Question

我有很多端点，但是它们使用相同的jsonapi模式。

POST /foods
{
    "data": {
        "type": "food",
        "attributes": {
            "name": "abc",
            "item": "food"
        }
    }
}

POST /chats
{
    "data": {
        "type": "food",
        "attributes": {
            "name": "bcd",
            "item": "chat"
        }
    }
}

POST /animals
{
    "data": {
        "type": "food",
        "attributes": {
            "name": "poi",
            "item": "animals"
        }
    }
}

我的模式是：

import marshmallow_jsonapi as jsonapi
ItemSchema(jsonapi.Schema):
    id = jsonapi.fields.UUID(dumps_only=True)
    name = jsonapi.fields.String(required=True)
    item = jsonapi.fields.String(required=True)

    class Meta:
        type_ = "items"
        strict = True

遵循jsonapi标准，type将需要匹配每个端点（foods，chats，animals）。

import marshmallow_jsonapi as jsonapi
FoodSchema(jsonapi.Schema):
    id = jsonapi.fields.UUID(dumps_only=True)
    name = jsonapi.fields.String(required=True)
    item = jsonapi.fields.String(required=True)

    class Meta:
        type_ = "foods"
        strict = True

import marshmallow_jsonapi as jsonapi
ChatSchema(jsonapi.Schema):
    id = jsonapi.fields.UUID(dumps_only=True)
    name = jsonapi.fields.String(required=True)
    item = jsonapi.fields.String(required=True)

    class Meta:
        type_ = "chats"
        strict = True

import marshmallow_jsonapi as jsonapi
AnimalSchema(jsonapi.Schema):
    id = jsonapi.fields.UUID(dumps_only=True)
    name = jsonapi.fields.String(required=True)
    item = jsonapi.fields.String(required=True)

    class Meta:
        type_ = "animals"
        strict = True

有什么方法可以减少冗余代码，而不必更改ItemSchema的一堆副本type？

Answer 1

这是基类（BaseClass.py）

from marshmallow_jsonapi import fields, Schema
    
class Base(Schema):
    id = jsonapi.fields.UUID(dumps_only=True)
    name = jsonapi.fields.String(required=True)
    item = jsonapi.fields.String(required=True)

您可以在其他架构中导入Base类

from BaseClass import Base

FoodSchema(Base):
    # add other fields if required

class Meta:
    type_ = "foods"
    strict = True

您需要具有相同字段的其他架构

from Base import Base

ChatSchema(Base):
    # add other fields if required
    
class Meta:
    type_ = "chats"
    strict = True

Answer 2

我太迟了一年。但是您可以从BaseSchema开始扩展，只需更改派生架构中的值

您可以在https://marshmallow.readthedocs.io/en/latest/extending.html#custom-class-meta-options

上查看文档以扩展架构并减少代码冗余

marshmallow-jsonapi：如何重用架构但更改`type_`？

2 个答案: