我在django中使用unordered_list标记。
我有以下列表：
foo = ['A', ['B', 'C', 'D'], 'E']
以下标记：
{{ foo|unordered_list }}
正如预期的那样产生以下内容：
<li>A
<ul>
<li>B</li>
<li>C</li>
<li>D</li>
</ul>
</li>
<li>E</li>
我想要做的是将 id 和 class 属性添加到每个“li”节点。 我知道我可以在JavaScript中完成它，或者在django中实现我自己的模板标记。但我想先问一下。可能在django模板中已经存在一种简单的内置方式吗？
答案 0 :(得分：5)
正如您在源代码中看到的那样，<li>是硬编码的，因此如果不创建自己的模板过滤器，则无法执行此操作。
def unordered_list(value, autoescape=None):
"""
Recursively takes a self-nested list and returns an HTML unordered list --
WITHOUT opening and closing <ul> tags.
The list is assumed to be in the proper format. For example, if ``var``
contains: ``['States', ['Kansas', ['Lawrence', 'Topeka'], 'Illinois']]``,
then ``{{ var|unordered_list }}`` would return::
<li>States
<ul>
<li>Kansas
<ul>
<li>Lawrence</li>
<li>Topeka</li>
</ul>
</li>
<li>Illinois</li>
</ul>
</li>
"""
if autoescape:
from django.utils.html import conditional_escape
escaper = conditional_escape
else:
escaper = lambda x: x
def convert_old_style_list(list_):
"""
Converts old style lists to the new easier to understand format.
The old list format looked like:
['Item 1', [['Item 1.1', []], ['Item 1.2', []]]
And it is converted to:
['Item 1', ['Item 1.1', 'Item 1.2]]
"""
if not isinstance(list_, (tuple, list)) or len(list_) != 2:
return list_, False
first_item, second_item = list_
if second_item == []:
return [first_item], True
old_style_list = True
new_second_item = []
for sublist in second_item:
item, old_style_list = convert_old_style_list(sublist)
if not old_style_list:
break
new_second_item.extend(item)
if old_style_list:
second_item = new_second_item
return [first_item, second_item], old_style_list
def _helper(list_, tabs=1):
indent = u'\t' * tabs
output = []
list_length = len(list_)
i = 0
while i < list_length:
title = list_[i]
sublist = ''
sublist_item = None
if isinstance(title, (list, tuple)):
sublist_item = title
title = ''
elif i < list_length - 1:
next_item = list_[i+1]
if next_item and isinstance(next_item, (list, tuple)):
# The next item is a sub-list.
sublist_item = next_item
# We've processed the next item now too.
i += 1
if sublist_item:
sublist = _helper(sublist_item, tabs+1)
sublist = '\n%s<ul>\n%s\n%s</ul>\n%s' % (indent, sublist,
indent, indent)
output.append('%s<li>%s%s</li>' % (indent,
escaper(force_unicode(title)), sublist))
i += 1
return '\n'.join(output)
value, converted = convert_old_style_list(value)
return mark_safe(_helper(value))
unordered_list.is_safe = True
unordered_list.needs_autoescape = True
重写过滤器以添加id / class支持应该相当容易，具体取决于您希望它如何工作。