我找不到类似的问题(也许是因为我不知道正确的用法),但这是我的难题。我有两个模型,一个显示另一个的关系。
我希望能够比较html模板中的“ user_list”和“ friends_list”,以便可以确定是否存在好友关系的情况下是否在“可用用户”部分中不显示用户。
但是,“ user_list”对象属于CustomUser模型类型,而“ friends_list”对象属于Friend模型类型。如何返回get_friends()方法的结果,以便可以从“ user_list”查询集中排除这些朋友(最好是pk),这样我就不必担心字符串比较了?
谢谢!
Models.py
class CustomUserModel(AbstractBaseUser, PermissionsMixin):
# Simplified for brevity
username = models.CharFiend(_('username'), unique=True)
class Friend(models.Model):
user1 = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete.models.CASCADE, related_name='user1')
user2 = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete.models.CASCADE, related_name='user2')
objects = FriendManager()
class Meta:
unique_together = ('user1', 'user2')
def __str__(self):
return str(self.user2)
Managers.py
class FriendManager(models.Manager):
def is_friend(self, user1, user2):
if not user1 or not user2:
return False
return super(FriendManager, self).get_queryset().filter(user1=user1).filter(user2=user2).exists()
def get_friends(self, user):
return super(FriendManager, self).get_queryset().filter(user1=user)
Views.py
class UserList(ListView):
model = get_user_model()
context_object_name = "object_list"
template_name = "users.html"
def dispatch(self, request, *args, **kwargs):
self.user = get_user(request)
return super(UserList, self).dispatch(request, *args, **kwargs)
def get_queryset(self):
return get_user_model().objects.order_by("-date_joined")
def get_friends(self):
if self.user:
return Friend.objects.get_friends(self.user).order_by("user2")
return None
def get(self, request, *args, **kwargs):
return render(request, self.template_name, {
'user_list': self.get_queryset(),
'friends_list': self.get_friends(),
'user': self.user
})
Users.html
{% extends 'base.html' %}
{% block content %}
<h1>Available Users</h1>
{% for user in user_list %}
{% if user not in friends_list %} <!-- This is my pain-point -->
<p>{{ user }}</p>
{% endif %}
{% endfor %}
<h1>Friends</h1>
{% for friend in friends_list %}
<p>{{ friend }}</p>
{% endfor %}
{% endblock content %}
答案 0 :(得分:1)
您只需更改查询即可获得CustomUser
。因此查询应该类似于CustomeUser.objects.filter(user2__user1=self.user)
,这将为您提供与CustomUser
成为朋友的self.user
。根据您对关系的解释方式,还需要查询CustomUser.objects.filter(user1__user2=self.user)
。这些查询应替换您视图的get_friends()
方法中的查询。