打开无效参数窗口的Python

Question

import datetime
import os
import sys
import time
if "ZLOG_FILE" not in globals():
    global ZLOG_FILE
    script_dir = os.path.realpath(sys.argv[0]).replace(sys.argv[0], "")
    ZLOG_FILE = os.path.join(script_dir, "log", datetime.datetime.fromtimestamp(time.time()).strftime("%m-%d-%Y %H:%M:%S"))
    with open(ZLOG_FILE, "w"):
        pass
def log(level, msg):
    with open(ZLOG_FILE, "a") as f:
        # Fix colors
        if level == "INFO":
            msg = "ESC[37m[INFO] " + msg
        elif level == "WARN":
            msg = "ESC[33m[WARN] " + msg
        elif level == "ERROR":
            msg = "ESC[31m[ERROR] " + msg
        if level != "ERROR":
            print(msg)
        else:
            sys.stderr.write(msg)
        f.write(msg)

我制作了这个简单的日志库，在第一次导入时，在与正在运行的主脚本相同位置的日志目录中创建一个具有当前时间和日期的新文件。然而，我得到以下错误:

Traceback (most recent call last):                                              
  File "test.py", line 1, in <module>                                           
    import zlog                                                                 
  File "C:UserszaneDesktopzlog.py", line 10, in <module>                    
    with open(ZLOG_FILE, "w"):                                                  
OSError: [Errno 22] Invalid argument: 'C:\Users\zane\Desktop\log\09-12-2016
 20:02:26'                                                                      

我做错了什么?

编辑:用script_dir变量去掉脚本名称，并固定if语句的格式。我仍然得到一个错误虽然

Answer 1

Windows不能在文件名中使用冒号，而且我忘记在获得目录后从路径中删除主脚本，它位于