A "slug" is a way of generating a valid URL, generally using data already obtained. For instance, a slug uses the title of an article to generate a URL. I advise to generate the slug by means of a function, given the title (or another piece of data), rather than setting it manually.
An example:
<title> The 46 Year Old Virgin </title>
<content> A silly comedy movie </content>
<slug> the-46-year-old-virgin </slug>
Now let's pretend that we have a Django model such as:
class Article(models.Model):
title = models.CharField(max_length=100)
content = models.TextField(max_length=1000)
slug = models.SlugField(max_length=40)
How would you reference this object with a URL and with a meaningful name? You could for instance use Article.id so the URL would look like this:
www.example.com/article/23
Or, you might want to reference the title like this:
www.example.com/article/The 46 Year Old Virgin
Since spaces aren't valid in URLs, they must be replaced by %20
, which results in:
www.example.com/article/The%2046%20Year%20Old%20Virgin
Both attempts are not resulting in very meaningful, easy-to-read URL. This is better:
www.example.com/article/the-46-year-old-virgin
In this example, the-46-year-old-virgin
is a slug: it is created from the title by down-casing all letters, and replacing spaces by hyphens -
.
Also see the URL of this very web page for another example.
Concatenating the querysets into a list is the simplest approach. If the database will be hit for all querysets anyway (e.g. because the result needs to be sorted), this won't add further cost.
from itertools import chain
result_list = list(chain(page_list, article_list, post_list))
Using itertools.chain
is faster than looping each list and appending elements one by one, since itertools
is implemented in C. It also consumes less memory than converting each queryset into a list before concatenating.
Now it's possible to sort the resulting list e.g. by date (as requested in hasen j's comment to another answer). The sorted()
function conveniently accepts a generator and returns a list:
result_list = sorted(
chain(page_list, article_list, post_list),
key=lambda instance: instance.date_created)
If you're using Python 2.4 or later, you can use attrgetter
instead of a lambda. I remember reading about it being faster, but I didn't see a noticeable speed difference for a million item list.
from operator import attrgetter
result_list = sorted(
chain(page_list, article_list, post_list),
key=attrgetter('date_created'))
Best Answer
I've used a simple technique that works nicely for small cases with no special tags and no additional context. Sometimes this comes in handy