Question

我是Django的新手。我尝试从字典中创建我的两个模型披萨和浇头的几个实例：

dictionary = dict()

dictionary[“Pizza1”] = (“A”,”B”,”C”)

dictionary[“Pizza2”] = () 

dictionary[“Pizza3”] = (“B”,”D”)

我想将它们与ManyToMany关系联系起来，所以Pizza1有A，B，C配料，但Topping B也可以在Pizza 1和3中。我的代码不是以某种方式工作，它给了我三个披萨实例（Pizza1， 2,3）但都有所有浇头（A，B，C，D）而不是只有相应的浇头。我该怎么办呢？谢谢！

for pizza in dictionary:
    pizza_instance, created1 = Pizza.objects.get_or_create(name=pizza)
    if (len(dictionary[pizza])>0):
       for topping in dictionary[pizza]:
           topping_instance, created2 = Topping.objects.get_or_create(name=topping)
           pizza_instance.has_toppings.add(topping_instance)
           pizza_instance.save()

模型：

  class Topping (models.Model):    
    name = models.CharField(max_length=20)  
 class Pizza (models.Model):     
    name = models.CharField (max_length=20)     
    has_toppings = models.ManyToManyField(Topping)

Answer 1

试试这个，对我有用：

for pizza in dictionary:
    pizza_instance, created1 = Pizza.objects.get_or_create(name=pizza)
    for topping in dictionary[pizza]:
        topping_instance, created2 = Topping.objects.get_or_create(name=topping)
        pizza_instance.has_toppings.add(topping_instance)
    pizza_instance.save()

测试：

>>> for p in Pizza.objects.all():
...     print(p.has_toppings.count())
... 
3
0
2

创建具有多对多关系的新实例迭代字典DJANGO

1 个答案: