如何序列化嵌套属性而不显示整个对象？

Question

我的目标如下：

{
  "name": "foo",
  "parent": {"id": "bar", "age": 27}
}

我希望得到以下输出：

{"name": "foo", "parent_id": "bar"}

这是我的Pydantic模型

型号：

class TeacherOutput(PropertyBaseModel):
    name: Optional[str]
    parent_id: Optional[str]

控制器：

@router.get('/teachers', response_model=List[TeacherOutput])
async def get_all():
    teachers = app.model.Teacher.all()
    return teachers

我不知道如何从嵌套对象映射所需的属性。我不需要其他属性

Answer 1

您可以使用的另一种方法（IMHO更干净）是使用多个模型，如Gino的回答所示，但是您可以简单地在path操作函数中转换数据，而不是覆盖__init__函数，如下所示：

from pydantic import BaseModel
from typing import Dict, List, Optional, Union

# I like to specify submodels too with Pydantic to use the dot notation
# but Dict type indicated by Gino works perfectly too 
class TeacherParent(BaseModel):
    id: str
    age: Union[str, int]

class TeacherInput(BaseModel):
    name: str
    parent: TeacherParent

class TeacherOutput(BaseModel):
    name: Optional[str] = None
    parent_id: Optional[str] = None


@router.get('/teachers', response_model=List[TeacherOutput])
async def get_all():
    """Here you still specify the response model for documentation purposes, 
       but transformation happens within the function"""

    teachers = app.model.Teacher.all()
    output = [ 
        TeacherOutput(name=t.name, parent_id=t.parent.id)
        for t in teachers
    ]
    # use parent_id=t.parent['id'] if you didn't specify the parent submodel

    return output