如何在Perl中将纪元时间转换为正常时间？

我正在尝试编写一个Perl脚本来解析日志,其中每行第二个值是日期.该脚本包含三个参数：输入日志文件,开始时间和结束时间.开始和结束时间用于解析每行之间的某个值,该值介于这两次之间.但为了正确运行这个,我将开始和结束时间转换为纪元时间.我遇到的问题是将循环’i’值转换回正常时间以与日志文件进行比较.运行localtime($i)后,我打印该值,只看到打印的参考值而不是实际值.

这是我到目前为止的脚本(这是一项正在进行的工作)：

#!/usr/bin/perl
use strict;
use warnings;
use Time::Local;
use Time::localtime;
use File::stat;

my $sec = 0;
my $min = 0;
my $hour = 0;
my $mday = 0;
my $mon = 0;
my $year = 0;
my $wday = 0;
my $yday = 0;
my $isdst = 0;

##########################
# Get the engine log date
##########################
my $date = `grep -m 1 'Metric' "$ARGV[0]" | awk '{print $2}'`;
($year,$mon,$mday) = split('-',$date);
$mon--;

#########################################
# Calculate the start and end epoch time
#########################################
($hour,$min,$sec) = split(':',$ARGV[1]);
my $startTime = timelocal($sec,$hour,$mday,$year);
($hour,$ARGV[2]);
my $endTime = timelocal($sec,$year);


my $theTime = 0;
for (my $i = $startTime; $i <= $endTime + 29; $i++) {
        #print "$startTime   $i n";

        $theTime = localtime($i);

        #my $DBInstance0 = `grep "$hour:$min:$sec" "$ARGV[0]"`;# | grep 'DBInstance-0' | awk '{print $9}'`;
        #print "$DBInstance0n";
        print "$theTimen";
}
print "$startTime   $endTime n";

输出如下：

Time::tm=ARRAY(0x8cbbd40)
Time::tm=ARRAY(0x8cbc1a0)
Time::tm=ARRAY(0x8cbbe80)
Time::tm=ARRAY(0x8cbc190)
Time::tm=ARRAY(0x8bbb170)
Time::tm=ARRAY(0x8cbc180)
Time::tm=ARRAY(0x8cbbf30)
Time::tm=ARRAY(0x8cbc170)
Time::tm=ARRAY(0x8cbc210)
Time::tm=ARRAY(0x8cbc160)
1275760356   1275760773

我只能访问核心Perl模块,无法安装任何其他模块.

use Time::Local; 
use Time::localtime; 
my $time=timelocal(1,2,3,24,6,2010);
print "$timen"; 
$theTime = ctime($time); 
print "$theTimen";

1279954921
Sat Jul 24 03:02:01 2010

use Time::Local; 
my $time=timelocal(1,2010); 
print "$timen"; 
$theTime = localtime($time); 
print "$theTimen";

1279954921
Sat Jul 24 03:02:01 2010