14

What is the best way to get previous and next elements in Django forloop? I'm printing a list of elements and want child element to be in div-block. So I want something like this:

{% for element in list %}
    {% if element.is_child and not list[forloop.counter-1].is_child %}
    <div class="children-block">
    {% endif %}
        {{ element.title }}
    {% if element.is_child and not list[forloop.counter+1].is_child %}
    </div>
    {% endif %}
{% endfor %}

As you can see my problem is list[forloop.counter-1]. How can I do it?

Rahul Gupta
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tetafro
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  • Please, look at this similar question: [Get list item dynamically in django templates](http://stackoverflow.com/questions/8948430/get-list-item-dynamically-in-django-templates) – Andrii Iushchuk Sep 26 '15 at 10:42

2 Answers2

20

You can create custom template filters next and previous which returns the next and the previous elements of the for loop respectively.

from django import template

register = template.Library()

@register.filter
def next(some_list, current_index):
    """
    Returns the next element of the list using the current index if it exists.
    Otherwise returns an empty string.
    """
    try:
        return some_list[int(current_index) + 1] # access the next element
    except:
        return '' # return empty string in case of exception

@register.filter
def previous(some_list, current_index):
    """
    Returns the previous element of the list using the current index if it exists.
    Otherwise returns an empty string.
    """
    try:
        return some_list[int(current_index) - 1] # access the previous element
    except:
        return '' # return empty string in case of exception

Then in your template you can do the following to access the next and previous elements.

{% with next_element=some_list|next:forloop.counter0 %} # assign next element to a variable
{% with previous_element=some_list|previous:forloop.counter0 %} # assign previous element to a variable

Final Code:

{% for element in list %}         
    {% with next_element=list|next:forloop.counter0 %} # get the next element 
    {% with previous_element=list|previous:forloop.counter0 %} # get the previous element 

        {% if element.is_child and not previous_element.is_child %}
            <div class="children-block">
        {% endif %}
            {{ element.title }}
        {% if element.is_child and not next_element.is_child %}
            </div>
        {% endif %}

    {% endwith %}
    {% endwith %}
{% endfor %}
wwl
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Rahul Gupta
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1

I modified the code from the accepted answer a little bit to get it though my linters.

from typing import Any

@register.filter(name="next")
def next(some_list: list[Any], current_index: int) -> Any:
    """Returns the next element of the list using the current index if it exists. Otherwise returns an empty string."""
    if not some_list or current_index == len(some_list) - 1:
        return ""
    return some_list[current_index + 1]  # access the next element

@register.filter(name="previous")
def previous(some_list: list[Any], current_index: int) -> Any:
    """Returns the previous element of the list using the current index if it exists. Otherwise returns an empty string."""
    if not some_list or current_index == 0:
        return ""
    return some_list[current_index - 1]  # access the previous element
Freiholtz
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