如何在不删除Django中的原始模型本身的情况下从queryset中删除

Question

因此，我有两个模型集，如下所示：

# User class
class User(AbstractBaseUser, PermissionsMixin):
objects = CustomUserManager()
...
email = models.EmailField(verbose_name='email', max_length=50, null=False, unique=True)
password = models.CharField(verbose_name="password", max_length=200)
name = models.CharField(verbose_name="name", max_length=50)
username = models.CharField(verbose_name="nickname", max_length=50, unique=True)
date_of_birth = models.DateField(verbose_name="birthday", max_length=8)
profile_pic = models.ImageField(verbose_name="profile picture", default="default_profile.png")
USERNAME_FIELD='email'
REQUIRED_FIELDS=['password', 'name', 'username', 'date_of_birth']
followers = models.ManyToManyField(
'self',
symmetrical=False,
through='Relationship',
related_name='followees',
through_fields=('to_user', 'from_user'),
)
@property
def followers(self):
follower_relationship = self.relationship_to_user.filter(relationship_type=Relationship.RELATIONSHIP_TYPE_FOLLOWING)
follower_list = follower_relationship.values_list('from_user', flat=True)
followers = User.objects.filter(pk__in=follower_list)
return followers
@property
def followees(self):
followee_relationship = self.relationship_from_user.filter(relationship_type=Relationship.RELATIONSHIP_TYPE_FOLLOWING)
followee_list = followee_relationship.values_list('to_user', flat=True)
followees = User.objects.filter(pk__in=followee_list)
return followees        

def follow(self, to_user):
self.relationship_from_user.create(
to_user = to_user,
relationship_type='f'
)

def __str__(self):
return self.username

# Relationships class
class Relationship(models.Model):
RELATIONSHIP_TYPE_FOLLOWING = 'f'
RELATIONSHIP_TYPE_BLOCKED = 'b'
CHOICE_TYPE = (
(RELATIONSHIP_TYPE_FOLLOWING, '팔로잉'),
(RELATIONSHIP_TYPE_BLOCKED, '차단'),
)
from_user = models.ForeignKey(
User,
related_name='relationship_from_user',
on_delete=models.CASCADE,
)
to_user = models.ForeignKey(
User,
related_name='relationship_to_user',
on_delete=models.CASCADE,
)
relationship_type=models.CharField(max_length=1, choices=CHOICE_TYPE)
def __str__(self):
return f"{self.from_user} follows {self.to_user}, type={self.relationship_type}"

在python manage.py shell中，我做了以下操作：

>>> user1 = User.objects.get(id=1)    # [e-mail] 
>>> user2 = User.objects.get(id=2)    # [e-mail]
>>> user3 = User.objects.get(id=3)    # [e-mail]
>>> user1.follow(user2)
>>> user1.follow(user3)
>>> user1.followees
<QuerySet [<User: [e-mail]>, <User: [e-mail]>]>

现在我的主要问题是关于user1.followees。我想从user1.followees查询集中删除<User: [e-mail]>，而不从数据库中删除用户。我试过以下几种：

>>> user1.followees[1].delete()                         # 1
>>> user1.followees.exclude(email='[e-mail]')    # 2

第一次尝试(#1(确实从查询集中删除了user3，但也从数据库中删除了用户本身，并打印出以下内容(打印时可能不一样，因为这是我自己第三次或第四次尝试(：

(4, {'accounts.Relationship': 3, 'accounts.User': 1})

第二次尝试(#2(确实排除了指定的用户，并返回了一个没有用户的新查询集，但当我访问它时，原始查询集没有改变，如下所示：

>>> user1.followees.exclude(email='[e-mail]')
<QuerySet [<User: [e-mail]>]>
>>> user1.followees
<QuerySet [<User: [e-mail]>, <User: [e-mail]>]>

Answer 1

Manytomany有添加和删除关系的特定方法。

user1.followees.remove(user3)

此外，除非您在follow()方法中需要自定义逻辑，否则您可能只需要使用user1.followees.add(user3)。

user1.followees.clear()可以移除所有的跟随者。

https://docs.djangoproject.com/en/3.1/topics/db/examples/many_to_many/

Answer 2

我似乎是从我正在做的事情访问用户本身：

>>> user1.followees[0] --> This accesses the user who's in the followees[0], not the relatioship itself

因此，当我尝试删除查询集时，我正在访问用户本身，所以我实际上是在删除查询集和用户本身。因此，我的解决方案是访问关系，而不是用户。我做了以下事情：

>>> Relationship.objects.all()
<QuerySet [<Relationship: [e-mail] follows [e-mail], type=f>, <Relationship: [e-mail] follows [e-mail] type=f>]>
>>> Relationship.objects.all()[0]
<QuerySet [<Relationship: [e-mail] follows [e-mail], type=f>]>
>>> Relationship.objects.all()[0].delete()
(1, {'accounts.Relationship': 1})

并检查关系模型和user1关系：

>>> Relationship.objects.all()
<QuerySet [<Relationship: [e-mail] follows [e-mail] type=f>]>
>>> user1.followees
>>> <QuerySet [<User: [e-mail]>]>

这就是答案！

**这个方法是可能的，但@user1464664给出的答案要好得多，所以我建议研究一下这个