如果没有命令行参数作为参数传递，则bash-throw用法警告

Question

示例脚本如下所示,

#!/bin/bash

function Usage() {
    echo "Usage: $0 [-q] [-d]"
    echo ""
    echo "where:"
    echo "     -q: Query info"
    echo "     -d: delete info"
    echo ""
}

# Exit if no argument is passed in
: ${1?Usage}

while getopts "qd" opt; do
    case $opt in
        q)
            echo "Query info"
            ;;
        d)
            echo "Delete info"
            ;;
        *)
            Usage;
            exit 1
            ;;
    esac
done

Answer 1

这个呢？

function Usage() {
    echo "Usage: $0 [-q] [-d]"
    echo ""
    echo "where:"
    echo "     -q: Query info"
    echo "     -d: delete info"
    echo ""
}

# Exit if no argument is passed in
: ${1?"$(Usage)"}

总之，我觉得这更有可读性：

if [ $# -lt 1 ] ; then
    Usage
fi

另一个想法是这样处理：

mode=""
while getopts "qdh" opt; do
    case $opt in
        q)
            mode="Query info"
            ;;
        d)
            mode="Delete info"
            ;;
        h)
            Usage
            exit 0
            ;;
        *)
            Usage >&2
            exit 1
            ;;
    esac
done

if [ -z "${mode}" ] ; then
    Usage >&2
    exit 1
fi