how to use custom django templatetag with django template if statement?

Question

I've made a django template tag that counts one of my custom user many-to-many field length:

from django import template

register = template.Library()

@register.simple_tag(takes_context=True)
def unread_messages_count(context):
    user = context['request'].user
    return len(user.messages_unread.all())

and within the template itself, I want to show it to user only if it's larger than zero, so I tried:

{% ifnotequal unread_messages_count 0 %}
   some code...
{% endifnotequal %}

but obviously it didn't work. not even with a 'with' statement:

{% with unread_messages_count as unread_count %}
    {% ifnotequal unread_count 0 %}
        some code...
    {% endifnotequal %}
{% endwith %}

How can I check that the variable is larger than 0 and only if it is, present some code to the user (including the number in the variable itself). thanks.

Yuji 'Tomita' Tomita · Accepted Answer · 2013-02-08 07:21:29Z

9

The easiest way would be to use an assignment tag..

https://docs.djangoproject.com/en/dev/howto/custom-template-tags/#assignment-tags

@register.assignment_tag(takes_context=True)
def unread_messages_count(context):
    user = context['request'].user
    return len(user.messages_unread.all())

{% unread_messages_count as cnt %}
{% if cnt %}
   foo
{% endif %}

answered Feb 8, 2013 at 7:21

Yuji 'Tomita' Tomita

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1 Comment

Kritz Over a year ago

Deprecated since version 1.9: "simple_tag can now store results in a template variable and should be used instead."

mabdrabo · Accepted Answer · 2013-02-08 13:05:13Z

2

you can use a django custom filter https://docs.djangoproject.com/en/dev/howto/custom-template-tags/#writing-custom-template-filters

def unread_messages_count(user_id):
  # x = unread_count   ## you have the user_id
  return x

and in the template

{% if request.user.id|unread_messages_count > 0 %}
  some code...
{% endif %}

answered Feb 8, 2013 at 13:05

mabdrabo

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