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";
posted on 2013-06-24 00:25 DaMengZhang 阅读(1004) 评论(0) 编辑 收藏 举报