1. 正确的方式

参考：https://stackoverflow.com/questions/52184686/dynamic-variable-assignment-in-shell-script

for ((i=1;i<=${#};i++))  ; do
    option="${*:${i}:1}"
    echo "arg: [${option}]"
done

使用了 shell 语法 ${parameter:offset:length}

• $* is a special variable which expands to the list of positional parameters (arguments to your script), separated by a space character.
• ${list: -1} expands to the last element of a list.

2. 错误的方式

for a in ${@} ; do  # 或者  for a in $*
    echo "${a}"
done

这种方式的错误在于，会把所有参数搞成一行，然后以空格作为分隔符进行拆分。这就回导致 "select count(*) from a" 这个被双引号包起来的语句被分割成了4个参数。