How to pass a variable to gawk program?

Asked May 31 '22 at 10:02

Active May 31 '22 at 10:13

Viewed 65 times

I am trying to make script to rename files in the directory, they are mess, because of spaces between words (e.g. Computer Networking Explained Cisco CCNA 200-301.mp4 etc). First I need to sort them by time and then print them with the help of gawk. But I need to pass variable with the number of each line to gawk.

var=1;
while [ $var -le $(wc file.txt | gawk '{print$1}') ]; do
    echo $var;
    gawk 'BEGIN{FS="\n";RS=""}{print $($var)}' file.txt;
    var=$[ $var + 1 ];
done

. But it prints all lines of file, not single line.

edited May 31 '22 at 10:13

Fravadona

13,917
1
23
35

asked May 31 '22 at 10:02

Untitled

1

the correct way to pass shell variables is `awk -v x="$var" 'BEGIN{...}{print $x}'` – Fravadona May 31 '22 at 10:15
https://stackoverflow.com/questions/19075671/how-do-i-use-shell-variables-in-an-awk-script – James Brown May 31 '22 at 10:25

How to pass a variable to gawk program?

0 Answers0