FastAPI 怎么更好的返回自定义的格式

holinhot May 5, 2020 4055 views

This topic created in 2198 days ago, the information mentioned may be changed or developed.

官方文档是用 response_model，但感觉这个好像不是很灵活。

我想要返回的统一格式, 把 pydantic 的验证错误也包进自己定义的格式里

pydantic 默认数据返回格式格式：

{ "email": "[email protected]", "id": 6, "is_active": true, "item": [] }

pydantic 默认验证错误格式：

{ "detail": [ { "loc": [ "body", "user", "email" ], "msg": "value is not a valid email address", "type": "value_error.email" } ] }

这种返回格式缺乏规范性，后面调用接口时就非常乱了。

我想要返回的统一格式统一成下面这样，方便调用和判断错误

{ "success": true, "time": "1588633541", "errors": "", "result": { "username": "bigbig", "created_at": "2020-05-04 21:57:22.815393" } } { "success": false, "time": "1588633541", "errors": { "msg": "Authorization failed", "errors": {"detail": "wrong user name or password"} }, "result": "" } { "success": false, "time": "1588633541", "errors": {"detail": [ { "loc": [ "body", "user", "email" ], "msg": "value is not a valid email address", "type": "value_error.email" } ]}, "result": "" }

errors

格式

msg

3 replies 2020-05-07 01:02:51 +08:00

holinhot

May 5, 2020

错误返回格式看文档可以改了

```
@app.exception_handler(RequestValidationError)
async def validation_exception_handler(request: Request, exc: RequestValidationError):
return JSONResponse(
status_code=status.HTTP_422_UNPROCESSABLE_ENTITY,
cOntent=jsonable_encoder(
{
"success": False,
"errors": exc.errors(),
"result": "",
"body": exc.body
}
)
)
```

```
{
"success": false,
"errors": [
{
"loc": [
"body",
"user",
"email"
],
"msg": "value is not a valid email address",
"type": "value_error.email"
}
],
"result": "",
"body": {
"email": "user522example.com",
"password": "string"
}
}
```

response_model 还在研究，不然就只能不用 response_model，完全自己返回 json 内容了

holinhot

May 5, 2020

仔细研究已搞定
```
class User(UserBase):
id: int
is_active: bool
item: List[Item] = []

class Config:
orm_mode = True

class Standard_result_format(BaseModel):
success: bool = True
errors: list = []
result: User
```

是可以在下面调用另一下 BaseModel class

holinhot

May 7, 2020

改了错误格式后自动生成的文档是一个应该是