在bash中存储命令输出到变量时如何保存换行符?

I’m using bash shell on Linux. I have this simple script …

我在Linux上使用bash shell。我有这个简单的脚本……

#!/bin/bash

TEMP=`sed -n '/'"Starting deployment of"'/,/'"Failed to start context"'/p' "/usr/java/jboss/standalone/log/server.log" | tac | awk '/'"Starting deployment of"'/ {print;exit} 1' | tac`
echo $TEMP

However, when I run this script

然而，当我运行这个脚本时

./temp.sh

all the output is printed without the carriage returns/new lines. Not sure if its the way I’m storing the output to $TEMP, or the echo command itself.

所有输出都是打印的，没有回车/新行。不确定是我将输出存储为$TEMP，还是echo命令本身。

How do I store the output of the command to a variable and preserve the line breaks/carriage returns?

如何将命令的输出存储到一个变量中，并保存换行/回车?

1 个解决方案

#1

Quote your variables. Here is it why:

引用变量。这里是原因:

$ f="fafafda
> adffd
> adfadf
> adfafd
> afd"

$ echo $f
fafafda adffd adfadf adfafd afd

$ echo "$f"
fafafda
adffd
adfadf
adfafd
afd

Without quotes, the shell replaces $TEMP with the characters it contains (one of which is a newline). Then, before invoking echo shell splits that string into multiple arguments using the Internal Field Separator (IFS), and passes that resulting list of arguments to echo. By default, the IFS is set to whitespace (spaces, tabs, and newlines), so the shell chops your $TEMP string into arguments and it never gets to see the newline, because the shell considers it a separator, just like a space.

没有引号，shell将$TEMP替换为它所包含的字符(其中之一是换行)。然后，在调用echo shell之前，使用内部字段分隔符(IFS)将该字符串分割成多个参数，并将产生的参数列表传递给echo。默认情况下，IFS被设置为空格(空格、制表符和换行符)，因此shell将$TEMP字符串分割为参数，并且它永远看不到换行符，因为shell将它视为分隔符，就像空间一样。