用空格循环数组字符串 [英] Loop through array of arrays of string with spaces

问题描述

我试图循环一个包含其他数组的数组，这些数组由包含空格的字符串组成。问题是，我似乎无法保留字符串中的间距。如果我将IFS更改为\\\
，或者将数组的所有元素都视为1项，如果我将IFS保持不变，则这些字符串会被分成多个项目：

 ＃！/ bin / sh 
 low1 =（AA QQBB LL）
 low2 =（CCDD） 
 low3 =（EEFF）
 high =（low1 low2 low3）
 
 for high_item in $ {high [@]} 
 do 
 eval arrayz = \ $ {$ high_item [@]} 
 #IFS = $'\\\
'
 for $ arrayz 
 
 echo $ item 
完成
完成
  

输出：

 
 AA 
 QQ 
 BB 
 LL 
 CC 
 DD 
 EE 
 FF 
 

您可以看到元素AA QQ和BB LL已被分割。

如果我取消注释将 IFS 设置为 \\\
的行，我得到以下结果：

 
 AA QQ BB LL 
 CC DD 
 EE FF 
 

现在将AA QQ和BB LL连接在一起！

有没有反正我可以保留这些元素，就像他们原来的一样...我需要的输出看起来像这样：

 
 AA QQ 
 BB LL 
 CC 
 DD 
 EE 
 FF 
 

解决方案

我认为你的意思是输出应该是这样的：

  AA QQ 
 BB LL 
 CC 
 DD 
 EE 
 FF 
  

  $ {low1 [0]} 
 $ {low1 [1]} 
 $ {low2 [0]} 
 $ {low2 [1]} 
 $ {low3 [0]} 
 $ {low3 [1]} 
  

可以使用以下方式来完成：

 ＃！/ bin / bash 
 
 low1 =（AA QQBB LL）
 low2 =（CCDD）
 low3 =（EEFF）
 high =（low1 low2 low3）
 
为high_item in $ {high [@]} 
 do 
x = $ {high_item} [@]＃ - > $ {array ['] 
 arrays =（$ {！x}）
 
 #IFS = $'\ n'
 @]}
 do 
 echo$ item
 done 
 done 
  

和请 总是使用＃！/ bin / bash 作为bash脚本。 b $ b

解释： $ {！x} 是间接变量扩展。它的计算结果是一个包含在 $ x 中的变量的值。

为了满足我们的需要， x 还需要使用 [@] 后缀进行数组扩展。尤其要注意的是它是 x = $ {high_item} [@] 和不是 x = $ {high_item [@] } 。

你必须在数组上下文中进行评估。否则，它不会像预期的那样工作（如果你做了 arrays = $ {！x} ）。

嗯，最后说明： IFS 在这里没有任何区别。只要你在引用的数组上工作， IFS 不会起作用。

I'm trying to loop through an array that contains other arrays and these arrays consist of strings with spaces. The problem is that I can't seem to preserve the spacing in the string. The string with spaces are either divided into multiple items if I change IFS to \n or all the elements of the array are seen as 1 item if I leave IFS unchanged here's some sample code:

#!/bin/sh
low1=("AA  QQ" "BB  LL")
low2=("CC" "DD")
low3=("EE" "FF")
high=(low1 low2 low3)

for high_item in ${high[@]}
do
   eval arrayz=\${$high_item[@]}
   #IFS=$'\n'
   for item in $arrayz
   do
      echo $item
   done
done

Output:

AA
QQ
BB
LL
CC
DD
EE
FF

As you can see the elements "AA QQ" and "BB LL" have been split.

If I uncomment the line that sets IFS to \n I get the following:

AA QQ BB LL
CC DD
EE FF

Now "AA QQ" and "BB LL" are concatenated!

Is there anyway I can preserve these elements just as they original are...I need the output to look like that:

AA QQ
BB LL
CC 
DD
EE 
FF

解决方案

I think you meant that the output should look like:

AA  QQ
BB  LL
CC
DD
EE
FF

i.e.:

${low1[0]}
${low1[1]}
${low2[0]}
${low2[1]}
${low3[0]}
${low3[1]}

This could be accomplished using:

#!/bin/bash

low1=("AA  QQ" "BB  LL")
low2=("CC" "DD")
low3=("EE" "FF")
high=(low1 low2 low3)

for high_item in ${high[@]}
do
    x="${high_item}[@]" # -> "low1[@]"
    arrays=( "${!x}" )

    #IFS=$'\n'
    for item in "${arrays[@]}"
    do
        echo "$item"
    done
done

And please always use #!/bin/bash for bash scripts.

Explanation: ${!x} is indirect variable expansion. It evaluates to the value of variable with a name contained in $x.

For our needs, x needs to have the [@] suffix for array expansion as well. Especially note that it is x=${high_item}[@] and not x=${high_item[@]}.

And you have to evaluate it in array context; otherwise, it wouldn't work as expected (if you do arrays=${!x}).

Ah, and as final note: IFS doesn't make any difference here. As long as you are working on quoted arrays, IFS doesn't come into play.

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