如何在 FastApi 中隐藏输入参数？

Question

如何在 OpenApi 中隐藏请求参数？我想从 OpenApi UI 中隐藏 user_agent。

我有一个简单的应用程序：

from typing import Optional
from fastapi import FastAPI, Header

app = FastAPI()
   
@app.get("/items/")
async def read_items(
       user_agent: Optional[str] = Header(None), 
       size: Optional[int] = Body(None)):
    return {"User-Agent": user_agent}

Answer 1

version 0.73.0 (related PR) 中的 FastAPI 原生支持此功能：https://fastapi.tiangolo.com/tutorial/query-params-str-validations/#exclude-from-openapi。

from typing import Optional

from fastapi import Body, FastAPI, Header

app = FastAPI()


@app.get("/items/")
async def read_items(
    user_agent: Optional[str] = Header(None, include_in_schema=False), 
    size: Optional[int] = Body(None)
):
    return {"User-Agent": user_agent}

Answer 2

您可以customize OpenAPI 模式来隐藏任意参数。下面的例子隐藏paramB：

from fastapi import FastAPI
from fastapi.openapi.utils import get_openapi

app = FastAPI()


@app.get("/")
def get_items(paramA: int, paramB: int):
    pass


def custom_openapi():
    if app.openapi_schema:
        return app.openapi_schema
    openapi_schema = get_openapi(
        title="Custom title",
        version="2.5.0",
        description="This is a very custom OpenAPI schema",
        routes=app.routes,
    )
    # Remove paramB
    params = openapi_schema["paths"]["/"]["get"]["parameters"]
    params = [param for param in params if param["name"] != "paramB"]
    openapi_schema["paths"]["/"]["get"]["parameters"] = params
    app.openapi_schema = openapi_schema
    return app.openapi_schema


app.openapi = custom_openapi