日期转换

Question

在Perl中，我怎么转换日期像

Thu Mar 06 02:59:39 +0000 2008 

到

2008-03-06T02:59:39Z 

尝试HTTP ::日期，如果这个问题没有在字符串中+0000它工作:(

Answer 1

DateTime::Format::Strptime将执行此转换。

#!/usr/bin/perl 
use strict; 
use warnings; 
use 5.012; 
use DateTime::Format::Strptime; 

my $date = 'Thu Mar 06 02:59:39 +0000 2008 '; 

my(@strp) = (
    DateTime::Format::Strptime->new(pattern => "%a %b %d %T %z %Y",), 
    DateTime::Format::Strptime->new(pattern => "%FY%T%Z",) 
); 

my $dt = $strp[0]->parse_datetime($date); 

print $strp[1]->format_datetime($dt); 

打印2008-03-06T02：59：39UTC

克里斯

Answer 2

所以，用一个正则表达式编辑和使用HTTP::Date：

(my $new_date_string = $old_state_string) =~ s/[+-]\d{4,}\s+//; 

Answer 3

如果你绝对，积极确保日期将始终以这种格式，你可以简单地使用正则表达式重新格式化它。唯一的问题是你必须有一种将月份转换为数字的方式。这样，您就不必下载任何额外的模块来做日期转换：

my $date = "Thu Mar 06 02:59:39 +0000 2008"; #Original String 

#Create the Month Hash (you might want all twelve months). 
my %monthHash (Jan => "01", Feb => 2, Mar => 3); 

# Use RegEx Matching to parse your date. 
# \S+ means one or more non-spaces 
# \s+ means one or more spaces 
# Parentheses save that part of the string in $1, $2, $3, etc. 
$date =~ m/\S+\s+(\S+)\s+(\S+)\s+(\S+)\s+\S+\s(.*)/; 

my $monthString = $1; 
my $day = $2; 
my $time = $3; 
my $year = $4; 

# Convert Month string to a number. 
my $month = $monthHash{$monthString}; 

#Reformat the string 
$fmtDate="$year-$month-$day" . "T" . "$time" . "Z"; 

否则我会说，你也可以尝试日期时间::格式:: Strptime，但Chris Charley打我吧。