Python – Django ORM’s “related data” loading behavior

djangodjango-modelslinqlinq-to-sqlpython

In LINQ (I come from a C# background), you can manually load data for related tables via the Include("xxx") method

from a in ctx.MainTable.Include("SubTable")
select a;

In the above code, every instance of MainTable is loaded and all the data for MainTable.SubTable is also loaded. If "Include" is not called, every returned MainTable's SubTable property will return null.

Are there equivalent methods for the Django ORM? If not, what's the default Django behavior for situtions like the above?

Best Answer

See this:

http://docs.djangoproject.com/en/dev/ref/models/querysets/#id4

You create a select_related query set to follow relationships and pre-fetch related rows.

Normally, you don't waste much time doing this until you know that the individual fetches done automagically by the ORM are too slow.

When you simply name a related object field:

x = MainTable.objects.get( id="some Object" )
y= x.subTable

The ORM will do the of the related subTable row. lazily.

You can't easily fetch an entire table into memory. QuerySets are "lazy" and don't fetch rows until there's no possible excuse. Consider this.

for m in MainTable.objects.all():
    y = m.subTable
    if y.someAttribute > 5: 
        break

Then you may not actually need all the objects in MainTable. The query set is ready to fetch them. It just doesn't actually fetch them all. That's why we don't generally do anything more than x.subTable and leave it to the ORM to fetch the row as needed.

Once in a while you can prove that it's too slow. Then you can do MainTable.objects.select_related()... and force navigation to rows in other tables.

Related Solutions

Python – a “slug” in Django

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.

Python – How to do a not equal in Django queryset filtering

You can use Q objects for this. They can be negated with the ~ operator and combined much like normal Python expressions:

from myapp.models import Entry
from django.db.models import Q

Entry.objects.filter(~Q(id=3))

will return all entries except the one(s) with 3 as their ID:

[<Entry: Entry object>, <Entry: Entry object>, <Entry: Entry object>, ...]

Best Answer

Related Solutions

Python – a “slug” in Django

Python – How to do a not equal in Django queryset filtering

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