Question

我正在尝试编写一个简单的sh脚本，必须使用2个参数调用：

sh myscript.sh --user "some user" --fizz "buzz"

在myscript.sh的顶部，我有：

#!/bin/sh
# VALIDATION
# 1. Make sure there are 5 positional arguments (that $4 exists).
die () {
    echo >&2 "$@"
    exit 1
}

[ "$#" -eq 5 ] || die "5 arguments required, $# provided"

# 2. Make sure $1 is "-u" or "--user".
# 3. Make sure $3 is "-f" or "--fizz".

如果验证失败，我想打印一条简单的用法消息，然后退出脚本。

我认为我有＃1正确（检查位置参数＃），但不知道如何实现＃2和＃3。想法？

Answer 1

# 2. Make sure $1 is "-u" or "--user".

if ! [ "$1" = -u -o "$1" = --user ]; then
    # Test failed. Send a message perhaps.
    exit 1
fi

# 3. Make sure $3 is "-f" or "--fizz".

if ! [ "$3" = -f -o "$3" = --fizz ]; then
    # Test failed. Send a message perhaps.
    exit 1
fi

用于测试两个可能的值的变量的其他形式：

[ ! "$var" = value1 -a ! "$var" = value2 ]
[ ! "$var" = value1 ] && [ ! "$var" = value2 ]
! [ "$var" = value1 && ! [ "$var" = value2 ]

对于Bash和类似语法的shell：

! [[ $var = value1 || $var = value2 ]]
[[ ! $var = value1 || ! $var = value2 ]]

除了使用if块的否定条件外，您还可以使用||

获得积极的条件

true_condition || {
    # Failed. Send a message perhaps.
    exit 1
}

true_condition || exit 1

另一方面，&&当然会适用于否定的条件。

使用案例陈述：

case "$var" in
value1|value2)
    # Valid.
    ;;
*)
    # Failed.
    exit 1
    ;;
esac

Answer 2

手动：

if [ -z "$1" ];then
fi

if [ -z "$2" ];then
fi

if [ -z "$3" ];then
fi
...

或检查getopt

while getopts "uf" OPTION
do
 case $OPTION in
     u)
         echo "-u"
         ;;
     f)
         echo "-f"
         ;;
 esac
done

如何验证shell参数？

2 个答案: