将命令的输出捕获到保留新行的变量中 [英] Capture output from command into variable retaining new lines

问题描述

我有两个脚本； parentScript.sh 和 childScript.sh 。

I希望能够在 parentScript.sh 内调用 childScript.sh 并返回在任何阶段发生的错误。即在 childScript.sh 中发现的错误如下：

  echo错误：缺少$ siteTag的供稿文件& 2 
  

我知道如何返回错误 out 回到父外壳。

但是我感觉它被篡改了，我不能再 printf 将结果转换为漂亮的变量。即

  error + = $（{./childScript.sh | sed's / Output / Useless /'2>& 4 1>& 3;} 2>& 1） 
 error + = $（{./childScript.sh | sed's / Output / Useless /'2>& 4 1>& 3 ;} 2>& 1） 
  

本质上应该两次调用脚本，并从两个脚本中获取错误并按照我的想法将它们存储在变量 error 中，但确实如此，但是通过使用 echo $ error 或 printf $ error 。

有人知道这里的解决方案吗？设法从多个命令中获取错误输出，但在 childScript.sh 命令中维护对 echo 的单独调用？ p>

编辑：输出应为..

 错误：缺少供稿文件（..）
错误：（..）
错误的Feed文件：（..）
  

但是相反

  ER ROR：供稿文件缺少（..）错误：供稿文件缺少（..）错误：供稿文件缺少（..）
  

解决方案

$（..）删除尾随换行符。这在大多数时候非常有用，例如在

  echo中显示欢迎使用$（hostname）。祝您逗留愉快。 
  

但是，在您的情况下，它有点毁了。您可以只添加一个：

  error + = $（{./childScript.sh | sed的/输出/无用/'2>& 4 1>& 3;} 2>& 1） $'\n'
  

I have two scripts; parentScript.sh and childScript.sh.

I want to be able to call childScript.sh inside parentScript.sh and return the errors that occur within at any stage. i.e. an error found within childScript.sh looks like:

echo "ERROR: Feed file missing for $siteTag" >&2

I know how to return the error out back towards the parent shell.

But I have a feeling it is being tampered with, I can no longer printf the result to a nice looking variable. i.e.

error+="$( { ./childScript.sh | sed 's/Output/Useless/' 2>&4 1>&3; } 2>&1 )"
error+="$( { ./childScript.sh | sed 's/Output/Useless/' 2>&4 1>&3; } 2>&1 )"

Should essentially call the script twice, get errors from both scripts and store them in the variable error as I thought, which it does but it somehow gets rid of the lines both with the use of echo "$error" or printf "$error".

Does anyone know a solution here to manage to grab error output from several commands but maintain the separate calls to echo within the childScript.sh commands?

Edit: Output should be..

ERROR: Feed file missing for (..)
ERROR: Feed file missing for (..)
ERROR: Feed file missing for (..)

But is instead

ERROR: Feed file missing for (..) ERROR: Feed file missing for (..) ERROR: Feed file missing for (..)

解决方案

$(..) strips trailing line feeds. This is very useful most of the time, like in

echo "Welcome to $(hostname). Enjoy your stay."

However, in your case, it ruins it a bit. You can just add one back:

error+="$( { ./childScript.sh | sed 's/Output/Useless/' 2>&4 1>&3; } 2>&1 )"$'\n'

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