名称退出未在python中定义

The following is the code. When I run, I got an error message, saying that "name exit is not defined". Could anyone tell me why? Thanks very much for your time and attention.

以下是代码。当我运行时,我收到一条错误消息,说“名称退出未定义”。谁能告诉我为什么?非常感谢您的时间和关注。

if len(sys.argv) == 4:
   ### do something
    pass
else:
    print
    "usage: #### something here"
    exit(-1)

2 个解决方案

#1

You need to import sys first, since exit (and argv) is in that module.

您需要先导入sys,因为exit(和argv)位于该模块中。

The error I get when I run your code is:

我运行代码时遇到的错误是:

File "", line 1, in NameError: name 'sys' is not defined

NameError中的文件“”,第1行:名称'sys'未定义

which is complaining about sys.argv instead of exit. But in either case, the solution -- import sys is the same.

这是抱怨sys.argv而不是退出。但在任何一种情况下,解决方案 - import sys都是一样的。

#2

If you imported sys then use sys.exit(0) - change 0 with any result code you want to exit with. I had the same issue that before compilation when running from .py file exit(0) worked well but after compiling to .exe it gave me an error. I had to change exit(0) to sys.exit(0) and it worked.

如果导入了sys,则使用sys.exit(0) - 使用要退出的任何结果代码更改0。我有同样的问题,在编译之前从.py文件exit(0)运行良好但编译到.exe后它给了我一个错误。我不得不将exit(0)更改为sys.exit(0)并且它工作正常。

#1