Question

我正在尝试编写一个bash脚本，允许我在一组不同的数组中选择一个数组。为此，我打算使用一个简单的变量来引用那个数组。

#!/bin/bash
#To get all the members of a given array as the output
#variables
FIRST=(A B C D)
SECOND=(planes cars trains bicycles gocarts)
THIRD=(werewolfs vampires zombies ghosts daemons)
FOURTH=(football soccer basketball rugby batmington zumo)
FIFTH=(handguns rifles machineguns bazookas slingshots)
SIXTH=(dogs cats turtles ferrets birds hamsters fish)
SEVENTH=(monday tuesday wednesday thursday friday saturday sunday)

#execution
select ARRAY in "FIRST" "SECOND" "THIRD" "FOURTH" "FIFTH" "SIXTH" "SEVENTH"; do 
    OUTPUT=eval '"${'${ARRAY}'[@]}"'
    echo $OUTPUT
    break 
done

#end

以上脚本不起作用。到目前为止，我已尝试用以下选项替换第9行：

OUTPUT=eval '$'{ARRAY'[@]'}
OUTPUT=eval ${"$ARRAY"[@]}
OUTPUT=eval ${'$ARRAY'[@]}
OUTPUT=eval ${'$'ARRAY[@]}
OUTPUT=eval '$'{"$ARRAY"[@]}
OUTPUT=eval \${${ARRAY}[@]}

我在这里缺少什么？

Answer 1

这对我有用：

#!/bin/bash
#To get all the members of a given array as the output
#variables
FIRST=(A B C D)
SECOND=(planes cars trains bicycles gocarts)
THIRD=(werewolfs vampires zombies ghosts daemons)
FOURTH=(football soccer basketball rugby batmington zumo)
FIFTH=(handguns rifles machineguns bazookas slingshots)
SIXTH=(dogs cats turtles ferrets birds hamsters fish)
SEVENTH=(monday tuesday wednesday thursday friday saturday sunday)

#execution
ARRAY="FIFTH"
select ARRAY in "FIRST" "SECOND" "THIRD" "FOURTH" "FIFTH" "SIXTH" "SEVENTH"; do
    eval "OUTPUT=\${$ARRAY[*]}"
    echo $OUTPUT
    break
done

eval可用于引入新变量。我们构造一个字符串，其中包含将所需值分配给OUTPUT的表达式，然后对其进行评估，从而引入一个具有所需值的新变量OUTPUT。

Answer 2

eval绝对没有必要解决这个问题。在使用eval之前，您应该始终三思，因为它的脆弱性。（也就是说，错误会带来灾难性的后果。）

这是“传统”解决方案，它使用!间接语法。它仍然有些脆弱，但没有eval那么糟糕：

select array in "FIRST" "SECOND" "THIRD" "FOURTH" "FIFTH" "SIXTH" "SEVENTH"; do
  if [[ $array ]]; then
    # Indirection requires the full subscript to be included
    # in the variable which is used to indirect. "${!array[@]}"
    # would be "0", because that is not indirect syntax; rather it
    # is "array keys" syntax.
    array_at="$array"[@]
    echo "${!array_at}"
    break
  else
    echo "Invalid input; try again" >> /dev/stderr
  fi
done

从bash 4.3开始，你可以使用引用声明，这使得上面的内容不那么笨重了：

select name in "FIRST" "SECOND" "THIRD" "FOURTH" "FIFTH" "SIXTH" "SEVENTH"; do
  if [[ $name ]]; then
    declare -n array=name
    echo "${array[@]}"
    break
  else
    echo "Invalid input; try again" >> /dev/stderr
  fi
done
# Unless the user exits the select by typing an EOF,
# then `array` is now effectively a synonym
# for whichever of the arrays was selected.

Answer 3

我明白了。以下适用于第9行：

OUTPUT=$(eval echo \${${ARRAY}[@]})

非常感谢你对这个可怜的小学徒的耐心：）

使用shell变量选择一个bash数组

3 个答案: