Django 1.8 ArrayField append & extend

Question

Django 1.8 will come with new advanced field types including ArrayField these rely on PostgreSQL and are implemented at a DB level.

PostgreSQL's array field implements an append method.

However I can't find any documentation on appending an item to an ArrayField. This would clearly be very useful as it would allow the field to be updated without transferring its entire contents from the db then back to it.

Is this possible? If not will it be possible in future? Any pointers to docs I've missed would be much appreciated.

To clarify what I'm asking about, this would be great:

Note: this is fantasy code and I assume won't work (I haven't tried)

# on model:
class Post(models.Model):
    tags = ArrayField(models.CharField(max_length=200))

# somewhere else:
p = Post.objects.create(tags=[str(i) for i in range(10000)])
p.tags.append('hello')

Is there currently any way of doing this without resorting to raw sql?

Answer 1

This works:

from django.db.models import F
from django.db.models.expressions import CombinedExpression, Value

post = Post.objects.get(id=1000)
post.tags = CombinedExpression(F('tags'), '||', Value(['hello']))
post.save()

or in an update clause:

Post.objects.filter(created_on__lt=now() - timespan(days=30))\
    .update(tags=CombinedExpression(F('tags'), '||', Value(['old'])))

Answer 2

I think the features you are looking for are currently not implemented (and may not be planned). Many of the Postgres contrib features originated based on this kickstarter project.

I find the most useful documentation for the new features come from the source code itself. Which includes a link to the original pull request for many of these features.

An important note in regards to the Array Functions mentioned, they are Functions and arguably outside the scope of a typical ORM.

I hope this information is useful and you find a nice solution to this issue.

Answer 3

You could use django_postgres_extensions. It supports a lot of functions like append, prepend, remove, concatenate.

But if you are using Django 1.8 like me, you should use only the required classes from this package. That way, you won't have to change database backend too. I've pasted the required classes here. Use them as described in first link.

Answer 4

Note: OP code will absolutely work. We just need to save the model (because these is just a model field, not relation). Let's see:

>>> p = Post.objects.create(tags=[str(i) for i in range(10000)])
>>> p.tags.append("working!")
>>> p.save()
>>> working_post = Post.objects.get(tags__contains=["working!"])
<Post: Post object>
>>> working_post.tags[-2:]
[u'9999', u'working!']

Going deeper

Django gets ArrayField as python list

Code reference

Everything you could do with list, you can do with ArrayField. Even sorting

Django saves ArrayField as python list

Code reference

These means that it saves structure and elements of python list.

Answer 5

Another solution is using a custom expression. I tested the following code with Django 1.11 and Python 3.6 (f-strings).

from django.db.models.expressions import Func

class ArrayAppend(Func):

    function = 'array_append'
    template = "%(function)s(%(expressions)s, %(element)s)"
    arity = 1

    def __init__(self, expression: str, element, **extra):
        if not isinstance(element, (str, int)):
            raise TypeError(
                f'Type of "{element}" must be int or str, '
                f'not "{type(element).__name__}".'
            )

        super().__init__(
            expression,
            element=isinstance(element, int) and element or f"'{element}'",
            **extra,
        )

The expression can be used in update():

Post.objects \
    .filter(pk=1) \
    .update(tags=ArrayAppend('tags', 'new tag'))