如何在python中抛出错误并使用自定义消息退出
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How to throw error and exit with a custom message in python
提问by Mostafa Talebi
I've seen people suggesting sys.exit() in Python. My question is that, is there any other way to exit the execution of current script, I mean termination, with an error.
我见过有人建议在 Python 中使用 sys.exit()。我的问题是，有没有其他方法可以退出当前脚本的执行，我的意思是终止，并出现错误。
Something like this:
像这样的东西：
sys.exit("You can not have three process at the same time.")
Currently my solution would be:
目前我的解决方案是：
print("You can not have three process at the same time.")
sys.exit()
采纳答案by lvc
回答by Osian
I know this is an old thread, however you can also raise an error like this:
我知道这是一个旧线程，但是您也可以引发这样的错误：
raise SystemExit('Error: 3 processes cannot run simultaneously.')
raise SystemExit('错误：3 个进程不能同时运行。')
One advantage of this approach is that you don't have to import the Python sys module. This works on Linux with Python 3 and Python 2. I have not tested it on Windows or Mac OS.
这种方法的一个优点是您不必导入 Python sys 模块。这适用于使用 Python 3 和 Python 2 的 Linux。我尚未在 Windows 或 Mac OS 上对其进行测试。
回答by nima moradi
There are 3 approaches, the first as lvc mentioned is using sys.exit
有 3 种方法，lvc 提到的第一种方法是使用 sys.exit
sys.exit('My error message')
The second way is using print, print can write almost anything including an error message
第二种方法是使用print, print 几乎可以写入任何内容，包括错误消息
print >>sys.stderr, "fatal error" # Python 2.x
print("fatal error", file=sys.stderr) # Python 3.x
The third way is to rise an exception which I don't like because it can be try-catch
第三种方法是提出一个我不喜欢的例外，因为它可以 try-catch
raise SystemExit('error in code want to exit')
it can be ignored like this
它可以像这样被忽略
try:
raise SystemExit('error in code want to exit')
except:
print("program is still open")
回答by Dasun
You have to use import sysfirst
你必须使用import sys第一
Then use sys.exit("your custom error message")
然后使用 sys.exit("your custom error message")
