python - 在中间件上下文中获取 starlette 请求正文

Question

我有这样的中间件

class RequestContext(BaseHTTPMiddleware):
    async def dispatch(self, request: Request, call_next: RequestResponseEndpoint):
        request_id = request_ctx.set(str(uuid4()))  # generate uuid to request
        body = await request.body()
        if body:
            logger.info(...)  # log request with body
        else:
            logger.info(...)  # log request without body
 
        response = await call_next(request)
        response.headers['X-Request-ID'] = request_ctx.get()
        logger.info("%s" % (response.status_code))
        request_ctx.reset(request_id)

        return response

所以行 body = await request.body()卡住所有有正文的请求，我有来自所有请求的 504。在这种情况下，如何安全地读取请求正文？我只想记录请求参数。

Answer 1

我不会创建一个继承自 BaseHTTPMiddleware 的中间件，因为它有一些 issues , FastAPI 让您有机会创建自己的路由器，根据我的经验，这种方法更好。

from fastapi import APIRouter, FastAPI, Request, Response, Body
from fastapi.routing import APIRoute

from typing import Callable, List
from uuid import uuid4


class ContextIncludedRoute(APIRoute):
    def get_route_handler(self) -> Callable:
        original_route_handler = super().get_route_handler()

        async def custom_route_handler(request: Request) -> Response:
            request_id = str(uuid4())
            response: Response = await original_route_handler(request)

            if await request.body():
                print(await request.body())

            response.headers["Request-ID"] = request_id
            return response

        return custom_route_handler


app = FastAPI()
router = APIRouter(route_class=ContextIncludedRoute)


@router.post("/context")
async def non_default_router(bod: List[str] = Body(...)):
    return bod


app.include_router(router)

按预期工作。

b'["string"]'
INFO:     127.0.0.1:49784 - "POST /context HTTP/1.1" 200 OK