在Django forloop中获取上一个和下一个元素的最佳方法是什么?我打印了一个元素列表,并希望子元素在div块中。所以我想要这样的东西:
{% 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 %}
正如您所看到的,我的问题是list[forloop.counter-1]
。我该怎么办?
答案 0 :(得分:6)
您可以创建custom template filters next
和previous
,它会返回for
循环的下一个和前一个元素。
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
然后在您的模板中,您可以执行以下操作来访问下一个和上一个元素。
{% 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
最终代码:
{% 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 %}