我的models.py
class Ingredient(models.Model):
name = models.CharField(max_length=16, unique=True)
price = models.SmallIntegerField()
def __str__(self):
return self.name
class Topping(models.Model):
name = models.CharField(max_length=32)
ingredient = models.ForeignKey(Ingredient, related_name='indole',
blank=True, null=True, default='base')
def __str__(self):
return self.nome
class Pizza(models.Model):
nome = models.CharField(max_length=32, unique=True)
toppings = models.ManyToManyField(Topping)
def __str__(self):
return self.nome在行政管理中,它是有效的!我可以添加配料,比萨饼等,但我想使用脚本来填充。
我的script
import os
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'recipt.settings')
import django
django.setup()
from core.models import *
def populate():
cheap = add_ingredient('Cheap', 3)
base = add_ingredient('Base', 5)
good = add_ingredient('Good', 10)
cheese = add_topping('Cheese', None)
tomato = add_topping('Tomato', None)
olive = add_topping('Olive', None)
simple = add_pizza('Simple', cheese) #just one toppings for now
complex = add_pizza('Complex', tomato)
def add_ingredient(name, price):
i = Ingredient.objects.get_or_create(name=name, price=price)[0]
i.save()
return i
def add_topping(name, ingredient):
t = Topping.objects.get_or_create(name=name, ingredient=ingredient)[0]
t.save()
return t
def add_pizza(name, toppings):
p = Pizza.objects.get_or_create(name=name, toppings=toppings)[0]
p.save()
return p
if __name__ == '__main__':
print ("Starting Core population script...")
populate()这个剧本适用于配料和配料,但不适用于比萨。
我的error (很抱歉格式化):
开始核心人口剧本..。 回溯(最近一次调用): 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\query.py",第465行,在get_or_create中 返回self.get(**查找),False 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\query.py",第387行,在get中 self.model._meta.object_name core.models.DoesNotExist:比萨饼匹配查询不存在。 在处理上述异常的过程中,发生了另一个异常: 回溯(最近一次调用): 文件"populate_core.py",第437行,模块中 人口() 文件"populate_core.py",第63行,在填充 simple = add_pizza('Simple',奶酪) 文件"populate_core.py",第307行,add_pizza P= Pizza.objects.get_or_create(name=name,toppings=toppings) 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\manager.py",第122行,在manager_method中 返回getattr(self.get_queryset(),name)(*args,**kwargs) 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\query.py",第467行,在get_or_create中 返回self._create_object_from_params(查找,参数) 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\query.py",第499行,在_create_object_from_params中 obj = self.create(**params) 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\query.py",第399行,在create中 obj = self.model(**kwargs) 文件"c:\Python34\Scripts\recipt\myvenv\lib\site-packages\django\db\mode ls\base.py",第443行,在init中 引发TypeError(“‘’%s‘’是该函数的无效关键字参数”% li st(kwargs)) TypeError:'toppings‘是该函数的无效关键字参数
有什么帮助吗?我在某个地方读到,我应该把配料留为空白,然后再加,但是.
发布于 2016-01-21 01:36:19
当您创建具有ManyToMany字段的数据库记录时,无法正常执行。您必须创建对象,然后向ManyToMany字段添加内容。就像这样。
class Author(models.Model):
name = models.CharField(max_length=100)
class Article(models.Model):
name = models.CharField(max_length=100)
authors = models.ManyToManyField(Author)
zach = Author("Zach Braff")
zach.save()
# Say Zach writes an article...
# You can't do this because the authors field could have many things in it.
a1 = Article(name="Scrubs remake coming?", authors=zach)
# Instead, you have to do this...
a1 = Article(name="Scrubs remake coming?")
a1.authors.add(zach)
a1.save()您可能要做的是用它的等效值替换get_or_create(),如下所示。
p = Pizza.objects.filter(name=name, toppings=toppings)
# This is faster than `if p`
if p.exists():
return p
else:
p = Pizza.objects.create(name=name)
p.toppings.add(toppings)
p.save()
return p我认为那应该有效。
发布于 2016-01-21 11:22:12
很管用,谢谢。
这里是我在script中的新代码
simple = add_pizza('Simple', [cheese, tomato])
complex = add_pizza('Complex', [cheese, tomato, olive])
def add_pizza(name, toppings):
p = Pizza.objects.filter(name=name)
if p.exists():
pass
else:
p = Pizza.objects.create(name=name)
for t in toppings:
p.toppings.add(t)
p.save()
return phttps://stackoverflow.com/questions/34913632
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