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python - 在 SQLAlchemy 中建模关系以使用 Marshmallow 显示嵌套的“集合”

问题描述

我试图弄清楚如何对（flask-）SQLAlchemy 和（flask-）Marshamallow 进行建模以提供以下 JSON 输出。

父级是产品，子级是“变体集”......它们是选项的类型，例如“颜色”或“尺寸”。嵌套在这些集合中，我想要选项本身（S、M、L 等）

好像我错过了一些明显的东西。

期望的输出：

{
  "skuid": "B1234",
  "name": "Test Product 1",
  "variant_sets": [
    {
        "variant_set": "B1234_1",
        "variants": [
            {
                "code": "S",
                "variant_type": "size",
                "description": "Small
            }, 
            {
                "code": "M",
                "variant_type": "size",
                "description": "Medium
            },
            {
                "code": "L",
                "variant_type": "size",
                "description": "Large
            }
        ]
    },
    {
        "variant_set": "B1234_2",
        "variants": [
            {
                "code": "RD",
                "variant_type": "color",
                "description": "Small
            }, 
            {
                "code": "GR",
                "variant_type": "color",
                "description": "Green
            },
            {
                "code": "YL",
                "variant_type": "color",
                "description": "Yellow
            }
        ]
    }
  ]
}

到目前为止，我拥有的棉花糖模式：

class ProductToOptionSchema(ma.ModelSchema):
    variants = ma.Nested(OptionSchema, many=True)

    class Meta:
        model = ProductToOption


class ProductSchema(ma.ModelSchema):
    variant_sets = ma.Nested(ProductToOptionSchema, many=True)

    class Meta:
        model = Product

当我尝试这段代码时：

product = Product.query.filter_by(skuid="B1234").first()
product_schema = ProductSchema()
result = product_schema.jsonify(product)

我得到的错误是：

TypeError: 'Option' object is not iterable

产品通过辅助表与变体（选项）相关。我到目前为止的模型是：

产品

-----------------------------
| skuid | name              |
-----------------------------
| B1234 | Test Product 1    |
-----------------------------
| B1235 | Test Product 2    |
-----------------------------

class Product(db.Model):
    __tablename__ = 'products'

    skuid = db.Column(db.String(16), primary_key=True)
    name = db.Column(db.String(128))
    variants = db.relationship("Option", secondary="products_to_options")

products_to_options

------------------------
| skuid | variant_set  |
------------------------
| B1234 | B1234_1      |
------------------------
| B1234 | B1234_2      |
------------------------
| B1235 | B1235_1      |
------------------------

class ProductToOption(db.Model):
    __tablename__ = 'products_to_options'

    skuid = db.Column(db.String(16), db.ForeignKey('products.skuid'), nullable=False)
    variant_set = db.Column(db.String(16), db.ForeignKey('options.variant_set'), nullable=False)

    products = db.relationship('Product', foreign_keys="ProductToOption.skuid")
    variants = db.relationship('Option', foreign_keys="ProductToOption.variant_set")

选项

-----------------------------------------------------
| variant_set | code | variant_type | description   |
-----------------------------------------------------
| B1234_1     | S    | size         | Small         |
-----------------------------------------------------
| B1234_1     | M    | size         | Medium        |
-----------------------------------------------------
| B1234_1     | L    | size         | Large         |
-----------------------------------------------------
| B1234_2     | RD   | color        | Red           |
-----------------------------------------------------
| B1234_2     | GR   | color        | Green         |
-----------------------------------------------------
| B1234_2     | YL   | color        | Yellow        |
-----------------------------------------------------
| B1235_1     | OK   | wood         | Oak           |
-----------------------------------------------------
| B1235_1     | CH   | wood         | Cherry        |
-----------------------------------------------------

class Option(db.Model):
    __tablename__ = 'options'

    variant_set = db.Column(db.String(16), nullable=False)
    code = db.Column(db.String(8), nullable=False)
    variant_type = db.Column(db.String(16), nullable=False)
    description = db.Column(db.String(16), nullable=False)

    product = db.relationship("Product", secondary="products_to_options")

标签： pythonsqlalchemyflask-sqlalchemymarshmallow

解决方案

想我想通了。我修改了表之间的 SQLAlchemy 关系并删除了“辅助”属性。现在的关系是：

skuid -> variant_set -> 变种

这似乎奏效了。或者至少，我得到了我想要的输出。

class Product(db.Model):
    __tablename__ = 'products'

    skuid = db.Column(db.String(16), primary_key=True)
    name = db.Column(db.String(128))
    variant_sets = db.relationship("ProductToOption")

class ProductToOption(db.Model):
    __tablename__ = 'products_to_options'

    id = db.Column(db.INTEGER, primary_key=True)
    skuid = db.Column(db.String(36), db.ForeignKey('products.skuid'), nullable=False)
    options_code = db.Column(db.String(36), nullable=False)

    variants = db.relationship('Option')

class Option(db.Model):
    __tablename__ = 'options'

    variant_set = db.Column(db.String(16), db.ForeignKey('products_to_options.options_code'), nullable=False)
    code = db.Column(db.String(8), nullable=False)
    variant_type = db.Column(db.String(16), nullable=False)
    description = db.Column(db.String(16), nullable=False)

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