Question

I have a newsletter application where a newsletter has multiple articles within each issue. I want to display a summary page online that lists the newsletter year, volume and label, and then in an unordered list display all the articles in the issue. I am quite new to Django so I am trying to determine the best way to do this.

I have the models defined (just the relevant parts):

Models.py:

class Newsletter(models.Model):
    volume = models.ForeignKey(Volume)
    year   = models.IntegerField()
    season = models.CharField(max_length=6, choices=VOLUME_SEASON)
    label  = models.CharField(max_length=20)
    number = models.IntegerField()

class Article(models.Model):
    newsletter = models.ForeignKey(Newsletter)
    section    = models.ForeignKey(Section)
    title      = models.CharField(max_length=200)

What I want to see on the web looks like:

<h2>Spring 2012</h2>
<p>Volume 14, Number 1</p>
<ul>
    <li>Foo</li>
    <li>Bar</li>
    <li>Baz</li>
</ul>

<h2>Winter 2011</h2>
<p>Volume 13, Number 4</p>
<ul>
  <li>Boffo</li>
</ul>

Pretty simple. However, I am confused by the best way to write my view. Whether to use:

• Two lists which I zip() and then iterate over in the template
• Use the select_related() queryset
• Use the prefetch_related() queryset

I have it working using the first option:

Views.py:

from django.shortcuts import render_to_response, get_object_or_404
from www.apps.newsletter.models import Newsletter, Article

def index(request):
    article_group = []
    newsletter = Newsletter.objects.all().order_by('-year', '-number')
    for n in newsletter:
        article_group.append(n.article_set.all())
    articles_per_newsletter = zip(newsletter, article_group)

    return render_to_response('newsletter/newsletter_list.html',
                              {'newsletter_list': articles_per_newsletter})

And then render it using the following template:

Newsletter_list.html:

{% block content %}
  {% for newsletter, articles in newsletter_list %}
    <h2>{{ newsletter.label }}</h2>
    <p>Volume {{ newsletter.volume }}, Number {{ newsletter.number }}</p>
    <p>{{ newsletter.article }}</p>
    <ul>
    {% for a in articles %}
      <li>{{ a.title }}</li>
    {% endfor %}
    </ul>
  {% endfor %}
{% endblock %}

Pretty straightforward, but as I am pretty new to Django I was wondering if what I am doing is completely inefficient in terms of its powerful ORM. I would love to not have to make a list on-the-fly and then zip() the two lists together if there is a faster way.

TIA.

Answer 1

The approach you are doing now will be heavily inefficient, because it will result in an 1+N number of queries. That is, 1 for the query of all your Newsletters, and then 1 for every single time you evaluate those n.article_set.all() results. So if you have 100 Newletter objects in that first query, you will be doing 101 queries.

This is an excellent reason to use prefetch_related. It will only result in 2 queries. One to get the Newsletters, and 1 to batch get the related Articles. Though you are still perfectly able to keep doing the zip to organize them, they will already be cached, so really you can just pass the query directly to the template and loop on that. :

view

newsletters = Newsletter.objects.prefetch_related('article_set').all()\
                    .order_by('-year', '-number')

return render_to_response('newsletter/newsletter_list.html',
                          {'newsletter_list': newsletters})

template

{% block content %}
  {% for newsletter in newsletter_list %}
    <h2>{{ newsletter.label }}</h2>
    <p>Volume {{ newsletter.volume }}, Number {{ newsletter.number }}</p>
    <p>{{ newsletter.article }}</p>
    <ul>
    {% for a in newsletter.article_set.all %}
      <li>{{ a.title }}</li>
    {% endfor %}
    </ul>
  {% endfor %}
{% endblock %}