如果有任何命令返回非零值，则中止shell脚本 [英] Aborting a shell script if any command returns a non-zero value

问题描述

我有一个Bash shell脚本，可以调用许多命令.

I have a Bash shell script that invokes a number of commands.

如果任何命令返回的非零值，我都希望shell脚本自动以返回值1退出.

I would like to have the shell script automatically exit with a return value of 1 if any of the commands return a non-zero value.

是否可以不显式检查每个命令的结果?

Is this possible without explicitly checking the result of each command?

例如，

dosomething1
if [[ $? -ne 0 ]]; then
    exit 1
fi

dosomething2
if [[ $? -ne 0 ]]; then
    exit 1
fi

推荐答案

将其添加到脚本的开头:

Add this to the beginning of the script:

set -e

如果简单命令以非零退出值退出，这将导致shell立即退出.简单命令是不属于if，while，或直到测试的一部分的任何命令，也不属于&&的一部分的任何命令.或||列表.

This will cause the shell to exit immediately if a simple command exits with a nonzero exit value. A simple command is any command not part of an if, while, or until test, or part of an && or || list.

有关"set"内部命令的更多信息，请参见 bash(1)手册页

See the bash(1) man page on the "set" internal command for more details.

我个人几乎所有的shell脚本都以"set -e"开头.当某个脚本在中间出现故障并顽固地破坏了其余脚本的假设时，固执地继续执行脚本确实很烦人.

I personally start almost all shell scripts with "set -e". It's really annoying to have a script stubbornly continue when something fails in the middle and breaks assumptions for the rest of the script.

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