使用{$ var}在for循环中进行变量替换 [英] Variable substitution in a for-loop using {$var}

问题描述

我对bash脚本非常陌生，我正在尝试通过制作一个只要求输入数字范围的小脚本来进行练习.我会输入ex.5..20，它应该显示范围，但是-无论输入什么，它都将回显(在本示例中为"5..20")，并且不扩展变量.有人可以告诉我我在做什么错吗?

I'm very new to bash scripting and I'm trying to practice by making this little script that simply asks for a range of numbers. I would enter ex. 5..20 and it should print the range, however - it just echo's back whatever I enter ("5..20" in this example) and does not expand the variable. Can someone tell me what I'm doing wrong?

脚本:

    echo -n "Enter range of number to display using 0..10 format: "
    read range

    function func_printrage
    {
         for n in {$range}; do
         echo $n
         done
    }

func_printrange

推荐答案

1. bash中的括号扩展不会扩展参数(与zsh不同)
2. 您可以通过使用 eval 和命令替换 $()
来解决此问题.
3. eval 是邪恶的，因为您需要清理输入内容，否则人们可以输入 rm -rf/; 和 eval 这样的范围
4. 请勿使用 function 关键字，它不是POSIX，已被弃用
5. 使用 read 的 -p 标志代替echo

1. Brace expansion in bash does not expand parameters (unlike zsh)
2. You can get around this through the use of eval and command substitution $()
3. eval is evil because you need to sanitize your input otherwise people can enter ranges like rm -rf /; and eval will run that
4. Don't use the function keyword, it is not POSIX and has been deprecated
5. use read's -p flag instead of echo

但是，出于学习目的，这是您要这样做的方式:

However, for learning purposes, this is how you would do it:

read -p "Enter range of number to display using 0..10 format: " range

func_printrange()
{
  for n in $(eval echo {$range}); do
    echo $n
  done
}

func_printrange

注意:在这种情况下，使用 eval 是可以的，因为您只是 echo 在范围内

Note: In this case the use of eval is OK because you are only echo'ing the range

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