Python 检查Django中的OneToOneField是否为None
我有两个这样的模型:Python 检查Django中的OneToOneField是否为None,python,django,django-models,one-to-one,Python,Django,Django Models,One To One,我有两个这样的模型: class Type1Profile(models.Model): user = models.OneToOneField(User, unique=True) ... class Type2Profile(models.Model): user = models.OneToOneField(User, unique=True) ... 如果用户有Type1或Type2配置文件,我需要执行以下操作: if request.user.ty
class Type1Profile(models.Model):
user = models.OneToOneField(User, unique=True)
...
class Type2Profile(models.Model):
user = models.OneToOneField(User, unique=True)
...
如果用户有Type1或Type2配置文件,我需要执行以下操作:
if request.user.type1profile != None:
# do something
elif request.user.type2profile != None:
# do something else
else:
# do something else
但是,对于没有type1或type2配置文件的用户,执行这样的代码会产生以下错误:
Type1Profile matching query does not exist.
如何检查用户的配置文件类型
谢谢使用try/except块怎么样
def get_profile_or_none(user, profile_cls):
try:
profile = getattr(user, profile_cls.__name__.lower())
except profile_cls.DoesNotExist:
profile = None
return profile
然后,像这样使用
u = request.user
if get_profile_or_none(u, Type1Profile) is not None:
# do something
elif get_profile_or_none(u, Type2Profile) is not None:
# do something else
else:
# d'oh!
我想您可以将此作为一个通用函数来获取任何反向OneToOne实例,给定一个原始类(此处:您的配置文件类)和一个相关实例(此处:request.user)。要检查(OneToOne)关系是否存在,您可以使用hasattr
函数:
if hasattr(request.user, 'type1profile'):
# do something
elif hasattr(request.user, 'type2profile'):
# do something else
else:
# do something else
使用
选择相关的
>>> user = User.objects.select_related('type1profile').get(pk=111)
>>> user.type1profile
None
只需测试模型上的相应字段的None
ness,就可以查看特定模型中可为空的一对一关系是否为空,但前提是您在一对一关系起源的模型上进行测试。例如,给定这两个类
class Place(models.Model):
name = models.CharField(max_length=50)
address = models.CharField(max_length=80)
class Restaurant(models.Model): # The class where the one-to-one originates
place = models.OneToOneField(Place, blank=True, null=True)
serves_hot_dogs = models.BooleanField()
serves_pizza = models.BooleanField()
…要查看餐厅
是否有场所
,我们可以使用以下代码:
>>> r = Restaurant(serves_hot_dogs=True, serves_pizza=False)
>>> r.save()
>>> if r.place is None:
>>> print "Restaurant has no place!"
Restaurant has no place!
要查看地点
是否有餐厅
,重要的是要了解在地点
的实例上引用餐厅
属性会引发一个餐厅。如果没有相应的餐厅,则不会出现异常。这是因为Django使用QuerySet.get()
在内部执行查找。例如:
>>> p2 = Place(name='Ace Hardware', address='1013 N. Ashland')
>>> p2.save()
>>> p2.restaurant
Traceback (most recent call last):
...
DoesNotExist: Restaurant matching query does not exist.
