如何抛出错误和退出自定义消息在 python

我见过有人在 Python 中建议使用 sys.exit ()。 我的问题是，是否有任何其他方式退出当前脚本的执行，我的意思是终止，一个错误。

就像这样:

sys.exit("You can not have three process at the same time.")

目前我的解决方案是:

print("You can not have three process at the same time.")
sys.exit()
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最佳答案

Calling sys.exit with a string will work. The docs mention this use explicitly:

In particular, sys.exit("some error message") is a quick way to exit a program when an error occurs.

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You can also raise an error like this:

raise SystemExit('Error: 3 processes cannot run simultaneously.')

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.

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There are 3 approaches, the first as lvc mentioned is using sys.exit 

sys.exit('My error message')

The second way is using print, print can write almost anything including an error message

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 

  raise SystemExit('error in code want to exit')

it can be ignored like this

try:
raise SystemExit('error in code want to exit')
except:
print("program is still open")
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You have to use import sys first

Then use sys.exit("your custom error message")


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