Django未处理的异常

Question

它在DEBUG = True模式下运行.有时它会在遇到错误时抛出带有回溯信息的错误消息,但有时它只显示以下行:

Unhandled Exception

An unhandled exception was thrown by the application.

我必须切换到开发服务器才能看到详细消息.

如何在遇到错误时始终显示回溯消息？

Answer 1

只需连接到got_request_exception信号并记录异常:

from django.core.signals import got_request_exception
import logging    

def log(*args, **kwargs):
    logging.exception('error')

got_request_exception.connect(log)

这将记录整个跟踪.在开发服务器中,它会在控制台中登录.

Answer 2

也许您可以使用此代码段,这将在apache的日志中记录异常:

utils.py:

def log_traceback(exception, args):
    import sys, traceback, logging
    exceptionType, exceptionValue, exceptionTraceback = sys.exc_info()
    logging.debug(exception)
    logging.debug(args)
    for tb in traceback.format_exception(exceptionType, exceptionValue, exceptionTraceback):
        logging.debug(tb)

site_logging.py:

import logging
import sys

logger = logging.getLogger('')
logger.setLevel(logging.DEBUG)
handler = logging.StreamHandler(sys.stderr)
handler.setLevel(logging.DEBUG)
formatter = logging.Formatter('%(levelname)-8s %(message)s')
handler.setFormatter(formatter)
logger.addHandler(handler)

把它放在你的settings.py:

import site_logging

在你的代码中:

from where.is.your.utils import log_traceback
try:
   `do something`
except Exception, args:
    log_traceback(Exception, args)