在这种情况下,Occam的剃须刀占了上风,确定<代码>场所<代码>是否有<代码>餐厅<代码>的最佳方法将是标准<代码>尝试<代码>/<代码>,除了所述的<代码>构造
虽然Jocee建议使用hasattr
在实践中有效,但它实际上只是偶然起作用,因为hasattr
抑制了所有异常(包括DoesNotExist
),而不是像它应该的那样只抑制AttributeError
。正如所指出的,在Python3.2中,实际上根据以下票证更正了此行为:。此外,尽管听起来有点固执己见,但我相信上面的try
/除外结构更能代表Django的工作方式,而使用hasattr
可能会给新手带来麻烦,这可能会造成FUD并传播坏习惯
编辑合理的妥协对我来说也是合理的。我喜欢,因为它太简单了
if hasattr(request.user, 'type1profile'):
# do something
elif hasattr(request.user, 'type2profile'):
# do something else
else:
# do something else
其他评论者担心它可能无法与某些版本的Python或Django一起使用,但将此技术作为选项之一:
您还可以使用hasattr来避免异常捕获的需要:
当然,文档还显示了异常捕获技术:
p2没有关联餐厅:
我同意捕捉异常可以让事情变得更清楚,但对我来说似乎更混乱。也许这是一个合理的妥协
>>> print(Restaurant.objects.filter(place=p2).first())
None
这只是按地点查询餐厅
对象。如果那个地方没有餐厅,它将返回None
这里有一个可执行代码片段供您使用。如果您安装了Python、Django和SQLite3,那么应该只运行它。我使用Python2.7、Python3.4、Django 1.9.2和SQLite3.8.2对其进行了测试
# Tested with Django 1.9.2
import sys
import django
from django.apps import apps
from django.apps.config import AppConfig
from django.conf import settings
from django.core.exceptions import ObjectDoesNotExist
from django.db import connections, models, DEFAULT_DB_ALIAS
from django.db.models.base import ModelBase
NAME = 'udjango'
def main():
setup()
class Place(models.Model):
name = models.CharField(max_length=50)
address = models.CharField(max_length=80)
def __str__(self): # __unicode__ on Python 2
return "%s the place" % self.name
class Restaurant(models.Model):
place = models.OneToOneField(Place, primary_key=True)
serves_hot_dogs = models.BooleanField(default=False)
serves_pizza = models.BooleanField(default=False)
def __str__(self): # __unicode__ on Python 2
return "%s the restaurant" % self.place.name
class Waiter(models.Model):
restaurant = models.ForeignKey(Restaurant)
name = models.CharField(max_length=50)
def __str__(self): # __unicode__ on Python 2
return "%s the waiter at %s" % (self.name, self.restaurant)
syncdb(Place)
syncdb(Restaurant)
syncdb(Waiter)
p1 = Place(name='Demon Dogs', address='944 W. Fullerton')
p1.save()
p2 = Place(name='Ace Hardware', address='1013 N. Ashland')
p2.save()
r = Restaurant(place=p1, serves_hot_dogs=True, serves_pizza=False)
r.save()
print(r.place)
print(p1.restaurant)
# Option 1: try/except
try:
print(p2.restaurant)
except ObjectDoesNotExist:
print("There is no restaurant here.")
# Option 2: getattr and hasattr
print(getattr(p2, 'restaurant', 'There is no restaurant attribute.'))
if hasattr(p2, 'restaurant'):
print('Restaurant found by hasattr().')
else:
print('Restaurant not found by hasattr().')
# Option 3: a query
print(Restaurant.objects.filter(place=p2).first())
def setup():
DB_FILE = NAME + '.db'
with open(DB_FILE, 'w'):
pass # wipe the database
settings.configure(
DEBUG=True,
DATABASES={
DEFAULT_DB_ALIAS: {
'ENGINE': 'django.db.backends.sqlite3',
'NAME': DB_FILE}},
LOGGING={'version': 1,
'disable_existing_loggers': False,
'formatters': {
'debug': {
'format': '%(asctime)s[%(levelname)s]'
'%(name)s.%(funcName)s(): %(message)s',
'datefmt': '%Y-%m-%d %H:%M:%S'}},
'handlers': {
'console': {
'level': 'DEBUG',
'class': 'logging.StreamHandler',
'formatter': 'debug'}},
'root': {
'handlers': ['console'],
'level': 'WARN'},
'loggers': {
"django.db": {"level": "WARN"}}})
app_config = AppConfig(NAME, sys.modules['__main__'])
apps.populate([app_config])
django.setup()
original_new_func = ModelBase.__new__
@staticmethod
def patched_new(cls, name, bases, attrs):
if 'Meta' not in attrs:
class Meta:
app_label = NAME
attrs['Meta'] = Meta
return original_new_func(cls, name, bases, attrs)
ModelBase.__new__ = patched_new
def syncdb(model):
""" Standard syncdb expects models to be in reliable locations.
