echo not printing next line unix bash

Question

In my shell script, I have using below code:

RECORD=`ps -ef | grep sysProj | cut -d" " -f1,2` echo $RECORD

on executing this, its printing

45646654 28770 45646654 32021 45646654 32176

but when I don't save command output in variable and executing as below:

ps -ef | grep sysProj | cut -d" " -f1,2

the output is:

45646654 28770 45646654 32021 45646654 32176

Why saving the command output in varible is suppressing new line character. Also, how can I retain it?

Also use lowercase for your script variables as full uppercase variables are usuallu system variables — sjsam
– sjsam, Commented Sep 1, 2016 at 5:06
See also Security implications of forgetting to quote a variable in bash/POSIX shells — glenn jackman
– glenn jackman, Commented Sep 1, 2016 at 20:23

whoan · Accepted Answer · 2016-09-01 03:55:03Z

4

Just enclose the variable within double quotes to avoid word splitting:

echo "$RECORD"

answered Sep 1, 2016 at 3:46

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