Avoid printing line breaks in shell script

Question

while read pos; do 
     string1=`echo $pos | cut -c 20-38`
     string2="=$pos" 
     string3="$string1 $string2" 
     echo "$string3" 
done < file

This is the code of my script. I want to display the output in one single line, but the output comes on different lines. How can I do this?

ex- pos = abcdefghi1234567890QWERTYUXY.tar.gz string1 = 1234567890QWERTYUXY string2 = abcdefghi1234567890QWERTYUXY.tar.gz

output wanted :

1234567890QWERTYUXY abcdefghi1234567890QWERTYUXY.tar.gz

Answer 1

At first, there is multiple typos in your script. Apart from that, you can use

echo -n "$string3 "

to print variable string3, without adding linebreak after that.

From man echo:

   -n     do not output the trailing newline

Answer 2

you want to use tr -d '\n' at the end of the $string1 assignment to remove the line break that is present in your file:

while read pos; do 
     string1=`echo $pos | cut -c 20-38 | tr -d '\n'`
     string2="=$pos" 
     string3="$string1 $string2" 
     echo "$string3" 
done < file

note that however, this will give the output:

QWERTYUXY.tar.gz =abcdefghi1234567890QWERTYUXY.tar.gz

with pos = abcdefghi1234567890QWERTYUXY.tar.gz and the script you gave.

Answer 3

There are lots of options, but the way I would recommend is printf.

It doesn't print a newline by default, so changing your echo to a printf should remove the unwanted newline.

while read pos; do 
    string1=`printf "%s" $pos | cut -c 20-38`
    string2="=$pos" 
    string3="$string1 $string2" 
    printf "%s\n" "$string3" 
done < file

Or perhaps just

while read pos; do 
    string1=`printf "%s" $pos | cut -c 20-38`
    string2="=$pos" 
    printf "%s=%s\n" "$string1" "$string2"
done < file

Output

1234567890QWERTYUXY=abcdefghi1234567890QWERTYUXY.tar.gz

That said, your original script works for me too. :-/

Output

1234567890QWERTYUXY abcdefghi1234567890QWERTYUXY.tar.gz