Re: [Kimchi-devel] [PATCH] [Wok 2/3] Fix issue #140 - Add original exception to user request log message

On 08/22/2016 11:58 AM, Lucio Correia wrote: > On 22-08-2016 11:13, Aline Manera wrote: >> >> >> On 08/09/2016 04:15 PM, Lucio Correia wrote: >>> Signed-off-by: Lucio Correia <luciojhc(a)linux.vnet.ibm.com&gt; >>> --- >>> src/wok/control/base.py | 9 +++++++++ >>> src/wok/reqlogger.py | 34 +++++++++++++++++++++++++++++----- >>> src/wok/root.py | 7 ++++++- >>> 3 files changed, 44 insertions(+), 6 deletions(-) >>> >>> diff --git a/src/wok/control/base.py b/src/wok/control/base.py >>> index f563aed..04cf2cb 100644 >>> --- a/src/wok/control/base.py >>> +++ b/src/wok/control/base.py >>> @@ -119,6 +119,7 @@ class Resource(object): >>> def wrapper(*args, **kwargs): >>> # status must be always set in order to request be >>> logged. >>> # use 500 as fallback for "exception not handled" cases. >>> + details = None >>> status = 500 >>> >>> method = 'POST' >>> @@ -149,6 +150,7 @@ class Resource(object): >>> status = cherrypy.response.status >>> return result >>> except WokException, e: >> >>> + details = e >>> status = e.getHttpStatusCode() >> >> 'details' and 'status' depends on which exception was raised. >> I'd suggest to pass only the exception (which represents the 'details' >> in this patch) to the RequestRecord() and let it work with the >> exception. >> That way we simplify the amount of parameters and IMO make it easier to >> work with. >> > > Hi Aline, we also log successful requests (no exception). > > That's why status is informed separately. > But you get the status code from the exception, right? status = e.getHttpStatusCode() So it seems always an exception exists there.