Regex capturing

anna_black has asked for the wisdom of the Perl Monks concerning the following question:

Hi,

I need to do a simple command line substitution of a string, but I'm obviously making some mistake, as it doesn't work:

> echo "Hello Dolly" | perl -lpe "s/l(lo)/$1/"
He Dolly
[download]

I'm probably missing something completely silly, but I can't understand why the $1 doesn't capture the lo in the parentheses... (I want the output to be Helo Dolly)

Thanks a lot!

Comment on Regex capturing Select or Download Code

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Re: Regex capturing by FunkyMonk (Chancellor) on Apr 06, 2009 at 12:00 UTC
It's all down to quoting. On *nix systems you need to use single quotes around your Perl to avoid the shell trying to expand $1: `echo "Hello Dolly" \| perl -lpe 's/l(lo)/$1/' #Helo Dolly` [download]	[reply] [d/l]
Re: Regex capturing by manoj_speed (Prior) on Apr 06, 2009 at 13:21 UTC
In Unix system the double quotes will expand the $1, so you may not get the actual output that you required. You can use single quotes or escape the $1(in double quotes) to solve this. But you can also achieve this using this simple method also.(substitute loo with lo (for this requirement/string. But it won't be nice for bigger match)). -- The wisest mind has something yet to learn.	[reply]
Re: Regex capturing by Grey Fox (Chaplain) on Apr 06, 2009 at 14:36 UTC
Hi Anna; There is a great php based tool called My REGEX Tester for checking out and experimenting with your REGEX. I have a link to it in Found new online Regex tool. Plus there is a nice website called Regular-Expressions Quickstart. This teaches all of the basics of REGEX. Hope this helps. -- Grey Fox "We are grey. We stand between the darkness and the light" B5	[reply]
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