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我有以下格式的字符串"2013-02-20T17:24:33Z"和"Mon Feb 25 02:42:27 +0000 2013".

有没有一种快速的方法可以将它们转换为日期时间格式，以便可以测试它们的相等性和/或排序。

clj-time 确实允许我使用这种格式 (date-time 1986 10 14 4 3 27 456)。然而，为了实现这一点，我将不得不解析上面的两个字符串。上述字符串是标准格式，有没有办法直接将它们转换为日期时间对象？

谢谢，穆尔塔萨

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2 回答 2

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clj-time 定义了标准格式化程序，请参阅clj-time.format/show-formatters，但就 clj-time 而言，您的第二种格式不是“标准”格式（尽管它看起来确实接近 rfc822）。您可以创建自定义格式化程序...

(use 'clj-time.format)

(parse (formatters :date-time-no-ms) "2013-02-20T17:24:33Z")
#<DateTime 2013-02-20T17:24:33.000Z>

(parse (formatter "E MMM dd hh:mm:ss Z YYYY") "Mon Feb 25 02:42:27 +0000 2013" )
#<DateTime 2013-02-25T02:42:27.000Z>

(show-formatters)
:basic-date                             20130228
:basic-date-time                        20130228T114047.213Z
:basic-date-time-no-ms                  20130228T114047Z
:basic-ordinal-date                     2013059
:basic-ordinal-date-time                2013059T114047.213Z
:basic-ordinal-date-time-no-ms          2013059T114047Z
:basic-t-time                           T114047.213Z
:basic-t-time-no-ms                     T114047Z
:basic-time                             114047.213Z
:basic-time-no-ms                       114047Z
:basic-week-date                        2013W094
:basic-week-date-time                   2013W094T114047.213Z
:basic-week-date-time-no-ms             2013W094T114047Z
:date                                   2013-02-28
:date-hour                              2013-02-28T11
:date-hour-minute                       2013-02-28T11:40
:date-hour-minute-second                2013-02-28T11:40:47
:date-hour-minute-second-fraction       2013-02-28T11:40:47.213
:date-hour-minute-second-ms             2013-02-28T11:40:47.213
:date-time                              2013-02-28T11:40:47.213Z
:date-time-no-ms                        2013-02-28T11:40:47Z
:hour                                   11
:hour-minute                            11:40
:hour-minute-second                     11:40:47
:hour-minute-second-fraction            11:40:47.213
:hour-minute-second-ms                  11:40:47.213
:ordinal-date                           2013-059
:ordinal-date-time                      2013-059T11:40:47.213Z
:ordinal-date-time-no-ms                2013-059T11:40:47Z
:rfc822                                 Thu, 28 Feb 2013 11:40:47 +0000
:t-time                                 T11:40:47.213Z
:t-time-no-ms                           T11:40:47Z
:time                                   11:40:47.213Z
:time-no-ms                             11:40:47Z
:week-date                              2013-W09-4
:week-date-time                         2013-W09-4T11:40:47.213Z
:week-date-time-no-ms                   2013-W09-4T11:40:47Z
:weekyear                               2013
:weekyear-week                          2013-W09
:weekyear-week-day                      2013-W09-4
:year                                   2013
:year-month                             2013-02
:year-month-day                         2013-02-28
于 2013-02-28T11:42:58.693 回答
3

使用SimpleDateFormat。

(let [input "Mon Feb 25 02:42:27 +0000 2013"
      fmt (java.text.SimpleDateFormat. "EEE MMM dd HH:mm:ss zzz yyyy")]
  (.parse fmt input))

请记住，解析月份和工作日的名称需要适当设置语言环境。

作为旁注，您问题中的第一个日期格式是ISO 8601。

于 2013-02-28T08:18:19.063 回答