How to get specific line from output and save it in one variable?

Question

When I run this command

user@ubuntu:~$ ./2input1output Jojo Ar Dublin

it gives me this output

Your First Name: Jojo 
Your Last Name : Ar 
Your Hometown : Dublin
Thank you, Jojo Ar From Dublin,
Welcome to Stackoverflow

Now I need to display the fourth and fifth lines from that output and then save it into a variable. I have tried

user@ubuntu:~$ (sed -n 4,5p ./2input1output)

but it gave me this output

echo"Thank you, $1 $2 From $3,
Welcome to stackoverflow"

Answer 1

Your command is actually working as expected; you are asking sed to return lines 4 and 5 from the file "2input1output". As you can see, the positional arguments (i.e. $1, $2 and $3) are returned because that is what is in the script.

What you need to do is actually run the script first in order to get the desired output (STDOUT) and then pipe that to the stream editor (sed). So, in one hit, to save the desired content to a variable:

$ MESSAGE=$(./2input1output Jojo Ar Dublin | sed -n 4,5p)

Please remember that if you need to preserve newline characters when accessing the variable, wrap it in double quotation marks first.

Answer 2

i think sed get fifth line from your script. you can send output to another file then sed lines you want from there or in your bash save output to a variable then grep the lines you want.

Answer 3

We can do it by below method too using awk

MESSAGE=`sh 2input1output Jojo Ar Dublin| awk 'NR >3 && NR<=5 {print $0}'