shell 获取指定参数
while [ -n "$1" ] do case "$1" in -a) echo "发现 -a 选项" ;; -b) echo "发现 -b 选项" echo "-b 选项的参数值是:$2" shift ;; -c) echo "发现 -c 选项" echo "-c 选项的参数值是:$2" shift ;; -d) echo "发现 -d 选项" ;; *) echo "$1 is not an option" ;; esac shift done
不想平凡,奈何太懒 T_T