为什么添加这个__init __（）方法会破坏我的Django模型？

Question

这个问题是我之前提到的问题的后续问题here。

我有一个Django模型如下：

class MyModel(models.Model):
    my_field1 = models.DateTimeField(default=datetime.utcnow, editable=False)
    my_field2 = models.DateTimeField()

有效：

>>> MyModel.objects.all()
[<MyModel: MyModel object>, <MyModel: MyModel object>,

现在我将以下构造函数添加到MyModel：

def __init__(self, **kwargs):
    super(MyModel, self).__init__(**kwargs)
    if self.my_field2 is None:
        self.my_field2 = self.my_field1

这打破了班级：

>>> MyModel.objects.all()
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "MYvirtualenv/lib/python2.7/site-packages/django/db/models/query.py", line 138, in __repr__
    data = list(self[:REPR_OUTPUT_SIZE + 1])
  File "MYvirtualenv/lib/python2.7/site-packages/django/db/models/query.py", line 162, in __iter__
    self._fetch_all()
  File "MYvirtualenv/lib/python2.7/site-packages/django/db/models/query.py", line 965, in _fetch_all
    self._result_cache = list(self.iterator())
  File "MYvirtualenv/src/django-cache-machine-master/caching/base.py", line 118, in __iter__
    obj = iterator.next()
  File "MYvirtualenv/lib/python2.7/site-packages/django/db/models/query.py", line 255, in iterator
    obj = model_cls.from_db(db, init_list, row[model_fields_start:model_fields_end])
  File "MYvirtualenv/lib/python2.7/site-packages/django/db/models/base.py", line 489, in from_db
    new = cls(*values)
TypeError: __init__() takes exactly 1 argument (2 given)

为什么在添加相对简单的构造函数时会发生这种情况？我该如何补救呢？我需要构造函数中的逻辑，所以我不能消除它。

Answer 1

尽管强烈建议您在文档中使用classmethod或自定义管理器，但由于您修改了superclass的调用签名，因此您的代码无效：

super(MyModel, self).__init__(*args, **kwargs)

您可以使用classmethod：

来避免这样做

class MyModel(models.Model):
    my_field1 = models.DateTimeField(default=datetime.utcnow, editable=False)
    my_field2 = models.DateTimeField()

    @classmethod
    def sync_dates(cls, myfield1):
         my_model = cls(my_field1=myfield1, my_field2=myfield1)
         # do something with my_model
         return my_model

MyModel.sync_dates(some_date)可以解决这个问题。

或者您可以使用自定义管理器，这是首选方式：

class MyModelManager(models.Manager):
    def create_with_sync_date(self, myfield1):
        my_model = self.create(my_field1=myfield1, my_field2=myfield1)
        # do something with my_model
        return my_model

class MyModel(models.Model):
    my_field1 = models.DateTimeField(default=datetime.utcnow, editable=False)
    my_field2 = models.DateTimeField()

    objects_synced = MyModelManager()

您可以致电MyModel.objects_synced.create_with_sync_date(some_date)

Answer 2

您不需要在Django模型中使用__init__，而是使用Meta类

class MyModel(models.Model):
    my_field1 = models.DateTimeField(default=datetime.utcnow, editable=False)
    my_field2 = models.DateTimeField()

    class Meta:
        verbose_name = 'My Model'
        verbose_name_plural = 'My Models' # and so on

Answer 3

不建议使用 init 。 根据Django的文档：

https://docs.djangoproject.com/en/1.9/ref/models/instances/#creating-objects

from django.db import models

class Book(models.Model):
    title = models.CharField(max_length=100)

    @classmethod
    def create(cls, title):
        book = cls(title=title)
        # do something with the book
        return book

 book = Book.create("Pride and Prejudice")