bash - 用于退出带有错误消息的 bash 脚本的更好的一行代码

Question

我想用一个更干净、不那么丑陋的单行代码来完成测试:

#!/bin/bash
test -d "$1" || (echo "Argument 1: '$1' is not a directory" 1>&2 ; exit 1) || exit 1

# ... script continues if $1 is directory...

基本上我追求的是不重复 exit 的东西，并且最好不产生子 shell(因此看起来也不那么难看)，但仍然适合一行。

Answer 1

没有子 shell 也没有复制 exit:

test -d "$1" || { echo "Argument 1: '$1' is not a directory" 1>&2 ; exit 1; }

您可能还想引用 Grouping Commands :

{}

{ list; }

Placing a list of commands between curly braces causes the list to be executed in the current shell context. No subshell is created. The semicolon (or newline) following list is required.