Django snippets: Capture template output as a variable

from django import template

register = template.Library()

@register.tag(name='captureas')
def do_captureas(parser, token):
    try:
        tag_name, args = token.contents.split(None, 1)
    except ValueError:
        raise template.TemplateSyntaxError("'captureas' node requires a variable name.")
    nodelist = parser.parse(('endcaptureas',))
    parser.delete_first_token()
    return CaptureasNode(nodelist, args)

class CaptureasNode(template.Node):
    def __init__(self, nodelist, varname):
        self.nodelist = nodelist
        self.varname = varname

    def render(self, context):
        output = self.nodelist.render(context)
        context[self.varname] = output
        return ''

Tags like url and trans provide no way to get the result as a context variable. But how would you get a computed URL into a blocktrans?

This snippet solves the general problem. Just put the template code whose output you want to capture within captureas tags. For example:

{% captureas login_url %}{% url login %}{% endcaptureas %}
{% blocktrans %}
<a href="{{login_url}}">login</a>
{%endblocktrans%}

Author:: kcarnold
Posted:: January 14, 2008
Language:: Python
Tags:: assign blocktrans capture template variable
Score:: 9 (after 9 ratings)

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cubes (on June 24, 2009):

Thanks, I found this incredibly useful. One suggestion for improvement is to call django.utils.safestring.mark_safe on output after it is rendered in line 21.

This is because the render call performs HTML escaping of any interpolated variables, which means the newly created capture variable will be doubly escaped if it is not passed through the safe filter.

Thoughts?

I found that the built in filter tag somewhat decreases my need for captureas, but it's still nifty. :)

Django snippets

Capture template output as a variable

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