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How to replace a comma with an underscore in Django template filter

Hi I have javascript as follows contains django variables. 

<script>

$(document).ready(function() {
    {% for a in all_results %}
            $('#{{a.set|slice:":7"}}_{{ a.item }}_{{ a.kidu|slice:":2" }}').html('<a class="a1" href="{% url 'req_detail_view' a.id %}">{{ a.id }}</a>');
            $('#{{a.set|slice:":7"}}_{{ a.item}}_{{ a.kidu|slice:":2" }}').addClass('{{ a.state__name.split|join:"_" }}');
    {% endfor %}
});
</script>

all results are a query result fro my view:

context['all_results'] = m.values('set', 'kidu', 'id', 'item', 'state__name')

The problem am facing is if the value of item is a comma separated value, then the id is not passing to the template I have. But if the value of item is a one word my code is working perfectly and the id is passing to the template. Any idea guys?

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Answer

In file where you have you custom filters and tags, add following function

 from django import template
 register = template.Library()

 @register.filter
 def replace_commas(string):
    return string.replace(',', '_')

For view how to create custom filters, view this documentation of django.

  1. Create a folder ‘templatetags’ in your app directory
  2. Create a file name ‘my_filter.py’ and '__init__.py' in that folder
  3. Copy above code in my_filter.py
  4. In you template, add {% load my_filter %} at the top.
  5. Now you can use your filter like this {{m.item|replace_commas}}

Note, you could make the template tag a little more flexible by making the find/replace an argument. e.g.:

from django import template
register = template.Library()

@register.filter
def find_replace(string, find_replace=",|_"):
    find, replace = find_replace.split("|")
    return string.replace(',', '_')

Then you would be able to use it like so in your template:

{{ m.item|find_replace:",|-" }}

(would replace ,‘s with -‘s

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