate html and styling for each item by distinguishing with the item type:
# home.html
{% for item in all_items_feed %}
{% if item.type == 'tender' %}
{% comment "html block for tender items" %}{% endcomment %}
{% elif item.type == 'news' %}
{% comment "html block for news items" %}{% endcomment %}
{% else %}
{% comment "html block for job items" %}{% endcomment %}
{% endif %}
{% endfor %}
You may avoid the annotation altogether by using the __class__ attribute of the model objects to distinguish and put them in the appropriate html block.
For a Tender object, item.__class__ will be app_name.models.Tender where app_name is the name of the Django application containing the model.
So without using annotations in your home view, your template will look:
{% for item in all_items_feed %}
{% if item.__class__ == 'app_name.models.Tender' %}
{% elif item.__class__ == 'app_name.models.News' %}
...
{% endif %}
{% endfor %}
With this, you save extra overhead on the annotations or having to modify your models.
A straight forward way is to use chain in combination with sorted:
View
# your_app/views.py
from django.shortcuts import render
from itertools import chain
from models import Tender, Job, News
def feed(request):
object_list = sorted(chain(
Tender.objects.all(),
Job.objects.all(),
News.objects.all()
), key=lambda obj: obj.pubdat)
return render(request, 'feed.html', {'feed': object_list})
Please note - the querysets mentioned above using .all() should be understood as placeholder. As with a lot of entries this could be a performance issue. The example code would evaluate the querysets first and then sort them. Up to some hundreds of records it likely will not have a (measurable) impact on performance - but in a situation with millions/billions of entries it is worth looking at.
To take a slice before sorting use something like:
Tender.objects.all()[:20]
or use a custom Manager for your models to off-load the logic.
class JobManager(models.Manager):
def featured(self):
return self.get_query_set().filter(featured=True)
Then you can use something like:
Job.objects.featured()
Template
If you need additional logic depending the object class, create a simple template tag:
#templatetags/ctype_tags.py
from django import template
register = template.Library()
#register.filter
def ctype(value):
return value.__class__.__name__.lower()
and
#templates/feed.html
{% load ctype_tags %}
<div>
{% for item in feed reversed %}
<p>{{ item|ctype }} - {{ item.title }} - {{ item.pubdat }}</p>
{% endfor %}
</div>
Bonus - combine objects with different field names
Sometimes it can be required to create these kind of feeds with existing/3rd party models. In that case you don't have the same fieldname for all models to sort by.
DATE_FIELD_MAPPING = {
Tender: 'pubdat',
Job: 'publish_date',
News: 'created',
}
def date_key_mapping(obj):
return getattr(obj, DATE_FIELD_MAPPING[type(obj)])
def feed(request):
object_list = sorted(chain(
Tender.objects.all(),
Job.objects.all(),
News.objects.all()
), key=date_key_mapping)
Do I make a separate model e.g. class Post and then use signal to
create a new post whenever a new object is created from Tender, or
Job, or News? I really hope there is a better way to achieve this. Or
do I use multi-table inheritance?
I don't want to show each of the model objects separately at the same
page. But like feeds of facebook or any other social media.
I personally don't see anything wrong using another model, IMHO its even preferable to use another model, specially when there is an app for that.
Why? Because I would never want to rewrite code for something which can be achieved by extending my current code. You are over-engineering this problem, and if not now, you are gonna suffer later.
An alternative solution would be to use Django haystack:
http://haystacksearch.org/
http://django-haystack.readthedocs.io/en/v2.4.1/
It allows you to search through unrelated models. It's more work than the other solutions but it's efficient (1 fast query) and you'll be able to easily filter your listing too.
In your case, you will want to define pubdate in all the search indexes.
I cannot test it right now, but you should create a model like:
class Post(models.Model):
pubdat = models.DateTimeField(default=timezone.now)
tender = models.ForeignKey('Tender')
job = models.ForeignKey('Job')
news = models.ForeignKey('News')
Then, each time a new model is created, you create a Post as well and relate it to the Tender/Job/News. You should relate each post to only one of the three models.
Create a serializer for Post with indented serializers for Tender, Job and News.
Sorry for the short answer. If you think it can work for your problem, i'll write more later.
I have my models like this:
class Personne(BaseModel):
# [skip] many fields then:
photos = models.ManyToManyField(Photo, blank=True,
through='PersonnePhoto',
symmetrical=False,
related_name='parent')
def photo_profil(self):
a = PersonnePhoto.objects.filter(
personne=self, photo_type=PersonnePhoto.PHOTO_PROFIL)
return a[0] if len(a) else None
# [skip] many fields then:
travels = models.ManyToManyField(
TagWithValue, blank=True,
through='PersonneTravel',
default=None, symmetrical=False,
related_name='personne_travel')
class PersonneRelation(BaseModel):
src = models.ForeignKey('Personne', related_name='src')
dst = models.ForeignKey('Personne', related_name='dst')
opposite = models.ForeignKey('PersonneRelation',
null=True, blank=True, default=None)
is_reverse = models.BooleanField(default=False)
I need to show all contacts of one person, and for each contact, show all his/her travels, and his/her photo.
