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Regular expression capturing a repeated group

Answer 1

A:

You can use strtok. Or best use string container, use the find_first_of algorithm to find the required position of \ or / and then extract the substring. Depends if you are using C or C++

DumbCoder 2010-06-28 14:25:48

Answer 2

+3 A:

When you repeat a capturing group in a regular expression, the capturing group only stores the text matched by its last iteration. If you need to capture multiple iterations, you'll need to use more than one regex. (.NET is the only exception to this. Its CaptureCollection provides the matches of all iterations of a capturing group.

Yousui 2010-06-28 14:27:09

My intention was to use boost::regex_search to achive this,so that I can loop, but loop executes only once as it matches last instance always, is there any way to get around this ?std::string::const_iterator start = str.begin(), end = str.end();while(regex_search(start,end,what,expr)){ cout << what[0]; cout << what[1]; ... start += what.position () + what.length ();}

omshanti 2010-06-28 15:06:53

Answer 3

A:

In python something like this will find them:

mystr = 'OS BENKL/LHRBA/MANQFL\r\n'
myre = re.compile('((\w{3})(\w{2,3}))')

myre.findall(mystr) # prints [('BENKL', 'BEN', 'KL'), ('LHRBA', 'LHR', 'BA'), ('MANQFL', 'MAN', 'QFL')]

g.d.d.c 2010-06-28 14:28:15

I have to do this in C++, I am using boost::regex_search to achive this, but not working:std::string::const_iterator start = str.begin(), end = str.end(); while(regex_search(start,end,what,expr)) { cout << what[0]; cout << what[1]; ... start += what.position () + what.length (); }

omshanti 2010-06-28 15:09:29

Your tags included Python. Sorry, but I don't know c++ or the boost libraries.

g.d.d.c 2010-06-28 15:12:26

thanks for the help.

omshanti 2010-06-28 15:18:40

Answer 4

+1 A:

If you want a Perl solution, i might be :

use strict;
use warnings;
use 5.10.1;

my @in = ('OS BENKL/LHRBA/MANQFL\r\n', 'OS BENKL/LHRBA/MANQFL/ABCDE/XYZTTT\r\n');
foreach(@in) {
  say "in = $_";
  my @l = $_ =~ m!((\w{3})(\w{2,3}))+!g;
  for(my $i=1; $i<@l; $i+=3) {
    say "($l[$i] $l[$i+1])";
  }
}

M42 2010-06-28 15:00:47

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Regular expression capturing a repeated group

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