Only showing year in django admin, a YearField instead of DateField?

Question

I've got a model where I need to store birth year. I'm using django admin. The person using this will be filling out loads of people every day, and DateField() shows too much (not interested in the day/month).

This is a mockup model showing how it is now:

class Person(models.Model):
  name = models.CharField(max_length=256)
  born = models.IntegerField(default=lambda: date.today().year - 17)

As you can see, most of the people is 17 years old, so I'm getting their birth year as default.

Can I do this better? How can I make a YearField out of the DateField? If making a YearField I can maybe even make some "easy tags" like the "now" that date has. (Of course, a specialized BornYearField would have easy buttons for 1989, 1990, 1991 and other common years)

Answer 1

I found this solution which solves the whole thing quite elegantly I think (not my code):

import datetime
YEAR_CHOICES = []
for r in range(1980, (datetime.datetime.now().year+1)):
    YEAR_CHOICES.append((r,r))

year = models.IntegerField(_('year'), choices=YEAR_CHOICES, default=datetime.datetime.now().year)

Edit the range start to extend the list :-)

Answer 2

Also, and this isn't that relevant for your case, but is useful being aware of, don't use datetime.date.today(), or datetime.datetime.now() as defaults. This is executed once, when the server is started up.

You are much better off passing the callables in:

date = models.DateField(default=datetime.date.today)

Note, you could use a lambda to make it relative:

date = models.DateField(default=lambda : datetime.date.today() - datetime.timedelta(days=6210))

Of course, this is naive, and assumes there have been 5 leap years in the past 17 years.

Answer 3

You can also do this:

YEARS = (
    ("1990", "1990"),
    ("1991", "1991"),
    ("1992", "1992"),
    # P.e. generate a list from 1960 to date.today().year
    # The reason why they choice key and text value are the
    # same is that if you generate from 1960 to 2060 it will collide.
    #
    # E.g
    # from datetime import datetime
    # def tuplify(x): return (x,x)   # str(x) if needed
    # current_year = datetime.now().year
    # YEARS = map(tuplify, range(1930, current_year + 1))  # range(1,4) gives [1,2,3]
)

class Whatever(models.Model):
    # Show a list with years
    birthdate = models.IntegerField(max_length=2, choices=YEARS)

I hope this will help you.

Answer 4

For a solution not involving Model choices:

from django.core.validators import MinValueValidator, MaxValueValidator

class Person(models.Model):
    year = models.PositiveIntegerField(
            validators=[
                MinValueValidator(1900), 
                MaxValueValidator(datetime.now().year)],
            help_text="Use the following format: <YYYY>")

That'll also create a placeholder in the input field with the value of help_text.