Fixed #3184 -- Changed the `unordered_list` template filter to use a more simple format, while maintaining backwards compatibility with the old format. `unordered_list` now works with a simple list of items. Thanks for the patch, SmileyChris.

git-svn-id: http://code.djangoproject.com/svn/django/trunk@6019 bcc190cf-cafb-0310-a4f2-bffc1f526a37
This commit is contained in:
Gary Wilson Jr 2007-08-26 01:11:20 +00:00
parent 32c729be10
commit 1a1a58c9f8
3 changed files with 85 additions and 11 deletions

View File

@ -355,8 +355,8 @@ def unordered_list(value):
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', []]]]``,
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
@ -371,14 +371,61 @@ def unordered_list(value):
</ul>
</li>
"""
def _helper(value, tabs):
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
if value[1]:
return u'%s<li>%s\n%s<ul>\n%s\n%s</ul>\n%s</li>' % (indent, force_unicode(value[0]), indent,
u'\n'.join([_helper(v, tabs+1) for v in value[1]]), indent, indent)
else:
return u'%s<li>%s</li>' % (indent, force_unicode(value[0]))
return _helper(value, 1)
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, force_unicode(title),
sublist))
i += 1
return '\n'.join(output)
value, converted = convert_old_style_list(value)
return _helper(value)
###################
# INTEGERS #

View File

@ -1301,9 +1301,14 @@ unordered_list
Recursively takes a self-nested list and returns an HTML unordered list --
WITHOUT opening and closing <ul> tags.
**Changed in Django development version**
The format accepted by ``unordered_list`` has changed to an easier to
understand format.
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::
``['States', ['Kansas', ['Lawrence', 'Topeka'], 'Illinois']]``, then
``{{ var|unordered_list }}`` would return::
<li>States
<ul>
@ -1317,6 +1322,9 @@ then ``{{ var|unordered_list }}`` would return::
</ul>
</li>
Note: the previous more restrictive and verbose format is still supported:
``['States', [['Kansas', [['Lawrence', []], ['Topeka', []]]], ['Illinois', []]]]``,
upper
~~~~~

View File

@ -266,6 +266,22 @@ u'bc'
>>> slice_(u'abcdefg', u'0::2')
u'aceg'
>>> unordered_list([u'item 1', u'item 2'])
u'\t<li>item 1</li>\n\t<li>item 2</li>'
>>> unordered_list([u'item 1', [u'item 1.1']])
u'\t<li>item 1\n\t<ul>\n\t\t<li>item 1.1</li>\n\t</ul>\n\t</li>'
>>> unordered_list([u'item 1', [u'item 1.1', u'item1.2'], u'item 2'])
u'\t<li>item 1\n\t<ul>\n\t\t<li>item 1.1</li>\n\t\t<li>item1.2</li>\n\t</ul>\n\t</li>\n\t<li>item 2</li>'
>>> unordered_list([u'item 1', [u'item 1.1', [u'item 1.1.1', [u'item 1.1.1.1']]]])
u'\t<li>item 1\n\t<ul>\n\t\t<li>item 1.1\n\t\t<ul>\n\t\t\t<li>item 1.1.1\n\t\t\t<ul>\n\t\t\t\t<li>item 1.1.1.1</li>\n\t\t\t</ul>\n\t\t\t</li>\n\t\t</ul>\n\t\t</li>\n\t</ul>\n\t</li>'
>>> unordered_list(['States', ['Kansas', ['Lawrence', 'Topeka'], 'Illinois']])
u'\t<li>States\n\t<ul>\n\t\t<li>Kansas\n\t\t<ul>\n\t\t\t<li>Lawrence</li>\n\t\t\t<li>Topeka</li>\n\t\t</ul>\n\t\t</li>\n\t\t<li>Illinois</li>\n\t</ul>\n\t</li>'
# Old format for unordered lists should still work
>>> unordered_list([u'item 1', []])
u'\t<li>item 1</li>'
@ -275,6 +291,9 @@ u'\t<li>item 1\n\t<ul>\n\t\t<li>item 1.1</li>\n\t</ul>\n\t</li>'
>>> unordered_list([u'item 1', [[u'item 1.1', []], [u'item 1.2', []]]])
u'\t<li>item 1\n\t<ul>\n\t\t<li>item 1.1</li>\n\t\t<li>item 1.2</li>\n\t</ul>\n\t</li>'
>>> unordered_list(['States', [['Kansas', [['Lawrence', []], ['Topeka', []]]], ['Illinois', []]]])
u'\t<li>States\n\t<ul>\n\t\t<li>Kansas\n\t\t<ul>\n\t\t\t<li>Lawrence</li>\n\t\t\t<li>Topeka</li>\n\t\t</ul>\n\t\t</li>\n\t\t<li>Illinois</li>\n\t</ul>\n\t</li>'
>>> add(u'1', u'2')
3