Hi, I want to replace few characters based on position. For example, I have the following string: 202204051121010000033333301000086963000000000000000000000000000000000000000000 in which the characters in 7th position to 12th are date. I want to replace the characters in 7th position to 12th position with today's date. That is replace 051121 with 051123. I will be having today's date in a variable. I need to assign that to 7th - 13th position. Can you please let me know how to change this. Thanks, Anil

I think the easiest way is to cut the string at columns 1 to 6 and postion 13 to the end of the line. Then glue them back together with today's date: #!/bin/ksh line="202204051121010000033333301000086963000000000000000000000000000000000000000000" l1=$(echo "$line"|cut -c1-6) length=${#line} # build today's date l2=$(date "+%m%d%y") l3=$(echo $line|cut -c13-${length}) newline="${l1}${l2}${l3}" echo $newline

Wouldn't a single regex be easier? #!/usr/bin/perl $str = '202204051121010000033333301000086963000000000000000000000000000000000000000000'; $date = `date "+%m%d%y"`; $str =~ s/^(.{6})(.{6})(.*)$/$1$date$3/; print $str;

Fishmonger: Yes, it is easier if I knew perl. -) I tried your perl script on Red Hat Linux and Solaris, and each case I had to chop the newline to get the output right: . $date = `date "+%m%d%y"`; chop($date); . Would you mind explaining the regex you used. Thanks Nails

It's a substitution regex; here's its break down. ^ anchors it to the beginning of the string. (.{6}) captures/puts the first 6 characters onto $1 like your first echo/cut statement (.{6}) captures the next 6 chars (the date) into $2 like your second echo/cut. Actually, since it's not being used, we don't need to capture the date into a var; we just need to do the matching. (.*) captures the remaining chars into $3 $ anchors to the end of the string That was the matching part, the rest just “glues” together the parts $1 $date and $3 and assigns it back to $str. The removing of the \n from $date can be done in its assignment. chomp($date = `date "+%m%d%y"`); chomp only removes the newline char chop removes any char I don't do any shell scripting, but you can use the same regex with sed, or possibly with awk.

Correction: (.*) is the same as your second echo/cut statement