linux - 在 bash 中获取参数

Question

我需要在 bash 中获取参数，我可以获取参数 h,c,p,s ,a 但我不能接受参数 b。

为什么我不能接受？这是我的脚本:

if [ ! -z $1 ]; then
    HOSTNAME=""
    CLIENT=""
    SUBSCRIPTIONS_GROUPS=""
    PROVIDER=""
    SERVER=""
    while getopts ":h:c:p:s:a:b" opt; do
            case $opt in 
                    h)  HOSTNAME=${OPTARG}
                            ;;
                    c)  CLIENT=${OPTARG}
                            ;;
                    p)      PROVIDER=${OPTARG}
                            ;;
                    s)  SUBSCRIPTIONS_GROUPS=${OPTARG}
                            ;;
                    a)  ALWAYS_ON="on"
                            ;;
                    b)  SERVER=${OPTARG}
                            ;;
                    ?)  ;;
            esac
    done
fi

Answer 1

我发现您的实现存在两个缺陷:

• 第一个字符是冒号
• 在b之后没有冒号

您的 getops 字符串应如下所示:

while getopts "h:c:p:s:ab:" opt; do
...

When you want getopts to expect an argument for an option, just place a : (colon) after the proper option flag

和

If the very first character of the option-string is a : (colon), which would normally be nonsense because there's no option letter preceding it, getopts switches to "silent error reporting mode". In productive scripts, this is usually what you want because it allows you to handle errors yourself without being disturbed by annoying messages.

摘自 http://wiki.bash-hackers.org/howto/getopts_tutorial