Issue
I want to replace the nth charater of a line to '.', with n being line dependent and can be provided on the line itself, e.g. a example file can look like this:
$cat test
0 ABCD
1 ABCD
2 ABCD
The desired output is
0 .BCD
1 A.CD
2 AB.D
3 ABC.
I've tried capturing the number and using \1 inside {}, but it doesn't work:
$ cat test | perl -pe 's/([0-9]) ([A-Z]{\1})./\1 \2./'
0 ABCD
1 ABCD
2 ABCD
whereas using a normal number inside {} gives the desired output:
$ cat test | perl -pe 's/([0-9]) ([A-Z]{2})./\1 \2./'
0 AB.D
1 AB.D
2 AB.D
Using perl (and not sed) because I know that using capturing groups in the pattern does in principle work:
$ echo 'KALLO' | perl -pe 's/([A-Z])\1/__/'
KA__O
So my questions:
Why doesn't my approach with {\1} work?
How can I achieve what I want without using a loop?
Solution
Perl expects a number after {, not a backslash.
You can build the pattern from variables, though, and use it in a substitution:
perl -pe '/^([0-9]+)/; $p = "(.* .{$1})."; s/$p/$1./'
Or even, on Perl 5.010+
perl -pe '/^([0-9]+)/; $p = ".* .{$1}\\K."; s/$p/./'
Answered By - choroba Answer Checked By - Mary Flores (WPSolving Volunteer)