Perl question

[Allanm]

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.

[James R. Ferguson · Accepted solution]

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...

[Allanm]

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

Allan.