Comma separated lists in django templates

Question

If fruits is the list [\'apples\', \'oranges\', \'pears\'],

is there a quick way using django template tags to produce \"apples, oranges, and p

Answer 1

All of the answers here fail one or more of the following:

• They rewrite something (poorly!) that's in the standard template library (ack, top answer!)
• They don't use and for the last item.
• They lack a serial (oxford) comma.
• They use negative indexing, which won't work for django querysets.
• They don't usually handle string sanitation properly.

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 }}

Answer 2

I would simply use ', '.join(['apples', 'oranges', 'pears']) before sending it to the template as a context data.

UPDATE:

data = ['apples', 'oranges', 'pears']
print(', '.join(data[0:-1]) + ' and ' + data[-1])

You will get apples, oranges and pears output.

Answer 3

First choice: use the existing join template tag.

http://docs.djangoproject.com/en/dev/ref/templates/builtins/#join

Here's their example

{{ value|join:" // " }}

Second choice: do it in the view.

fruits_text = ", ".join( fruits )

Provide fruits_text to the template for rendering.

Answer 4

If you want a '.' on the end of Michael Matthew Toomim's answer, then use:

{% if not forloop.last %}{% ifequal forloop.revcounter 2 %} and {% else %}, {% endifequal %}{% else %}{% endif %}{% if forloop.last %}.{% endif %}

Answer 5

I would suggest a custom django templating filter rather than a custom tag -- filter is handier and simpler (where appropriate, like here). {{ fruits | joinby:", " }} looks like what I'd want to have for the purpose... with a custom joinby filter:

def joinby(value, arg):
    return arg.join(value)

which as you see is simplicity itself!

Answer 6

On the Django template this all you need to do for establishing a comma after each fruit. The comma will stop once its reached the last fruit.

{% if not forloop.last %}, {% endif %}