通过循环回显变量集 [英] echo the set of variables while going through a loop

问题描述

有人可以帮我吗.我一直在尝试替代方法，但仍然得到相同的输出.

Can someone help me. I've been trying alternatives but I still get the same output.

这是我的剧本

  #!/bin/sh

  field1="a"
  field2="s"
  field3="d"
  field4="f"
  field5="g"

  for i in {1..5}
  do

    var1="field"$i
    echo $var1

    var2="$"$var1
    echo $var2 

    echo $field1

  done

这是我的输出

  field1
  $field1
  a
  field2
  $field2
  a
  field3
  $field3
  a
  field4
  $field4
  a
  field5
  $field5
  a

我想做的是得到类似的输出

what I'm trying to do is get an output like

a
s
d
f
g

推荐答案

您在寻找

You're looking for ${!var1} — a Bash-specific shell parameter expansion.

field1="a"
field2="s"
field3="d"
field4="f"
field5="g"

for i in {1..5}
do
    var1="field"$i
    echo $var1=${!var1}
done

输出:

field1=a
field2=s
field3=d
field4=f
field5=g

在bash和sh都运行脚本的情况下，我明白了这一点.显然，如果您没有在echo中包含$var1=，则只会得到您要求的单个字母.

I got that with both bash and sh running the script. Clearly, if you don't include $var1= in the echo, you get just the single letters you ask for.

您还应该考虑是否将字段存储在Bash 数组中.

You should also consider whether your fields should be stored in a Bash array.

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