regex - 仅当模式存在时才匹配模式

Question

我有以下代码:

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

my $SourceStr='Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: Prelim 3  Optional: Some stuff here';
#my $SourceStr='Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: Prelim 3';

my $RegEx = qr/Name: (.+)  Time: (.+)  State: (.+)  Optional: (.+?)(  |$)/;

if ($SourceStr =~ m/$RegEx/) {
   print "1=[$1]\n";
   print "2=[$2]\n";
   print "3=[$3]\n";
   print "4=[$4]\n";
}

当使用第一个 $SourceStr 运行时，它按预期工作。但是，对于第二个被注释掉的，有没有办法让 $4 填充空字符串？

第一个字符串结果:

1=[Rob]
2=[11/2/2011 1:47:30 PM]
3=[3]
4=[Some stuff here]

第二个字符串结果:不匹配

想要:

1=[Rob]
2=[11/2/2011 1:47:30 PM]
3=[3]
4=[]

Answer 1

您可以使用更具体的正则表达式:

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

my @SourceStrA=('Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: 3  Optional: Some stuff here',
                'Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: 3');

my $RegEx = qr!Name:\s*(\w+)\s*Time:\s*([\d/]*\s*[\d:]*)\s*State:\s*(\d+)\s*(?:Optional:\s*(.*))?!;

for my $SourceStr (@SourceStrA) {
  print "$SourceStr\n";
  if ($SourceStr =~ m/$RegEx/) {
    print "1=[$1]\n";
    print "2=[$2]\n";
    print "3=[$3]\n";
    print "4=[$4]\n" if defined $4; 
  }
}

输出:

Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: 3  Optional: Some stuff here
1=[Rob]
2=[11/2/2011 13:47:30]
3=[3]
4=[Some stuff here]
Foo - Name: Rob  Time: 11/2/2011 13:47:30  State: 3
1=[Rob]
2=[11/2/2011 13:47:30]
3=[3]