以 bash 中的错误消息退出（oneline）
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24597818/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
exit with error message in bash (oneline)
提问by branquito
Is it possible to exit on error, with a message, withoutusing ifstatements?
是否可以在不使用if语句的情况下错误退出并显示消息？
[[ $TRESHOLD =~ ^[0-9]+$ ]] || exit ERRCODE "Threshold must be an integer value!"
Of course the right side of ||won't work, just to give you better idea of what I am trying to accomplish.
当然右侧的||不起作用，只是为了让您更好地了解我要完成的工作。
Actually, I don't even mind with which ERR code it's gonna exit, just to show the message.
实际上，我什至不介意它退出哪个 ERR 代码，只是为了显示消息。
EDIT
编辑
I know this will work, but how to suppress numeric arg requiredshowing
after my custom message?
我知道这会起作用，但是如何numeric arg required在我的自定义消息之后抑制显示？
[[ $TRESHOLD =~ ^[0-9]+$ ]] || exit "Threshold must be an integer value!"
回答by P.P
exitdoesn't take more than one argument. To print any message like you want, you can use echoand then exit.
exit不超过一个论点。要打印您想要的任何消息，您可以使用echo然后退出。
[[ $TRESHOLD =~ ^[0-9]+$ ]] || \
{ echo "Threshold must be an integer value!"; exit $ERRCODE; }
回答by konsolebox
You can use a helper function:
您可以使用辅助函数：
function fail {
printf '%s\n' "" >&2 ## Send message to stderr. Exclude >&2 if you don't want it that way.
exit "${2-1}" ## Return a code specified by or 1 by default.
}
[[ $TRESHOLD =~ ^[0-9]+$ ]] || fail "Threshold must be an integer value!"
Function name can be different.
函数名称可以不同。
回答by noonex
Using exitdirectly may be tricky as the script may be sourced from other places. I prefer instead using subshell with set -e(plus errors should go into cerr, not cout) :
exit直接使用可能会很棘手，因为脚本可能来自其他地方。我更喜欢使用带有set -e（加上错误应该进入cerr，而不是cout）的子shell ：
set -e
[[ $TRESHOLD =~ ^[0-9]+$ ]] || \
(>&2 echo "Threshold must be an integer value!"; exit $ERRCODE)
