如何将变量名传递给函数并使其在 bash 之外可用？

Question

如何将变量名传递给函数并在其作用域外使用它？

我试过了，但没用：

questionPrompt() {
    while [ -z "$answer" ]; do
    echo 
    read answer
    =$answer
    done
}

questionPrompt "Which color do you like?" "COLOR"

echo $COLOR

它说：COLOR=red: command not found

Answer 1

使用declare:

declare -g "=$answer"

或

declare -gn var=
var="$answer"

然而，macOS (3.2) 附带的 bash 版本不支持 declare -g；您可以改用 printf。

printf -v "" '%s' "$answer"

如何将变量名传递给函数并使其在 bash 之外可用？

How to pass variable name to a function and get it available outside in bash?

variables

macos

bash

scope