在中间件上下文中获取 starlette 请求正文

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

所以这条线body = await request.body()冻结所有有正文的请求，我收到了所有请求的 504 请求。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。

我不会创建一个继承自 BaseHTTPMiddleware 的中间件，因为它有一些issuesFastAPI 为您提供了创建自己的路由器的机会，根据我的经验，这种方法要好得多。

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