如何在Perl中将纪元时间转换为正常时间？ [英] How do I convert epoch time to normal time in Perl?

问题描述

我试图编写一个Perl脚本来分析日志，其中每一行的第二个值是日期。该脚本采用三个参数：输入日志文件，开始时间和结束时间。开始时间和结束时间用于解析介于两次之间的每行上的某个值。但是要正确运行，我将开始时间和结束时间转换为纪元时间。我遇到的问题是将循环的 i值转换回正常时间以与日志文件进行比较。运行 localtime（$ i）之后，我将打印该值，并且只看到打印的引用而不是实际值。

I am attempting to write a Perl script that parses a log where on each line the second value is the date. The script takes in three arguments: the input log file, the start time, and the end time. The start and end time are used to parse out a certain value on each line that that falls between those two times. But to properly run this I am converting the start and end time to epoch time. The problem I am having is that to convert the loops 'i' value back to normal time to compare against the log file. After running localtime($i) I print the value and only see a reference printed not the actual value.

这是我到目前为止的脚本（正在开发中）：

Here is the script I have so far (it is a work in progress):

#!/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,$min,$hour,$mday,$mon,$year);
($hour,$min,$sec) = split(':', $ARGV[2]);
my $endTime = timelocal($sec,$min,$hour,$mday,$mon,$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" "$ARGV[0]"`;# | grep 'DBInstance-0' | awk '{print \$9}'`;
        #print "$DBInstance0\n";
        print "$theTime\n";
}
print "$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Perl核心模块，无法安装其他任何模块。

I only have access to the core Perl modules and am unable to install any others.

推荐答案

您可以使用 ctime ，具体取决于您对正常时间的定义：

You can use ctime, depending on your definition of "Normal time":

示例代码：

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

结果：

1279954921
Sat Jul 24 03:02:01 2010

您不需要使用Time :: Localtime（这就是为什么您获得Time :: tm ，而不是Perl内部的 localtime ）中的标准数组/字符串：

Also, you don't need to use Time::Localtime (which is why you get Time::tm instead of a standard array/string from Perl's internal localtime):

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

1279954921
Sat Jul 24 03:02:01 2010

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