Django 3 - 为主键提供默认值时的 Model.save()

Question

我正在将我的项目从 Django 2 升级到 Django 3，我已经阅读了他们的 Django 3 发行说明，有一点我并不真正了解它会对我当前的项目产生什么影响。他们在这里说：

据我所知，如果我们尝试调用Model.save()，如果模型是现有记录，它总是会创建一个新记录而不是更新。例如：

car = Car.objects.first()
car.name = 'Honda'
car.save() # does it INSERT or UPDATE? I suspect it is an "INSERT" statement as their explanation and "UPDATE" statement in Django 2.

我进行了实验，它仍然与 Django 2 的行为相同，不确定它们的含义。

In [5]: u = User.objects.first()                                                                                                                                                                          
(0.001) SELECT "accounts_user"."id", "accounts_user"."password", "accounts_user"."last_login", "accounts_user"."is_superuser", "accounts_user"."username", "accounts_user"."first_name", "accounts_user"."last_name", "accounts_user"."is_staff", "accounts_user"."is_active", "accounts_user"."date_joined", "accounts_user"."email", "accounts_user"."avatar", "accounts_user"."last_location"::bytea, "accounts_user"."uuid", "accounts_user"."country", "accounts_user"."city", "accounts_user"."phone" FROM "accounts_user" ORDER BY "accounts_user"."id" ASC LIMIT 1; args=()

In [6]: u.save()                                                                                                                                                                                          
(0.006) UPDATE "accounts_user" SET "password" = 'pbkdf2_sha256_sha512$180000$FbFcNuPMrOZ6$GwIftEo+7+OpsORwn99lycye46aJn/aJNAtc50N478Y=', "last_login" = NULL, "is_superuser" = false, "username" = 'email0@mail.com', "first_name" = 'Noah', "last_name" = 'Spencer', "is_staff" = false, "is_active" = true, "date_joined" = '2020-05-12T07:06:20.605650+00:00'::timestamptz, "email" = 'email0@mail.com', "avatar" = 'account/user_avatar/example_HseJquC.jpg', "last_location" = NULL, "uuid" = 'f6992866-e476-409e-9f1b-098afadce5b7'::uuid, "country" = NULL, "city" = NULL, "phone" = NULL WHERE "accounts_user"."id" = 1; args=('pbkdf2_sha256_sha512$180000$FbFcNuPMrOZ6$GwIftEo+7+OpsORwn99lycye46aJn/aJNAtc50N478Y=', False, 'email0@mail.com', 'Noah', 'Spencer', False, True, datetime.datetime(2020, 5, 12, 7, 6, 20, 605650, tzinfo=<UTC>), 'email0@mail.com', 'account/user_avatar/example_HseJquC.jpg', UUID('f6992866-e476-409e-9f1b-098afadce5b7'), 1)

更新：

In [38]: u1 = User.objects.first()                                                                                                                                                                        
(0.000) SELECT "accounts_user"."id", "accounts_user"."password", "accounts_user"."last_login", "accounts_user"."is_superuser", "accounts_user"."username", "accounts_user"."first_name", "accounts_user"."last_name", "accounts_user"."is_staff", "accounts_user"."is_active", "accounts_user"."date_joined", "accounts_user"."email", "accounts_user"."avatar", "accounts_user"."last_location"::bytea, "accounts_user"."uuid", "accounts_user"."country", "accounts_user"."city", "accounts_user"."phone" FROM "accounts_user" ORDER BY "accounts_user"."id" ASC LIMIT 1; args=()

In [39]: u1.pk                                                                                                                                                                                            
Out[39]: 1

In [40]: u2 = User(pk=1)                                                                                                                                                                                  

In [41]: u2.email = 'email@email.com'                                                                                                                                                                     

In [42]: u2.save()                                                                                                                                                                                        
(0.006) UPDATE "accounts_user" SET "password" = '', "last_login" = NULL, "is_superuser" = false, "username" = 'email@email.com', "first_name" = '', "last_name" = '', "is_staff" = false, "is_active" = true, "date_joined" = '2020-05-13T01:20:47.718449+00:00'::timestamptz, "email" = 'email@email.com', "avatar" = '', "last_location" = NULL, "uuid" = '89ba0924-03a7-44d2-bc6d-5fd2dcb0de0b'::uuid, "country" = NULL, "city" = NULL, "phone" = NULL WHERE "accounts_user"."id" = 1; args=('', False, 'email@email.com', '', '', False, True, datetime.datetime(2020, 5, 13, 1, 20, 47, 718449, tzinfo=<UTC>), 'email@email.com', '', UUID('89ba0924-03a7-44d2-bc6d-5fd2dcb0de0b'), 1)

Answer 1

考虑这个例子。假设我们有一个简单的模型

CONSTANT = 10


def foo_pk_default():
    return CONSTANT


class Foo(models.Model):
    id = models.IntegerField(primary_key=True, default=foo_pk_default)
    name = models.CharField(max_length=10)

我在这个例子中所做的主要事情是，我确实为 Primary Keys 设置了一个默认的可调用函数。另外，为了演示，我只从函数返回了一个值。

## Django 2.2
In [5]: foo_instance_1 = Foo(name='foo_name_1')

In [6]: foo_instance_1.save()

In [7]: print(foo_instance_1.__dict__)
{'_state': , 'id': 10, 'name': 'foo_name_1'}

In [8]: foo_instance_2 = Foo(name='foo_name_2')

In [9]: foo_instance_2.save()

In [10]: print(foo_instance_2.__dict__)
{'_state': , 'id': 10, 'name': 'foo_name_2'}

## Django 3.X
In [6]: foo_instance_1 = Foo(name='foo_name_1')

In [7]: foo_instance_1.save()

In [8]: print(foo_instance_1.__dict__)
{'_state': , 'id': 10, 'name': 'foo_name_1'}

In [9]: foo_instance_2 = Foo(name='foo_name_2')

In [10]: foo_instance_2.save()
# Raised "IntegrityError: UNIQUE constraint failed: music_foo.id"

结论

在 中Django<3.0，如果存在与模型实例关联的 PK 值，Model.save()则将执行更新或插入操作，而在 中Django>=3.0，仅执行插入操作，因此UNIQUE constraint failed例外。

此更改对当前项目的影响

由于此 Django 更改仅在创建新实例时适用并且我们通常不会为主键设置任何默认值函数。

简而言之，除非您在模型实例创建期间提供默认值，否则此更改不会产生任何问题。