Based on https://github.com/django/django/blob/1.9.3
/django/core/management/commands/migrate.py#L285
"""
connection = connections[DEFAULT_DB_ALIAS]
with connection.schema_editor() as editor:
editor.create_model(model)
main()
我使用的是has_attr和is None的组合:
class DriverLocation(models.Model):
driver = models.OneToOneField(Driver, related_name='location', on_delete=models.CASCADE)
class Driver(models.Model):
pass
@property
def has_location(self):
return not hasattr(self, "location") or self.location is None
如果你有模型的话
class UserProfile(models.Model):
user = models.OneToOneField(User, unique=True)
对于任何用户,您只需要知道UserProfile是否存在,这是从数据库角度使用现有查询的最有效方法
Exists查询将只返回布尔值,而不是像hasattr(request.user,'type1profile')
那样的反向属性访问-这将生成获取查询,并返回完整的对象表示
为此,您需要向用户模型添加一个属性
class User(AbstractBaseUser)
@property
def has_profile():
return UserProfile.objects.filter(user=self.pk).exists()
智能方法之一是添加自定义字段OneToOneOrNoneField和使用it[适用于Django>=1.9]
from django.db.models.fields.related_descriptors import ReverseOneToOneDescriptor
from django.core.exceptions import ObjectDoesNotExist
from django.db import models
class SingleRelatedObjectDescriptorReturnsNone(ReverseOneToOneDescriptor):
def __get__(self, *args, **kwargs):
try:
return super().__get__(*args, **kwargs)
except ObjectDoesNotExist:
return None
class OneToOneOrNoneField(models.OneToOneField):
"""A OneToOneField that returns None if the related object doesn't exist"""
related_accessor_class = SingleRelatedObjectDescriptorReturnsNone
def __init__(self, *args, **kwargs):
kwargs.setdefault('null', True)
kwargs.setdefault('blank', True)
super().__init__(*args, **kwargs)
实施
class Restaurant(models.Model): # The class where the one-to-one originates
place = OneToOneOrNoneField(Place)
serves_hot_dogs = models.BooleanField()
serves_pizza = models.BooleanField()
用法
r = Restaurant(serves_hot_dogs=True, serves_pizza=False)
r.place # will return None
感谢您提供此解决方案。不幸的是,这并不总是有效。如果您现在或将来想使用select_related()
,或者甚至想确保您还可以处理其他地方可能发生的其他类型的魔法,您必须按如下方式扩展测试:if hasattr(object,'onetoonerevrelatetr')和object.onetoonerevrelatetr!=无
请注意,在Python<3.2中,hasattr
将吞噬数据库查找过程中发生的所有异常,而不仅仅是DoesNotExist
。这可能已损坏,而不是您想要的。不使用Python2.7。即使OneToOne不存在,它也会返回一个django.db.models.fields.related.RelatedManager对象。@alartur您使用的是哪个django版本?django 1.5。但我用一种完全不同的方式实现了我想做的事情,从而解决了我的特殊问题。我知道它是这样工作的,但select_的这种行为真的被记录下来了吗?我只是在Django 1.9.2中试过了,对于django 1.8,您需要使用SingleRelatedObjectDescriptor
,而不是像这样从django.db.models.fields.related导入SingleRelatedObjectDescriptor
from django.db.models.fields.related_descriptors import ReverseOneToOneDescriptor
from django.core.exceptions import ObjectDoesNotExist
from django.db import models
class SingleRelatedObjectDescriptorReturnsNone(ReverseOneToOneDescriptor):
def __get__(self, *args, **kwargs):
try:
return super().__get__(*args, **kwargs)
except ObjectDoesNotExist:
return None
class OneToOneOrNoneField(models.OneToOneField):
"""A OneToOneField that returns None if the related object doesn't exist"""
related_accessor_class = SingleRelatedObjectDescriptorReturnsNone
def __init__(self, *args, **kwargs):
kwargs.setdefault('null', True)
kwargs.setdefault('blank', True)
super().__init__(*args, **kwargs)
class Restaurant(models.Model): # The class where the one-to-one originates
place = OneToOneOrNoneField(Place)
serves_hot_dogs = models.BooleanField()
serves_pizza = models.BooleanField()
r = Restaurant(serves_hot_dogs=True, serves_pizza=False)
r.place # will return None