Django - Getting last object created, simultaneous filters

Question

Apologies, I am completely new to Django and Python.

I have 2 questions. First, how would I go about getting the last object created (or highest pk) in a list of objects? For example, I know that I could use the following to get the first object:

list = List.objects.all()[0]

Is there a way to get the length of List.objects? I've tried List.objects.length but to no avail.

Second, is it possible to create simultaneous filters or combine lists? Here is an example:

def findNumber(request, number)
    phone_list = Numbers.objects.filter(cell=number)

I want something like the above, but more like:

def findNumber(request, number)
    phone_list = Numbers.objects.filter(cell=number or home_phone=number)

What is the correct syntax, if any?

Answer 1

You can use the count() method on a query set the get the number of items.

list = List.objects.all()
list.count()

Arguments to filter are "AND"ed together. If you need to do OR filters look at Q objects. http://docs.djangoproject.com/en/dev/topics/db/queries/#complex-lookups-with-q-objects

Answer 2

I haven't tried this yet, but I'd look at the latest() operator on QuerySets:

latest(field_name=None)

Returns the latest object in the table, by date, using the field_name provided as the date field.

This example returns the latest Entry in the table, according to the pub_date field:

Entry.objects.latest('pub_date')

If your model's Meta specifies get_latest_by, you can leave off the field_name argument to latest(). Django will use the field specified in get_latest_by by default.

Like get(), latest() raises DoesNotExist if an object doesn't exist with the given parameters.

Note latest() exists purely for convenience and readability.

And the model docs on get_latest_by:

get_latest_by

Options.get_latest_by

The name of a DateField or DateTimeField in the model. This specifies the default field to use in your model Manager's latest method.

Example:

get_latest_by = "order_date"

See the docs for latest() for more.

Edit: Wade has a good answer on Q() operator.

Answer 3

For the largest primary key, try this:

List.objects.order_by('-pk')[0]

Note that using pk works regardless of the actual name of the field defined as your primary key.

Answer 4

this works!

Model.objects.latest('field') - field can be id. that will be the latest id