Catch `Exception` globally in FastAPI - python-3.x

Question

I am trying to catch unhandled exceptions at global level. So somewhere in main.py file I have the below:
#app.exception_handler(Exception)
async def exception_callback(request: Request, exc: Exception):
logger.error(exc.detail)
But the above method is never executed. However, if I write a custom exception and try to catch it (as shown below), it works just fine.
class MyException(Exception):
#some code
#app.exception_handler(MyException)
async def exception_callback(request: Request, exc: MyException):
logger.error(exc.detail)
I have gone through Catch exception type of Exception and process body request #575. But this bug talks about accessing request body. After seeing this bug, I feel it should be possible to catch Exception.
FastAPI version I am using is: fastapi>=0.52.0.
Thanks in advance :)
Update
There are multiple answers, I am thankful to all the readers and authors here.
I was revisiting this solution in my application. Now I see that I needed to set debug=False, default it's False, but I had it set to True in
server = FastAPI(
title=app_settings.PROJECT_NAME,
version=app_settings.VERSION,
)
It seems that I missed it when #iedmrc commented on answer given by #Kavindu Dodanduwa.

Answer 1

In case you want to capture all unhandled exceptions (internal server error), there's a very simple way of doing it. Documentation
from fastapi import FastAPI
from starlette.requests import Request
from starlette.responses import Response
app = FastAPI()
async def catch_exceptions_middleware(request: Request, call_next):
try:
return await call_next(request)
except Exception:
# you probably want some kind of logging here
return Response("Internal server error", status_code=500)
app.middleware('http')(catch_exceptions_middleware)
Make sure you place this middleware before everything else.

Answer 2

You can do something like this. It should return a json object with your custom error message also works in debugger mode.
from fastapi import FastAPI
from fastapi.responses import JSONResponse
app = FastAPI()
#app.exception_handler(Exception)
async def validation_exception_handler(request, err):
base_error_message = f"Failed to execute: {request.method}: {request.url}"
# Change here to LOGGER
return JSONResponse(status_code=400, content={"message": f"{base_error_message}. Detail: {err}"})

Answer 3

Adding a custom APIRoute can be also be used to handle global exceptions. The advantage of this approach is that if a http exception is raised from the custom route it will be handled by default Starlette's error handlers:
from typing import Callable
from fastapi import Request, Response, HTTPException, APIRouter, FastAPI
from fastapi.routing import APIRoute
from .logging import logger
class RouteErrorHandler(APIRoute):
"""Custom APIRoute that handles application errors and exceptions"""
def get_route_handler(self) -> Callable:
original_route_handler = super().get_route_handler()
async def custom_route_handler(request: Request) -> Response:
try:
return await original_route_handler(request)
except Exception as ex:
if isinstance(ex, HTTPException):
raise ex
logger.exception("uncaught error")
# wrap error into pretty 500 exception
raise HTTPException(status_code=500, detail=str(ex))
return custom_route_handler
router = APIRouter(route_class=RouteErrorHandler)
app = FastAPI()
app.include_router(router)
Worked for me with fastapi==0.68.1.
More on custom routes: https://fastapi.tiangolo.com/advanced/custom-request-and-route/

Answer 4

It is a known issue on the Fastapi and Starlette.
I am trying to capture the StarletteHTTPException globally by a following simple sample.
import uvicorn
from fastapi import FastAPI
from starlette.requests import Request
from starlette.exceptions import HTTPException as StarletteHTTPException
from starlette.responses import JSONResponse
app = FastAPI()
#app.exception_handler(StarletteHTTPException)
async def exception_callback(request: Request, exc: Exception):
print("test")
return JSONResponse({"detail": "test_error"}, status_code=500)
if __name__ == "__main__":
uvicorn.run("test:app", host="0.0.0.0", port=1111, reload=True)
It works. I open the browser and call the endpoint / and try to access http://127.0.0.1:1111/ , it will return the json {"detail":"test_error"} with HTTP code "500 Internal Server Error" .
However, when I only changed StarletteHTTPException to Exception in the #app.exception_handler,
import uvicorn
from fastapi import FastAPI
from starlette.requests import Request
from starlette.exceptions import HTTPException as StarletteHTTPException
from starlette.responses import JSONResponse
app = FastAPI()
#app.exception_handler(Exception)
async def exception_callback(request: Request, exc: Exception):
print("test")
return JSONResponse({"detail": "test_error"}, status_code=500)
if __name__ == "__main__":
uvicorn.run("test:app", host="0.0.0.0", port=1111, reload=True)
The method exception_callback could not capture the StarletteHTTPException when I accessed the http://127.0.0.1:1111/ . It reported 404 error.
The excepted behaviour should be: StarletteHTTPException error could be captured by the method exception_handler decorated by Exception because StarletteHTTPException is the child class of Exception.
However, it is a known issue reported in Fastapi and Starlette
https://github.com/tiangolo/fastapi/issues/2750
https://github.com/tiangolo/fastapi/issues/2683
https://github.com/encode/starlette/issues/1129
So we are not able to acheieve the goal currently.

Answer 5

First I invite to get familiar with exception base classes in python. You can read them in the document Built-in Exceptions
Secondly, read through fastApi default exception overriding behaviour Override the default exception handlers
What you must understand is that #app.exception_handler accepts any Exception or child classes derived from Exception. For example RequestValidationError is a subclass of python built in ValueError which itself a subclass of Exception.
So you must design your own exceptions or throw available exceptions with this background. I guess what went wrong is with your logger logger.error(exc.detail) by either not having a detail field or not having a proper logger configuration.
Sample code :
#app.get("/")
def read_root(response: Response):
raise ArithmeticError("Divide by zero")
#app.exception_handler(Exception)
async def validation_exception_handler(request, exc):
print(str(exc))
return PlainTextResponse("Something went wrong", status_code=400)
Output :
A stdout entry and a response with Something went wrong

Answer 6

I found a way to catch exceptions without the "Exception in ASGI application_" by using a middleware. Not sure if this has some other side effect but for me that works fine! #iedmrc
#app.middleware("http")
async def exception_handling(request: Request, call_next):
try:
return await call_next(request)
except Exception as exc:
log.error("Do some logging here")
return JSONResponse(status_code=500, content="some content")

Answer 7

I was searching for global handler for fast api for giving custome message for 429 status code i found and implemented, working fine for me
#app.exception_handler(429)
async def ratelimit_handler(request: Request, exc: Exception):
return JSONResponse({'message': "You have exceeded your request quota. Kindly try after some time.", 'status': 'failed'})