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 pears"?

I know it's not difficult to do this using a loop and {% if counter.last %} statements, but because I'm going to use this repeatedly I think I'm going to have to learn how to write custom tags filters, and I don't want to reinvent the wheel if it's already been done.

As an extension, my attempts to drop the Oxford Comma (ie return "apples, oranges and pears") are even messier.

Answer 1

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 2

Here's a super simple solution. Put this code into comma.html:

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

And now wherever you'd put the comma, include "comma.html" instead:

{% for cat in cats %}
Kitty {{cat.name}}{% include "comma.html" %}
{% endfor %}

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Update: @user3748764 gives us a slightly more compact version, without the deprecated ifequal syntax:

{% if not forloop.first %}{% if forloop.last %} and {% else %}, {% endif %}{% endif %}

Note that it should be used before the element, not after.

Answer 3

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 %}

Answer 4

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!