7 messages in org.freebsd.freebsd-questionsRe: Perl question| From | Sent On | Attachments |
|---|---|---|
| Christopher Rued | Oct 2, 2000 11:00 am | |
| Mark Ovens | Oct 2, 2000 11:15 am | |
| Mark Ovens | Oct 2, 2000 11:25 am | |
| Andresen,Jason R. | Oct 2, 2000 11:36 am | |
| Christopher Rued | Oct 2, 2000 2:09 pm | |
| Mark Ovens | Oct 2, 2000 3:34 pm | |
| Dirk Myers | Oct 2, 2000 7:21 pm |
| Subject: | Re: Perl question | |
|---|---|---|
| From: | Mark Ovens (mar...@freebsd.org) | |
| Date: | Oct 2, 2000 11:15:15 am | |
| List: | org.freebsd.freebsd-questions | |
On Mon, Oct 02, 2000 at 02:01:11PM -0400, Christopher Rued wrote:
I realize that this is not really a FreeBSD question, but I like this mailing list and the answers that are sent to it :).
My question is: How do I have perl return me the first and only the first match to a regular expression? I know that I can return /all/ matches to an array, and get the first one, but I only want to have it evaluate as far as the first match and then stop.
For example, given the string xayxbyxcy and the regular expression x.y, I want it to return to me only xay and not evaluate any further.
/x.?y/
From Learning Perl:
"We can force any multiplier to be non-greedy (or lazy) by following it with a question mark:
$_ = "a xxx c xxxxxxxxx c xxx d"; /a.?c.*d/;
Here, the a.*?c now matches the fewest characters between the a and he c, not the most characters. This means that the leftmost c ismatched, not the rightmost....."
Thanks in advance.
-- -Chris Rued
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