编辑：
你需要somemagic是一个具有当前视图名称的变量。
尝试这个：
{% with request.path_info|resolve_url_name as current_view %}
{% url current_view page_obj.previous_page_number object.id %}
{% endwith %}
您可以使用 django-snippets 中的一些代码来实现这一点：
- 变量解析 URL 模板标记使 {% url %} 标记从上下文中解析变量。
- 将 URL 解析为视图名称该函数
resolve_to_name(path)返回 的视图名称path。您只需要创建一个使用此功能的过滤器。
此解决方案不适用于以下网址：
'alias/param1_regexp/param2_regexp/page;(?P<page>[0-9]+)/(?P<id>.*)$'
因为你不知道 param1 和 param2。
可以对上面的 django-snippets 进行修改以使这种 url 工作：
第一个片段修改：
from django.template import defaulttags, VariableDoesNotExist, Variable
class ResolvingURLNode(defaulttags.URLNode):
def render(self, context):
original_view_name = self.view_name
try:
resolved = Variable(self.view_name).resolve(context)
if len(resolved) > 1:
self.view_name = resolved[0]
if resolved[1]:
self.args = [Variable(arg) for arg in resolved[1]]
elif len(resolved) > 0:
self.view_name = resolved[0]
else:
self.view_name = resolved
except VariableDoesNotExist:
pass
ret = super(defaulttags.URLNode, self).render(context)
# restore view_name in case this node is reused (e.g in a loop) in
# which case the variable might resolve to something else in the next iteration)
self.view_name = original_view_name
return ret
defaulttags.URLNode = ResolvingURLNode
第二个片段修改
from django.core.urlresolvers import RegexURLResolver, RegexURLPattern, Resolver404, get_resolver
__all__ = ('resolve_to_name',)
def _pattern_resolve_to_name(self, path):
match = self.regex.search(path)
if match:
name = ""
if self.name:
name = self.name
elif hasattr(self, '_callback_str'):
name = self._callback_str
else:
name = "%s.%s" % (self.callback.__module__, self.callback.func_name)
if len(match.groups()) > 0:
groups = match.groups()
else:
groups = None
return name, groups
def _resolver_resolve_to_name(self, path):
tried = []
match = self.regex.search(path)
if match:
new_path = path[match.end():]
for pattern in self.url_patterns:
try:
resolved = pattern.resolve_to_name(new_path)
if resolved:
name, groups = resolved
else:
name = None
except Resolver404, e:
tried.extend([(pattern.regex.pattern + ' ' + t) for t in e.args[0 ['tried']])
else:
if name:
return name, groups
tried.append(pattern.regex.pattern)
raise Resolver404, {'tried': tried, 'path': new_path}
# here goes monkeypatching
RegexURLPattern.resolve_to_name = _pattern_resolve_to_name
RegexURLResolver.resolve_to_name = _resolver_resolve_to_name
def resolve_to_name(path, urlconf=None):
return get_resolver(urlconf).resolve_to_name(path)
基本上，resolve_to_name 将视图的名称及其参数作为元组返回，而新 {% url myvar %}的则采用此元组并使用它来反转具有视图名称及其参数的路径。
如果您不喜欢过滤器方法，也可以使用自定义中间件来完成。
上一个答案
您应该检查 django-pagination，它是一个非常好的 django 应用程序，易于使用并完成工作。
使用 django 分页，对可迭代对象进行分页的代码将是：
{% load pagination_tags %}
{% autopaginate myiterable 10 %} <!-- 10 elements per page -->
{% for item in myiterable %}
RENDERING CONTENT
{% endfor %}
{% paginate %} <!-- this renders the links to navigate through the pages -->
myiterable可以是任何可迭代的对象：列表、元组、查询集等
googlecode 上的项目页面：
http ://code.google.com/p/django-pagination/