如何用选项编写Unix shell脚本？

Question

我不知道是否可能,但我想编写shell脚本,其作用类似于带有选项的常规可执行文件.作为一个非常简单的示例,请考虑配置为可执行的shell脚本foo.sh:

    ./foo.sh
    ./foo.sh -o

而代码的foo.sh作用就像

    #!/bin/sh
    if  ## option -o is turned on
        ## do something
    else
        ## do something different
    endif

有可能,怎么做？谢谢.

Answer 1

$ cat stack.sh 
#!/bin/sh
if  [[ $1 = "-o" ]]; then
    echo "Option -o turned on"
else
    echo "You did not use option -o"
fi

$ bash stack.sh -o
Option -o turned on

$ bash stack.sh
You did not use option -o

供参考:

$1 = First positional parameter
$2 = Second positional parameter
.. = ..
$n = n th positional parameter

有关更多整洁/灵活的选项,请阅读此其他线程:在bash shell脚本中使用getopts以获取长和短命令行选项

Answer 2

这是在一个脚本中如何做到这一点的方式:

#!/usr/bin/sh
#
# Examlple of using options in scripts
#

if [ $# -eq 0 ]
then
        echo "Missing options!"
        echo "(run $0 -h for help)"
        echo ""
        exit 0
fi

ECHO="false"

while getopts "he" OPTION; do
        case $OPTION in

                e)
                        ECHO="true"
                        ;;

                h)
                        echo "Usage:"
                        echo "args.sh -h "
                        echo "args.sh -e "
                        echo ""
                        echo "   -e     to execute echo \"hello world\""
                        echo "   -h     help (this output)"
                        exit 0
                        ;;

        esac
done

if [ $ECHO = "true" ]
then
        echo "Hello world";
fi

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