Question

这是事情。分支机构可能包含许多产品。表单会以字符串格式将path和branch发布到产品模型。

获得发布的数据后，我怎么能像这样Product.objects.create(path="path1", branch="branch1")一样使用？或者必须在forms.py？

中创建分支实例

这是错误的版本：它会引发ValueError: Cannot assign "'branch1'": "Product.branch" must be a "Branch" instance.

class Branch(models.Model):
    name = models.CharField(max_length=63, unique=True, blank=True)


class Product(models.Model):
    path = models.CharField(max_length=255)
    branch = models.ForeignKey(Branch, on_delete=models.CASCADE)

    def save(self, *args, **kwargs):
        kwargs['branch'], _ = Branch.objects.get_or_create(name=kwargs['branch'])
        super(Product, self).save(*args, **kwargs)

Answer 1

这不是保存问题。在将字符串赋值给分支名称期间会出错。如果要实现逻辑，请在保存之前执行

您可以使用python property来实现此目的，只需进行少量修改即可。不需要覆盖save方法。

models.py

class Branch(models.Model):
    name = models.CharField(max_length=63, unique=True)


class Product(models.Model):
    path = models.CharField(max_length=255)
    branch = models.ForeignKey(Branch, on_delete=models.CASCADE)

    @property
    def branch_name(self):
        return self.branch.name
    @branch_name.setter
    def branch_name(self, value):
        self.branch, _ = Branch.objects.get_or_create(name=value)

你的创建功能应该是

Product.objects.create(path="path1", branch_name="branch1")

注意：它是branch_name而不是branch。另外product.branch仍为branch对象，poduct.branch_name返回分支的名称。这也适用于更新。新product.branch_name branch更新了product

的

const users = [
  {
    name: 'Samir',
    age: 27,
    favoriteBooks:[
      {title: 'The Iliad'},
      {title: 'The Brothers Karamazov'}
    ]
  },
  {
    name: 'Angela',
    age: 33,
    favoriteBooks:[
      {title: 'Tenth of December'},
      {title: 'Cloud Atlas'},
      {title: 'One Hundred Years of Solitude'}
    ]
  },
  {
    name: 'Beatrice',
    age: 42,
    favoriteBooks:[
      {title: 'Candide'}
    ]
  }
];

    // Result: ['The Iliad', 'The Brothers Karamazov', 'Tenth of December', 'Cloud Atlas', 'One Hundred Years of Solitude', 'Candide'];

var books = users
  .map(function(user) {
    return user.favoriteBooks;
  })
  .map(function(book) {
    return book.title;
  })
  .reduce(function(arr, titles) {
    return [ ...arr, ...titles ];
  }, []);

console.log(books);

Django：根据字符串参数保存外键？

1 个答案: