将字符串转换为特定日期时间格式？

Question

串

"2011-05-19 10:30:14"

至

Thu May 19 10:30:14 UTC 2011

如何将特定字符串转换为此日期格式？

Answer 1

require 'date'

date = DateTime.parse("2011-05-19 10:30:14")
formatted_date = date.strftime('%a %b %d %H:%M:%S %Z %Y')

有关格式化日期的详细信息,请参阅strftime().

Answer 2

"2011-05-19 10:30:14".to_time

Answer 3

无需申请任何东西.只需在分配了日期的变量末尾添加此代码即可.例如

   @todaydate = "2011-05-19 10:30:14"
   @todaytime.to_time.strftime('%a %b %d %H:%M:%S %Z %Y')

您可以根据自己的喜好获得适当的格式.您可以在Rails控制台上查看

Loading development environment (Rails 3.0.4)
ruby-1.9.2-p136 :001 > todaytime = "2011-05-19 10:30:14"
 => "2011-05-19 10:30:14" 
ruby-1.9.2-p136 :002 > todaytime
 => "2011-05-19 10:30:14" 
ruby-1.9.2-p136 :003 > todaytime.to_time
 => 2011-05-19 10:30:14 UTC
ruby-1.9.2-p136 :008 > todaytime.to_time.strftime('%a %b %d %H:%M:%S %Z %Y')
 => "Thu May 19 10:30:14 UTC 2011"

尝试'date_format'gem以不同的格式显示日期.

Answer 4

更多格式:

 require 'date'

 date = "01/07/2016 09:17AM"
 DateTime.parse(date).strftime("%A, %b %d")
 #=> Friday, Jul 01

 DateTime.parse(date).strftime("%m/%d/%Y")
 #=> 07/01/2016

 DateTime.parse(date).strftime("%m-%e-%y %H:%M")
 #=> 07- 1-16 09:17

 DateTime.parse(date).strftime("%b %e")
 #=> Jul  1

 DateTime.parse(date).strftime("%l:%M %p")
 #=>  9:17 AM

 DateTime.parse(date).strftime("%B %Y")
 #=> July 2016
 DateTime.parse(date).strftime("%b %d, %Y")
 #=> Jul 01, 2016
 DateTime.parse(date).strftime("%a, %e %b %Y %H:%M:%S %z")
 #=> Fri,  1 Jul 2016 09:17:00 +0200
 DateTime.parse(date).strftime("%Y-%m-%dT%l:%M:%S%z")
 #=> 2016-07-01T 9:17:00+0200
 DateTime.parse(date).strftime("%I:%M:%S %p")
 #=> 09:17:00 AM
 DateTime.parse(date).strftime("%H:%M:%S")
 #=> 09:17:00
 DateTime.parse(date).strftime("%e %b %Y %H:%M:%S%p")
 #=>  1 Jul 2016 09:17:00AM
 DateTime.parse(date).strftime("%d.%m.%y")
 #=> 01.07.16
 DateTime.parse(date).strftime("%A, %d %b %Y %l:%M %p")
 #=> Friday, 01 Jul 2016  9:17 AM

Answer 5

这是另一个有用的代码：

"2011-05-19 10:30:14".to_datetime.strftime('%a %b %d %H:%M:%S %Z %Y')