bash - 为什么for循环中组合变量的回显不起作用？

Question

循环中的“echo”行有什么问题？它应该在第一个输入后回显 v1 的值，在第二个输入后回显 v2，在第三个输入后回显 v3。相反，它只是回应 $i。为什么忽略 $v ？循环完成后，我对保护程序变量进行了“手动检查”，它们显然存在。

脚本:

#!/bin/bash
for i in {1..3}
 do
  read -p "Insert value $i: " v$i
  echo "you wrote " $v$i          #I have problem in this line
 done

echo
echo "check of saved values:"
echo $v1
echo $v2
echo $v3

控制台输出:

./input_2.sh
Insert value 1: what
you wrote  1              #output should be: you wrote what
Insert value 2: the
you wrote  2              #output should be: you wrote the
Insert value 3: hell
you wrote  3              #output should be: you wrote hell

check of saved values:
what
the
hell

谢谢

Answer 1

这一行:

echo "你写了 "$v$i

尝试写入变量 $v 后跟 $i。

你需要的是将变量名存储到一个变量中，然后使用${!}，例如:

varname=v$i
echo "you wrote " ${!varname}

一个好的做法是将此变量名也用于 read 操作:

#!/bin/bash
for i in {1..3}
 do
  varname=v$i
  read -p "Insert value $i: " $varname
  echo "you wrote " ${!varname}
 done

echo
echo "check of saved values:"
echo $v1
echo $v2
echo $v3