将旧的Unix日期转换为Perl并进行比较 [英] Convert Old Unix Date to Perl and compare

问题描述

要求-我的文件名为 Rajesh.1202242219。数字只不过是日期 date’+％y％m’％d’％H’％M’格式。
现在，我试图编写一个perl脚本以从文件名中提取数字，并与当前系统日期和时间进行比较，并根据此比较的输出，使用perl打印一些值。

Requirement - I have file name called "Rajesh.1202242219". Numbers are nothing but a date "date '+%y''%m''%d''%H''%M'" format. Now I am trying to write a perl script to extract the numbers from file name and compare with current system date and time and based on output of this comparison, print some value using perl.

方法：

从文件名中提取数字：

if ($file =~ /Rajesh.(\d+).*/) {
print $1;
        }

将此时间转换为perl可读的时间

my $sec  =  0;  # Not Feeded
my $min  =  19;
my $hour =  22;
my $day  =  24;
my $mon  = 02   - 1;
my $year = 2012 - 1900;
my $wday = 0;   # Not Feeded
my $yday = 0;   # Not Feeded

my $unixtime = mktime ($sec, $min, $hour, $day, $mon, $year, $wday, $yday);
print "$unixtime\n";
my $readable_time = localtime($unixtime);
print "$readable_time\n";

查找当前时间并进行比较...

my $CurrentTime = time();
my $Todaydate = localtime($startTime);

但是这里的问题是，我没有得到如何从<$ c中提取2位数字的解决方案$ c> $ 1 并分配给 $ sec ， $ min 等。任何帮助？

But the problem here is, I am not getting solution of how to extract 2 digit from $1 and assign to $sec, $min, etc. Any help?

此外，如果您对此问题陈述有好的解决方法，请与我分享

Also, if you have good approach for this problem statement, Please share with me

推荐答案

我认为 解包 可能更合适。

I think unpack might be a better fit.

if ( my ( $num ) = $file =~ /Rajesh.(\d+).*/ ) {
    my ( $year, $mon, $day, $hour, $min ) = unpack( 'A2 A2 A2 A2 A2', $num ); 
    my $ts = POSIX::mktime( 0, $min, $hour, $day, $mon - 1, $year + 100 );
    ...
}

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