I'm missing something here probably but the raven docs have pretty
straightforward support for configuring it via the ini file. Have you tried
this?

https://docs.getsentry.com/hosted/clients/python/integrations/pyramid/#pastedeploy-filter

On Sat, Mar 19, 2016 at 5:44 PM, Zsolt Ero <zsolt....@gmail.com> wrote:

> Thanks, I was able to make a working client, but not using the pipeline +
> filter part in the ini, but by wrapping app by:
>
> from raven import Client
>
> from raven.middleware import Sentry
>
> client = Client('https://...:...@app.getsentry.com/
> <http://app.getsentry.com/69865>...')
>
> app = Sentry(app, client=client)
>
> This way client is configured on init and captureException works
> automatically.
>
>
>
>
> Friday, March 18, 2016 at 1:26:39 AM UTC+1, Zsolt Ero wrote:
>>
>> Thanks, I see. I believe this way I do not need the .ini part, do I?
>>
>> On Thursday, March 17, 2016 at 3:40:28 AM UTC+1, Jonathan Vanasco wrote:
>>>
>>>
>>>
>>> On Wednesday, March 16, 2016 at 9:36:44 PM UTC-4, Zsolt Ero wrote:
>>>>
>>>> David Cramer from Sentry replied to me that if Sentry is used with the
>>>> middleware, then it should automatically receive the WSGI context:
>>>> from sentry.middleware import Sentry
>>>>
>>>> application = Sentry(application, client=Client(dsn, ...))
>>>>
>>>> My problem is that in Pyramid I have no idea where could I get an
>>>> application, or if this would work at all.
>>>>
>>>
>>> That looks like the initial app setup in your `project/__init__.py`
>>>
>>> `application` would be what is returned from config.make_wsgi_app()
>>>
>>> Some people do "return config.make_wsgi_app()"
>>>
>>> Others prefer "app = config.make_wsgi_app()", then wrap it in middleware
>>>
>>>
>>> --
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