简单的格式化日期 [英] Simple way to format date

问题描述

我写了一个perl脚本来获取datetime。
它做的工作，但我想知道是否有任何更简单的方式格式化日期作为输出。

 ＃！/ usr / bin / perl 
使用DateTime; 
 
我的$ dt = DateTime-> now（time_zone =>'local'）; 
 $ myTimeStamp = $ dt-> subtract（days => 1）; 
 $ myYear = $ myTimeStamp-> year; 
 $ myMonth = $ myTimeStamp-> month; 
 if（length（$ myMonth）== 1）{
 $ myMonth =0 $ myMonth; 
} 
 $ myDay = $ myTimeStamp-> day; 
 if（length（$ myDay）== 1）{
 $ myDay =0 $ myDay; 
} 
 $ myHour = $ myTimeStamp->小时; 
 if（length（$ myHour）== 1）{
 $ myHour =0 $ myHour; 
} 
＃$ myDir =/ var / tmp / logs / $ myYear / $ myMonth / $ myYear $ myMonth-; 
 
 print - > $ myYear $ myMonth $ myDay $ myHour\\\
; 
＃ - > 2012 02 28 02 
 exit 0; 
  

解决方案

日期时间为此提供 format_cldr 方法：

 使用DateTime; 
 
 my $ myTimeStamp = DateTime-> now-> subtract（days => 1）; 
 
 printf - >％s\\\
，$ myTimeStamp-> format_cldr（'yyyy MM dd HH'）; 
＃ - > 2012 02 28 02 
  

I wrote a perl script to get datetime. It do work but I do wonder if there any easier way to format date as output.

#!/usr/bin/perl
use DateTime;

my $dt          = DateTime->now( time_zone => 'local' );
$myTimeStamp    = $dt->subtract( days => 1 );
$myYear         = $myTimeStamp->year;
$myMonth        = $myTimeStamp->month;
if ( length( $myMonth ) == 1 ) {
    $myMonth = "0$myMonth";
}
$myDay          = $myTimeStamp->day;
if ( length( $myDay ) == 1 ) {
    $myDay = "0$myDay";
}
$myHour         = $myTimeStamp->hour;
if ( length( $myHour ) == 1 ) {
    $myHour = "0$myHour";
}
#$myDir          = "/var/tmp/logs/$myYear/$myMonth/$myYear$myMonth-";

print "--> $myYear $myMonth $myDay $myHour\n";
# --> 2012 02 28 02
exit 0;

解决方案

DateTime provides the format_cldr method for this:

use DateTime;

my $myTimeStamp = DateTime->now->subtract( days => 1 );

printf "--> %s\n", $myTimeStamp->format_cldr('yyyy MM dd HH');
# --> 2012 02 28 02

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