CODEWITHSUNDEEP
linuxbashshellline-breaks
Dave
Dave

Reputation: 19090

How to preserve line breaks when storing command output to a variable?

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

#!/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

./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.

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

Upvotes: 295

Views: 151906

Answers (2)

jaypal singh
jaypal singh

Reputation: 77085

With shell scripting, one needs to always quote variables, especially when working with strings.

Here is an example of the problem:

Example variable:

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

Output without quoting the variable:

$ echo $f
fafafda adffd adfadf adfafd afd

Output WITH quoting the variable:

$ echo "$f"
fafafda
adffd
adfadf
adfafd
afd

Explaination:

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.

Upvotes: 557

Bob
Bob

Reputation: 319

I have ran into the same problem, a quote will help

ubuntu@host:~/apps$ apps="abc
> def"
ubuntu@host:~/apps$ echo $apps
abc def
ubuntu@host:~/apps$ echo "$apps"
abc
def

Upvotes: 8

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