在Django中,模型的save()方法是延迟执行的吗?
例如,在以下代码示例中,Django将在哪一行访问数据库?
my_model = MyModel()
my_model.name = 'Jeff Atwood'
my_model.save()
# Some code that is independent of my_model...
model_id = model_instance.id
print (model_id)
在Django中,模型的save()方法是延迟执行的吗?
例如,在以下代码示例中,Django将在哪一行访问数据库?
my_model = MyModel()
my_model.name = 'Jeff Atwood'
my_model.save()
# Some code that is independent of my_model...
model_id = model_instance.id
print (model_id)
拥有一个懒惰的保存并没有太多意义,对吗? Django的QuerySets
是懒加载的,但模型的save
方法却不是。
以下是Django源代码:
django/db/models/base.py
, 424-437行:
def save(self, force_insert=False, force_update=False, using=None):
"""
Saves the current instance. Override this in a subclass if you want to
control the saving process.
The 'force_insert' and 'force_update' parameters can be used to insist
that the "save" must be an SQL insert or update (or equivalent for
non-SQL backends), respectively. Normally, they should not be set.
"""
if force_insert and force_update:
raise ValueError("Cannot force both insert and updating in \
model saving.")
self.save_base(using=using, force_insert=force_insert,
force_update=force_update)
save.alters_data = True
接下来,save_base
承担了繁重的工作(同一文件,第439-545行):
...
transaction.commit_unless_managed(using=using)
...
在django/db/transaction.py
文件的第167行到178行,您会发现:
def commit_unless_managed(using=None):
"""
Commits changes if the system is not in managed transaction mode.
"""
...
补充说明:所有行号适用于django版本(1, 3, 0, 'alpha', 0)
。