perl 获取系统时间

最近需要将字符串转换成时间，找了下资料，实战如下，发现时timelocal费了些时间

strftime也可在 c / c++ / awk / php 中使用，用法基本一致。

这个也不错

$time = strftime("%Y-%m-%d %H:%M:%S", localtime($time));

print "$time \n";
print str2time($time);

use POSIX qw(strftime);
use Time::Local;
use HTTP::Date;

my $time = "{\"date\":\"2013-06-24\",\"time\":\"01:53:40\"}";
print "$time \n";

print str2time($time);

formattime($time);

sub formattime{
    my ($srctime) = @_;
    my ($year,$mon,$day,$hour,$min,$sec) = $srctime=~ /.+?(\d+)-(\d+)-(\d+).+?(\d+):(\d+):(\d+)/ ;
    print "\n $year,$mon,$day,$hour,$min,$sec \n";

    $time = str2time("$year-$mon-$day $hour:$min:$sec");
    #print "$time \n";

    #print strftime("%Y%m%d%H%M%S",localtime($time));
    return $time;
}

use strict;
use POSIX;
use Time::Local;

my $time = time();
print "$time \n";

my @time = localtime($time);
print "@time \n";

$time = strftime("%Y-%m-%d %H:%M:%S", localtime($time));
print "$time \n";

my ($year,$mon,$day,$hour,$min,$sec) = $time=~/(\d+)-(\d+)-(\d+) (\d+):(\d+):(\d+)/ ;
print "$year,$mon,$day,$hour,$min,$sec \n";

$time = timelocal($sec,$min,$hour,$day,$mon,$year);
print "$time \n";