python异常消息捕获

Question

import ftplib
import urllib2
import os
import logging
logger = logging.getLogger('ftpuploader')
hdlr = logging.FileHandler('ftplog.log')
formatter = logging.Formatter('%(asctime)s %(levelname)s %(message)s')
hdlr.setFormatter(formatter)
logger.addHandler(hdlr)
logger.setLevel(logging.INFO)
FTPADDR = "some ftp address"

def upload_to_ftp(con, filepath):
    try:
        f = open(filepath,'rb')                # file to send
        con.storbinary('STOR '+ filepath, f)         # Send the file
        f.close()                                # Close file and FTP
        logger.info('File successfully uploaded to '+ FTPADDR)
    except, e:
        logger.error('Failed to upload to ftp: '+ str(e))

这似乎不起作用，我收到语法错误，将所有类型的异常记录到文件的正确方法是什么

Answer 1

您必须定义要捕获的异常类型。所以写 except Exception, e: 而不是 except, e: 作为一般异常(无论如何都会被记录)。

另一种可能性是用这种方式编写整个 try/except 代码:

try:
    with open(filepath,'rb') as f:
        con.storbinary('STOR '+ filepath, f)
    logger.info('File successfully uploaded to '+ FTPADDR)
except Exception, e: # work on python 2.x
    logger.error('Failed to upload to ftp: '+ str(e))

在 Python 3.x 和现代版本的 Python 2.x 中使用 except Exception as e 而不是 except Exception, e:

try:
    with open(filepath,'rb') as f:
        con.storbinary('STOR '+ filepath, f)
    logger.info('File successfully uploaded to '+ FTPADDR)
except Exception as e: # work on python 3.x
    logger.error('Failed to upload to ftp: '+ str(e))