将 object 传递给 Django 自定义过滤器

Question

I am using Django 3.2

I am writing a moderation app, and I want to be able to display only approved values in my template.

I want to be able to use the new filter like this:

{{ moderated_object.field_name | approved }}

This is what I have so far:

from django import template

register = template.Library()


@register.filter(name='approved')
def moderator_approved_field_value(moderated_object, fieldname):

    return moderated_object.approved_value(fieldname)

As I have written the filter above, I can only use it like this:

{{ moderated_object| approved: fieldname }}

Which is ugly. Is there a way that I can pass the object to the function behind the scenes, so that I can use the cleaner way of using the filter in my template?

Answer 1

im my opinion django allow a bad deсigion with template filters.
You have a template and render function. You have a context. You can send in context already the result of moderated_object.approved_value(fieldname).

def get_context(...):
    ...
    context[moderated_object] = moderated_object.approved_value(fieldname)
    ...

template

{{ moderated_object }}

two }} , not three }}}

you can tell me: i create a loop with some additional elements in template.

answer - additional logic in template is bad, but, if you want - you can create a generator before:

def get_context(...):
    ...
    context[moderated_objects] = ((obj, obj.approved_value(fieldname)) for obj in context[moderated_objects])
    ...

template

{% for obj, approvement in moderated_objects %}
   {{ approvement }} 
   {{ moderated_object }}
{% endfor %}

all is possible. Try to think completely without template filter. And you find the best way, i am shure.