If fruits
is the list [\'apples\', \'oranges\', \'pears\']
,
is there a quick way using django template tags to produce \"apples, oranges, and p
All of the answers here fail one or more of the following:
and
for the last item.Here's my entry into this canon. First, the tests:
class TestTextFilters(TestCase):
def test_oxford_zero_items(self):
self.assertEqual(oxford_comma([]), '')
def test_oxford_one_item(self):
self.assertEqual(oxford_comma(['a']), 'a')
def test_oxford_two_items(self):
self.assertEqual(oxford_comma(['a', 'b']), 'a and b')
def test_oxford_three_items(self):
self.assertEqual(oxford_comma(['a', 'b', 'c']), 'a, b, and c')
And now the code. Yes, it gets a bit messy, but you'll see that it doesn't use negative indexing:
from django.utils.encoding import force_text
from django.utils.html import conditional_escape
from django.utils.safestring import mark_safe
@register.filter(is_safe=True, needs_autoescape=True)
def oxford_comma(l, autoescape=True):
"""Join together items in a list, separating them with commas or ', and'"""
l = map(force_text, l)
if autoescape:
l = map(conditional_escape, l)
num_items = len(l)
if num_items == 0:
s = ''
elif num_items == 1:
s = l[0]
elif num_items == 2:
s = l[0] + ' and ' + l[1]
elif num_items > 2:
for i, item in enumerate(l):
if i == 0:
# First item
s = item
elif i == (num_items - 1):
# Last item.
s += ', and ' + item
else:
# Items in the middle
s += ', ' + item
return mark_safe(s)
You can use this in a django template with:
{% load my_filters %}
{{ items|oxford_comma }}