Perl正则表达式-动态匹配

Question

Let's say I want to match and capture 5 integers separated by one or more spaces - example input:

1111 234 3333 456 7890

I could do this:

my $input = '1111        234            3333          456    7890';
if($input =~ /^\s*([0-9]+)\s+([0-9]+)\s+([0-9]+)\s+([0-9]+)\s+([0-9]+)/)
{
  #$1 = '1111', $2 = '234', $3 = '3333', $4= '456', $5 = '7890'
}

But I want to do something like this to keep the regex simpler, rather than repeating each int 5 times explicitly:

my $input = '1111        234            3333          456    7890';
if($input =~ /^((\s*[0-9]+){5})/)
{
  #$1 = '1111        234            3333          456    7890';
  #$2 = ' 7890'
  #all other capture variables are undefined
}

However, the captures don't seem to work out.

Is there a way I can I do this and still access my 5 captures?

Even better would be an unknown number of captures:

my $input = '1111        234            3333          456    7890';
if($input =~ /^((\s*[0-9]+)+)/)
{
   #foreach capture 1..N do something...
}

Answer 1

my @numbers = $input =~ /\d+/g;

全局标志将返回列表上下文中的所有匹配项，这些匹配项将存储在您的数组中。

Answer 2

If you know what your delimiter is (in this case, one or more spaces), then you don't need a regex to capture what you want. You can use split .

use strict;
use warnings;

my $input = "1111        234            3333          456    7890";
my @ints=split /\s+/,$input;
print "$_\n" foreach(@ints);

Which produces the output:

1111
234
3333
456
7890

Answer 3

Is the pattern of the line always digit groups separated by spaces? If so, rather than the regex, why not split into array based on whitespace

@outArray = split (/ +/,$input);

Answer 4

The following would capture the first 5 integers and ignore any after that if thats what you're after. I may not be entirely clear.

#!/usr/bin/perl
use strict;
use warnings;

my $in = '1111        234            3333          456    7890 12 13';

my @ints = (split ' ', $in)[0 .. 4];

print "@ints\n";

Prints:

1111 234 3333 456 7890

Chris