如何记录Python异常?
try:
do_something()
except:
# How can I log my exception here, complete with its traceback?
当前回答
也许没有那么时尚，但更简单:
#!/bin/bash
log="/var/log/yourlog"
/path/to/your/script.py 2>&1 | (while read; do echo "$REPLY" >> $log; done)
其他回答
下面是一个使用sys.excepthook的版本
import traceback
import sys
logger = logging.getLogger()
def handle_excepthook(type, message, stack):
logger.error(f'An unhandled exception occured: {message}. Traceback: {traceback.format_tb(stack)}')
sys.excepthook = handle_excepthook
使用exc_info选项可能更好，保留警告或错误标题:
try:
# coode in here
except Exception as e:
logging.error(e, exc_info=True)
也许没有那么时尚，但更简单:
#!/bin/bash
log="/var/log/yourlog"
/path/to/your/script.py 2>&1 | (while read; do echo "$REPLY" >> $log; done)
我要找的是:
import sys
import traceback
exc_type, exc_value, exc_traceback = sys.exc_info()
traceback_in_var = traceback.format_tb(exc_traceback)
看到的:
https://docs.python.org/3/library/traceback.html
要关闭其他可能在这里丢失的内容，在日志中捕获它的最佳方法是使用traceback.format_exc()调用，然后将该字符串拆分为每行，以便在生成的日志文件中捕获:
import logging
import sys
import traceback
try:
...
except Exception as ex:
# could be done differently, just showing you can split it apart to capture everything individually
ex_t = type(ex).__name__
err = str(ex)
err_msg = f'[{ex_t}] - {err}'
logging.error(err_msg)
# go through the trackback lines and individually add those to the log as an error
for l in traceback.format_exc().splitlines():
logging.error(l)
