topic Perl question in Operating System - HP-UX

Perl question

Allanm — Fri, 05 Feb 2010 20:06:56 GMT

perl -pi.back -e "s:(ap\d+) :ap1 \$1 :" filename

Can you explain in detail the above command, I understand the part abt ap\d+ but dont quite get the rest of colons and $1.

Thanks,
Allan.

Re: Perl question

James R. Ferguson — Fri, 05 Feb 2010 20:22:55 GMT

HI Allan:

The colons are used as delimiters. As written, the whitespace is considered part of the characters to match (which seems odd).

The parenthesized part captures the match which is then available in $1 and used as a backreference in the substitution.

Thus:

# echo "I am ap3 "|perl -pe 's:(ap\d+) :ap1 $1 :'
I am ap1 ap3

Yet, if we don't have trailing whitespace there is no match:

# echo "I am ap3"|perl -pe 's:(ap\d+) :ap1 $1 :'
I am ap3

Unless we use the 'x' modified to write more readable expressions where whitespace can be used to spread out the pieces:

# echo "I am ap3"|perl -pe 's:(ap\d+) :ap1 $1 :'x
I am ap1 ap3

Regards!

...JRF...

Re: Perl question

Allanm — Sat, 06 Feb 2010 02:01:49 GMT

awesome explaination and analysis.Is there something more than 10!! :)

Allan.