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Description
I cannot get SentryAsgiMiddleware to work with our FastAPI app. We tried to follow the example in the Sentry docs, so the app module basically looks like this:
from fastapi import FastAPI
from sentry_sdk.integrations.asgi import SentryAsgiMiddleware
...
app = FastAPI()
@app.post()
...
app = SentryAsgiMiddleware(app)This gives an error on all requests, see the following stack trace:
Traceback (most recent call last):
File "/usr/local/lib/python3.7/site-packages/uvicorn/protocols/http/httptools_impl.py", line 385, in run_asgi
result = await app(self.scope, self.receive, self.send)
File "/usr/local/lib/python3.7/site-packages/uvicorn/middleware/proxy_headers.py", line 45, in __call__
return await self.app(scope, receive, send)
File "/usr/local/lib/python3.7/site-packages/uvicorn/middleware/asgi2.py", line 7, in __call__
await instance(receive, send)
File "/usr/local/lib/python3.7/site-packages/sentry_sdk/integrations/asgi.py", line 54, in run_asgi2
scope, lambda: self.app(scope)(receive, send)
File "/usr/local/lib/python3.7/site-packages/sentry_sdk/integrations/asgi.py", line 93, in _run_app
raise exc from None
File "/usr/local/lib/python3.7/site-packages/sentry_sdk/integrations/asgi.py", line 90, in _run_app
return await callback()
File "/usr/local/lib/python3.7/site-packages/sentry_sdk/integrations/asgi.py", line 54, in <lambda>
scope, lambda: self.app(scope)(receive, send)
TypeError: __call__() missing 2 required positional arguments: 'receive' and 'send'Library versions:
- python==3.7.5
- sentry-sdk==0.13.2
- uvicorn==0.10.8
- fastapi==0.42.0
- starlette==0.12.9
petterhh, dnshio, antnieszka, louisguitton, shweta-pml and 2 more