Custom template filters in Django - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What is the output of this custom filter usage?

Given this custom filter that reverses a string, what will be the output of {{ 'hello'|reverse_string }} in a Django template?

Django

from django import template
register = template.Library()

@register.filter
def reverse_string(value):
    return value[::-1]

A olleh

B h e l l o

C Error: filter not found

D hello

💡 Hint

Think about what slicing with [::-1] does to a string.

The filter reverse_string returns the input string reversed. So 'hello' becomes 'olleh'.

📝 Syntax

intermediate

Which option correctly registers a custom filter?

Which of the following code snippets correctly registers a custom Django template filter named double that doubles a number?

A

from django import template
register = template.Library()

@register.filter(name='double')
def double(value):
    return value * 2

B

from django.template import Library
register = Library()

def double(value):
    return value * 2
register.filter(double)

C

import django.template
register = django.template.Library()

@register.filter
def double(value):
    return value + value

D

from django import template
register = template.Library()

def double(value):
    return value * 2
register.filter('double', double)

💡 Hint

Look for the correct decorator usage and import style.

Option A uses the correct import, creates a Library instance, and registers the filter with the decorator and name argument.

🔧 Debug

advanced

Why does this custom filter cause a TemplateSyntaxError?

Consider this filter code:

from django import template
register = template.Library()

@register.filter
def add_prefix(value, prefix):
    return prefix + value

Using {{ 'world'|add_prefix:'Hello ' }} in a template causes a TemplateSyntaxError. Why?

A The filter function must accept exactly one argument, but this one has two.

B The filter is missing the <code>is_safe=True</code> parameter in the decorator.

C The filter function must accept the value and an optional argument, but the template syntax is incorrect.

D The filter function must be registered with <code>@register.simple_tag</code> instead.

💡 Hint

Check how to pass arguments to filters in Django templates.

Custom filters can accept an argument, but the template syntax must use a colon to pass it, like {{ value|filter:arg }}. The code is correct, but the error is due to incorrect template usage or missing quotes.

❓ state_output

advanced

What is the output of this chained custom filter usage?

Given these filters:

from django import template
register = template.Library()

@register.filter
def shout(value):
    return value.upper() + '!!!'

@register.filter
def repeat(value, times=2):
    return value * times

What is the output of {{ 'hi'|shout|repeat:3 }} in a Django template?

A hi!!!hi!!!hi!!!

B HI!!!HI!!!HI!!!

C HIHIHI!!!

D Error: repeat filter requires integer argument

💡 Hint

Remember the order filters are applied and how arguments are passed.

The string 'hi' is first converted to uppercase with '!!!' added by shout, resulting in 'HI!!!'. Then repeat multiplies this string 3 times, producing 'HI!!!HI!!!HI!!!'.

🧠 Conceptual

expert

Which statement about custom template filters is TRUE?

Choose the correct statement about Django custom template filters.

A Filters can accept multiple arguments by defining the function with *args and **kwargs.

B Filters can modify the template context directly to affect other variables.

C Custom filters must always return strings; returning other types causes errors.

D Filters must be registered with a Library instance and can be used in templates after loading the tag library.

💡 Hint

Think about how filters are registered and used in templates.

Filters must be registered with a Library instance and the tag library must be loaded in the template to use them. Filters do not modify context or require returning strings only.