我的模型看起来像这样:
class ListEntry(models.Model):
STATUS_CHOICES = (
('PL', _('Playing')),
('CO', _('Completed')),
('OH', _('On-hold')),
('DR', _('Dropped')),
('WA', _('Want to play')),
)
user = models.ForeignKey(User)
game = models.ForeignKey(Game)
status = models.CharField(_('Status'), max_length=2,
choices=STATUS_CHOICES)
在我的观看中,我正在按用户过滤条目:
class GameListByUserView(ListView):
model = ListEntry
template_name = 'game_list_by_user.html'
def get_queryset(self):
self.user_profile = get_object_or_404(User, username=self.kwargs['slug'])
return ListEntry.objects.filter(user=self.user_profile)
def get_context_data(self, **kwargs):
context = super(GameListByUserView, self).get_context_data(**kwargs)
context['user_profile'] = self.user_profile
return context
我现在要做的是将此查询(ListEntry.objects.filter(user=self.user_profile)
)拆分为子组,具体取决于status
属性,因此呈现的模板如下所示:
UserFoo's list
=========================
Playing
Game 1
Game 2
Game 3
Completed
Game 4
Game 5
Game 6
我知道我可以这样做:
q = ListEntry.objects.filter(user=self.user_profile)
games_dict = {}
games_dict['playing'] = q.filter(status='PL')
games_dict['completed'] = q.filter(status='CO')
依此类推,并迭代模板中的键。或(在模板中):
{% for s in status_choices %}
{% for entry in object_list %}
{% if entry.status == s %}
Render things
{% endif %}
{% endfor %}
{% endfor %}
但是,如果没有按状态获取子查询,并且没有多次遍历对象列表,那么有没有更好的,优化的方法来执行此操作而不会访问数据库?
答案 0 :(得分:2)
您正在寻找regroup
filter
{% regroup object_list by status as games_list %}
<ul>
{% for game in games_list %}
<li>{{ game.grouper }}
<ul>
{% for item in game.list %}
<li>{{ item}}</li>
{% endfor %}
</ul>
</li>
{% endfor %}
</ul>
您可能需要自定义元素渲染的方式,但我会让您自己解决这个问题。