在我的方法之一中,中断异常和执行异常即将到来。我像这样输入了 try catch 。
try{
//my code
}catch(InterruptedException|ExecutionException e)
Log.error(" logging it");
throw new MonitoringException("it failed" , e)
//monitoringexception extends RunTimeException
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同样在我的方法中,我抛出了 InterruptedException,ExecutionException
我在声纳中遇到严重错误 - 重新中断此方法或重新抛出“ InterruptedException”
有人知道怎么修这个东西吗。
请立即帮助。
函数名是否应该main()为空并且在函数本身内调用参数,或者将它们作为函数的输入是否可接受main(arg1, arg2, arg3)?
我知道它有效,但我想知道它是不是很糟糕的编程习惯.如果这是重复的道歉,但我看不出专门针对Python的问题.
我是编程新手,所以我遇到了这个麻烦。我升级了我的 pip,然后使用 pip list、ipython Notebook、jupyter Notebook 来使用 anaconda,所以在所有这些命令之后我得到了一个答案:
Traceback (most recent call last):
File "/home/bakhytgul/anaconda3/bin/pip", line 7, in <module>
from pip._internal import main
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/__init__.py", line 40, in <module>
from pip._internal.cli.autocompletion import autocomplete
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/cli/autocompletion.py", line 8, in <module>
from pip._internal.cli.main_parser import create_main_parser
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/cli/main_parser.py", line 12, in <module>
from pip._internal.commands import (
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/commands/__init__.py", line 6, in <module>
from pip._internal.commands.completion import CompletionCommand
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/commands/completion.py", line 6, in <module>
from pip._internal.cli.base_command import Command
File "/home/bakhytgul/anaconda3/lib/python3.6/site-packages/pip/_internal/cli/base_command.py", line 19, in <module>
from …Run Code Online (Sandbox Code Playgroud) 如果用户已经输入'tea'或't'为自己的选择,我该如何定义变量beverage.lower()为'tea'?因为
if beverage.lower()=='t'or beverage.lower()=='tea':
beverage.lower()=='tea'
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line对整个程序本身绝对没有任何影响?
while True:
beverage=raw_input("What is your preferred beverage: coffee, or tea?")
if beverage.lower() not in ('coffee','tea','t','c'):
if beverage.lower()=='t'or beverage.lower()=='tea':
beverage.lower()=='tea'
elif beverage.lower()=='c'or beverage.lower()=='coffee':
beverage.lower()=='coffee'
print("Sorry! I didn't quite catch that. Please try again! (Note that you can use the letter 'c' or the letter 't' to denote coffee or tea respectively!)")
continue
else:
print("Ah! Fantastic choice!")
break
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同样,我如何定义beverage.lower()作为另一个变量输入的用户?