PydantiC语言 在response_model中使用带有"Field"的"Union"时无法序列化/验证

Question

• 操作系统：Ubuntu 18.04.4 LTS
• Python版本：3.7.7
• Pydantic版本：1.4

我正试图将class TableSetting设为BaseModel，并将其设为response body。但是这个类中似乎存在一些验证或序列化错误。我想可能是Union或Optional[str]，但我不确定。

源代码

from fastapi import FastAPI, Query, Body, Path
from pydantic import BaseModel, Field
from typing import Union, Optional
from enum import Enum

app = FastAPI()

class Tableware(str, Enum):
SPOON= "SPOON"
FORK= "FORK"
class TableSetting(BaseModel):
kitchen: Union[Tableware, Optional[str]] = Field(..., title="I am title")
numberOfknife: int = Field(None, ge=0, le=4)
@app.post(
"/{tableNumber}/provide-table-setting",
response_model= TableSetting,    
)
def provide_table_setting(
*,
tableNumber: str= Path(...),
):
results = {"tableNumber": tableNumber}
return results

class TableSetting的json模式应该看起来像这个

TableSetting{
kitchen*        I am title{
anyOf ->    string  
Enum:
[ SPOON, FORK ]
string
}
numberOfknife   integer
maximum: 4
minimum: 0
}

执行curl时

curl -X POST "http://localhost:8000/2/provide-table-setting" -H  "accept: application/json"

它返回的错误如下，响应代码为500，内部服务器错误，似乎kitchen存在一些验证问题，但我不知道为什么。

INFO:     127.0.0.1:53558 - "POST /2/provide-table-setting HTTP/1.1" 500 Internal Server Error
ERROR:    Exception in ASGI application
Traceback (most recent call last):
File "/home/*****/miniconda3/envs/fastAPI/lib/python3.7/site-packages/uvicorn/protocols/http/httptools_impl.py", line 385, in run_asgi
result = await app(self.scope, self.receive, self.send)
.
.
.
File "/home/*****/miniconda3/envs/fastAPI/lib/python3.7/site-packages/fastapi/routing.py", line 126, in serialize_response
raise ValidationError(errors, field.type_)
pydantic.error_wrappers.ValidationError: 1 validation error for TableSetting
response -> kitchen
field required (type=value_error.missing)

但是，当从代码中删除= Field(..., title="I a title")时，如下所示。它与响应代码200配合使用很好，但由于我需要Item kitchen，所以这不是我想要的。

class TableSetting(BaseModel):
kitchen: Union[Tableware, Optional[str]]
numberOfknife: int = Field(None, ge=0, le=4)

我该怎么解决这个问题？

感谢

Answer 1

根据单据联合，它接受不同的类型，但不能同时接受，相反，您应该将tuple与(必需，可选(一起使用类似的东西

from typing import Optional, Tuple
class TableSetting(BaseModel):
kitchen: Tuple[Tableware, Optional[str]] = (None, Field(..., title="I am title"))
#kitchen: Union[Tableware, Optional[str]] 
numberOfknife: int = Field(None, ge=0, le=4)