Getting multiple matches within a string using regex in Perl

Question

After having read this similar question and having tried my code several times, I keep on getting the same undesired output.

Let's assume the string I'm searching is "I saw wilma yesterday". The regex should capture each word followed by an 'a' and its optional 5 following characters or spaces.

The code I wrote is the following:

$_ = "I saw wilma yesterday";

if (@m = /(\w+)a(.{5,})?/g){
    print "found " . @m . " matches\n";

    foreach(@m){
        print "\t\"$_\"\n";
    }
}

However, I kept on getting the following output:

found 2 matches
    "s"
    "w wilma yesterday"

while I expected to get the following one:

found 3 matches:
    "saw wil"
    "wilma yest"
    "yesterday"

until I found out that the return values inside @m were $1 and $2, as you can notice.

Now, since the /g flag is on, and I don't think the problem is about the regex, how could I get the desired output?

I don't understand why you have day in your expected result and not yesterday? — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Jul 10, 2013 at 21:09

Casimir et Hippolyte · Accepted Answer · 2013-07-10 22:20:16Z

3

You can try this pattern that allows overlapped results:

(?=\b(\w+a.{1,5}))

or

(?=(?i)\b([a-z]+a.{0,5}))

example:

use strict;
my $str = "I saw wilma yesterday";
my @matches = ($str =~ /(?=\b([a-z]+a.{0,5}))/gi);
print join("\n", @matches),"\n";

more explanations:

You can't have overlapped results with a regex since when a character is "eaten" by the regex engine it can't be eaten a second time. The trick to avoid this constraint, is to use a lookahead (that is a tool that only checks, but not matches) which can run through the string several times, and put a capturing group inside.

For another example of this behaviour, you can try the example code without the word boundary (\b) to see the result.

edited Jul 10, 2013 at 22:20

answered Jul 10, 2013 at 21:12

Casimir et Hippolyte

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4 Comments

PP. Over a year ago

I tested this and it worked. Well done. while ( m/(?=\b(\w+a.{1,5}))/g ) { print "$1\n"; }

Acsor Over a year ago

Yeah, It works properly. But wasn't the "?=" the lookhahead option? Why is it necessary?

Casimir et Hippolyte Over a year ago

@none: the lookahead allows overlapped matches, then you can have overlapped results ( as saw wil and wilma yest) in one shot. see the example.

Acsor Over a year ago

Yeah, I noticed it. Thanks anyway!

PP. · Accepted Answer · 2013-07-10 21:03:33Z

1

Firstly you want to capture everything inside the expression, i.e.:

/(\w+a(?:.{5,})?)/

Next you want to start your search from one character past where the last expression's first character matched.

The pos() function allows you to specify where a /g regex starts its search from.

answered Jul 10, 2013 at 21:03

PP.

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Comments

perreal · Accepted Answer · 2013-07-10 21:17:48Z

1

$s = "I saw wilma yesterday";    
while ($s =~ /(\w+a(.{0,5}))/g){
    print "\t\"$1\"\n";
    pos($s) = pos($s) - length($2); 
}

Gives you:

"saw wil"
"wilma yest"
"yesterday"

But I don't know why you should get day and not yesterday.

answered Jul 10, 2013 at 21:17

perreal

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