Here's my code in my view:
class IndexView(LoginRequiredMixin, generic.TemplateView):
template_name = 'my_home/contacts/index.html'
def get_context_data(self, **kwargs):
context = super(IndexView, self).get_context_data(**kwargs)
context['contacts'] = Personne.objects.get(
user=self.request.user).relations.all()
return context
Pretty simple.
The problem is in my template. my_home/contacts/index.html includes contact_detail.html for each contact:
{% for c in contacts %}
{% with c as contact %}
{% include 'includes/contact_detail.html' %}
{% endwith %}
{% endfor %}
In contact_detail.html, I call photo_profil, which makes a query to get the value of the picture of the contact.
{{ contact.photo_profil }}
This implies that if a user has 100 contacts, I will do one query for all his contacts, then 100 queries for each contact. How is it possible to optimize this? I have the same problem for travels, and the same problem for each ManyToMany fields of the contact, actually.
Looks like you need some prefetch_related goodness:
context['contacts'] = (Personne.objects
.get(user=self.request.user)
.relations
.all()
.prefetch_related('travels', 'photos'))
Note, however, that it will fetch all the contacts' photos, which is not you seem to expect. Basically you have two options here. Either add some hairy raw SQL to the Prefetch object's queryset parameter. Or (as I did in one of the projects) designate a separate main_photo field for storing the user's avatar (a separate ForeignKey to the Photo, I mean, not the completely independent one). Yes, it's clearly a denormalization, but queries become much more simple. And your users get a way to set a main photo, after all.
I have constructed a view that subclasses Django's ListView. Currently it works fine. It provides a list of objects from my Artwork class that I can iterate over in my template.
However, I've now included a (not required) foreign key (series) and I would like to group the Artwork objects by series when I display them. I would also like to order them by date while interspersing any Artwork objects with series=None amongst them (according to the date also).
Like this:
artwork(12/2013) (series=None)
artwork(12/2013) artwork(12/2013) artwork(11/2013) (series='series1' 11/2013)
artwork(10/2013) (series=None)
artwork(12/2013) artwork(12/2013) artwork(11/2013) (series='series2' 09/2013)
My models look like this:
class Series(models.Model):
title = models.CharField(max_length=200, unique=True)
date = models.DateTimeField(default=timezone.now(), editable=False)
def __unicode__(self):
return self.title
class Artwork(models.Model):
title = models.CharField(max_length=200, unique=True)
series = models.ForeignKey(Series, blank=True, null=True, on_delete=models.SET_NULL)
date = models.DateTimeField(default=timezone.now(), editable=False)
image = models.ImageField(upload_to='artwork_images/%Y/%m/')
def __unicode__(self):
return self.title
I'm guessing I could somehow accomplish what I want if my view were to provide the template with something like a nested list. Is this possible with the ListView? Or should I maybe be constructing a function based view?
Apologies if this seems like a naive question. I'm new to the Django/Python world (and Stackoverflow!). Any advice or suggestions would be very much appreciated.
I would just use a normal function based view for this, but if you want you can use ListView's get_context_data() method to return arbitrary data.
Create a key function for sorting:
def key_function(artwork):
if artwork.series is None:
return (artwork.date, None)
else:
return (artwork.series.date, artwork.series)
The function uses serie's date if avialiable and artwork's date otherwise. Series id is a second part of the key (when available), so artworks belonging to the same series are kept next to each other, even if there are more than one serie with the same date.
Sort your data using the key:
def your_view(request):
artwork = Artwork.objects.select_related()
artwork_list_sorted = sorted(artwork, key=key_function)
return render(request, "your_app/artwork_list.html", {'artwork_list_sorted': artwork_list_sorted})
or using ListView:
def get_context_data(self, **kwargs):
context = super(ArticleListView, self).get_context_data(**kwargs)
queryset = self.get_queryset()
context['artwork_list_sorted'] = sorted(queryset, key=key_function)
return context
Use the regroup template tag to group your data:
{% regroup artwork_list_sorted by series as artwork_grouped %}
{% for artwork in artwork_grouped %}
{% for item in artwork.list %}
{{ item.title }}({{ item.date }})
{% endfor %}
<strong>(series={{ artwork.grouper }})</strong>
{% endfor %}
or alternatively just iterate through the original sorted list, displaying series when needed - using the ifchanged template tag:
{% for artwork in artwork_list_sorted %}
{{ artwork.title}}({{ artwork.date }})
{% ifchanged %}<strong>(series={{ artwork.series.title }})</strong>{% endifchanged %}
{% endfor %}