perl获取日起和时间

注意：localtime获取的年份是相对于1900的偏移，需要加上1900，而localtime获取的month范围是0-11,需要加1。

my ($sec,$min,$hour,$day,$mon,$year,$wday,$yday,$isdst) = localtime();    
$year += 1900;    
$mon++;
my $date = "$year-$mon-$day";    
print $date, "\n";

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