Python 如何忽略 Sentry 的一些错误(不发送它们)

发布于06月29日

我有一个基于django(3.2.10)和sentry sdk(1.16.0)的项目

from os import getenv


SENTRY_URL = getenv('SENTRY_URL')

if SENTRY_URL:
    from sentry_sdk import init
    from sentry_sdk.integrations.django import DjangoIntegration
    from sentry_sdk.integrations.redis import RedisIntegration
    from sentry_sdk.integrations.celery import CeleryIntegration

    init(
        dsn=SENTRY_URL,
        integrations=[DjangoIntegration(), RedisIntegration(), CeleryIntegration()],
        traces_sample_rate=1.0,
        send_default_pii=True,
        debug=True,
    )

我有一个从Exception继承的CustomError

每次我引发CustomError sentry时，sdk都会将其发送到dsn url.

我想忽略某类错误或类似的东西.

推荐答案

您可以传递一个过滤要发送的错误的函数:

from os import getenv


SENTRY_URL = getenv('SENTRY_URL')

if SENTRY_URL:
    from sentry_sdk import init
    from sentry_sdk.integrations.django import DjangoIntegration
    from sentry_sdk.integrations.redis import RedisIntegration
    from sentry_sdk.integrations.celery import CeleryIntegration
  
    def before_send(event, hint):
       if 'exc_info' in hint:
          exc_type, exc_value, tb = hint['exc_info']
          if isinstance(exc_value, CustomError):  # Replace CustomError with your custom error 
             return None
       return event

    init(
        dsn=SENTRY_URL,
        integrations=[DjangoIntegration(), RedisIntegration(), CeleryIntegration()],
        traces_sample_rate=1.0,
        send_default_pii=True,
        debug=True,
        before_send=before_send
    )

您可以在documentation中找到更多信息.