如何手动发送django异常日志? [英] How to send django exception log manually?
问题描述
HttpResponseBadRequest() c HttpResponse('')
/ code>否则。 此视图适用于外部数据,因此可以提出一些意外的异常。
当然我需要知道发生了什么。所以我有这样的样子:
def my_view(request):
try:
#的代码
return HttpResponse('')
除了:
import logging
logger = logging.getLogger('django.request')
#logger.error(extra = {'request':request})或logger.exception()
返回HttpResponseBadRequest('')
我有一个默认的日志记录配置(django 1.5)。
logger.exception
发送只有一个小回溯,没有请求数据。
logger.error
只发送请求数据。
所以问题是如何完全相同回溯作为django发送(与相同的主题/身体))。
我认为必须有一些干净简单的解决方案,我找不到。
更新
所以使用修补的异常
方法我最终得到以下代码:
logger.exception('Internal Server Error:%s',request.path,
extra = {'status_code':500,'request':request})
其中内置django日志记录产生等效的邮件追溯。
此问题是由于最近版本的Python 2.7中已经修复的错误。 此错误意味着异常
方法不接受
如果您无法升级到适合最近的Python 2.7,那么您可以补丁在我链接到的问题中引用了修复的Python。如果你不能这样做,你需要对 Logger
进行子类化,并覆盖异常
方法来做正确的事情(基本上是将 ** kwargs
添加到异常
方法的签名中,该签名被传递到其调用到错误
方法。
I have a specific view in my app which should return HttpResponse('')
if everything was done successfully and smth like HttpResponseBadRequest()
otherwise.
This view works with external data so some unexpected exceptions can be raised. Of course I need to know what happened. So I have smth like that:
def my_view(request):
try:
# a lot of code
return HttpResponse('')
except:
import logging
logger = logging.getLogger('django.request')
# logger.error(extra={'request': request}) or logger.exception()
return HttpResponseBadRequest('')
I have a default logging config if it matters (django 1.5).
logger.exception
sends only a small traceback and no request data.
logger.error
sends only request data.
So the question is how to get exactly the same traceback as django sends (with the same subject/body)). I think there must be some clean and simple solution that i just can't find.
UPDATE
So with the patched exception
method i ended up with the following code:
logger.exception('Internal Server Error: %s', request.path,
extra={'status_code': 500, 'request': request})
which produces equal email traceback as built-in django logging.
This problem is due to a bug which has been fixed in recent versions of Python 2.7. This bug meant that the exception
method did not accept the extra
argument.
If you can't upgrade to a suitably recent Python 2.7, then you can perhaps patch your Python with the fix referenced in the issue I've linked to. If you can't do that, you'll need to subclass Logger
and override the exception
method to do the right thing (which is basically to add a **kwargs
to the exception
method's signature, which is passed through to its call to the error
method.
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