这些是我们源文件标题中的有效年份,在版权行中:
2016
2007,2016
2007-2010,2016
2010,2012-2016
当前年份(2016 年)应该是其中的一部分。
无效的:
2016,2007 # not in order
2010-2007,2016 #range not in order.
2010-2015 # current year missing.
我制作了如下脚本,仍在处理它,有没有更好的方法来检查它?
use strict;
use warnings;
use List::Util 'first';
use Time::Piece;
my $t = Time::Piece->new();
my $currentYear=$t->year;
my $filename="sample.txt"; # this file name will be passed-in to the script.
my $fileContent=`cat $filename`;
my $copyrightFound = first { /copyright .* Shakespeare/i } $fileContent;
if (!$copyrightFound) {
print "\nERROR: Copyright missing\n";
exit;
}
#copyright Found
print $fileContent;
if ($fileContent =~ /Copyright \(c\) (.*) by Bill Shakespeare\s+.*All rights reserved./) {
print "\nCopyright is good: $1\n";
my $years = $1;
print "\n$years, $currentYear\n";
#e.g: 2016
if ($years !~ ', ' && $years !~ '-') {
print "\nerror\n" if ($years !~ /$currentYear/);
}
#e.g: 2010-2016
elsif ($years !~ ', ') {
print "\nError\n" if ($years !~ /\-$currentYear$/);
}
#eg: 2010, 2016
elsif ($years !~ '-') {
print "\nError\n" if ($years !~ /\, $currentYear$/);
}
#e.g: 2008, 2010, 2011-2016
elsif ($years =~ '-' && $years =~ ', ') {
print "\nError 5\n" if ($years !~ /\d\d\d\d, \d\d\d\d\-$currentYear$/);
}
else {
print "invalid format"
}
} else {
print "\nCopyright needs to be fixed\n";
}
sample.txt 有:
Copyright (c) 2008, 2010, 2011-2016 by Bill Shakespeare
All rights reserved.
最佳答案
您可以使用 Set::IntSpan验证日期:
use warnings;
use strict;
use Set::IntSpan qw();
my $currentYear = 2016;
while (<DATA>) {
s/\s+//g;
my $ok = 0;
if (Set::IntSpan->valid($_)) { # Checks order
my $set = Set::IntSpan->new($_);
my @years = $set->elements();
$ok = 1 if grep { $_ == $currentYear } @years;
}
print "$ok : $_\n";
}
__DATA__
2007
2007, 2016
2007-2010,2016
2010,2012-2016
2016,2007
2010-2007,2016
2010-2015
这打印:
0 : 2007
1 : 2007,2016
1 : 2007-2010,2016
1 : 2010,2012-2016
0 : 2016,2007
0 : 2010-2007,2016
0 : 2010-2015
简化使用
member
而不是 grep
,正如 pilcrow
所推荐的在评论中:my $ok = 0;
if (Set::IntSpan->valid($_)) {
my $set = Set::IntSpan->new($_);
$ok = 1 if $set->member($currentYear);
}
print "$ok : $_\n";
关于regex - 使用 Perl 验证年份范围,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36987918/