Question

我想要一个像这样工作的方法/闭包

 println TimeDifference.format( System.currentMilliSeconds()-1000*60*60*24*7*6 )

并打印

 1 month, 3 weeks, 2 days, 45 hours and 3 minutes ago.

这是否有现成的解决方案？

Answer 1

这是一个快速端口。你可以试试：

def distance_of_time_in_words = { fromTime, toTime, includeSeconds, options ->
    if(toTime==null) toTime = 0
    if(includeSeconds==null) includeSeconds = false
    if(options==null) options = []

    def distance_in_mins = (int)Math.round(Math.abs(toTime - fromTime)/60.0)
    def distance_in_seconds = (int)Math.round(Math.abs(toTime - fromTime))

    switch (distance_in_mins) {
        case 0..1:
            if(distance_in_mins == 0) {
                return "less than 1 minute"
            } else {
                if (includeSeconds == false) return "${distance_in_mins} minute"
            }

            switch (distance_in_seconds) {
                case 0..4: return "less than 5 seconds"
                case 5..9: return "less than 10 seconds"
                case 10..19: return "less than 20 seconds"
                case 20..39: return "half a minute"
                case 40..59: return "less than 1 minute"
                default: return "1 minute"
            }

        case 2..44: return "${distance_in_mins} minutes"
        case 45..89: return "about an hour"
        case 90..1439: return "about ${Math.round(distance_in_mins / 60.0)} hours"
        case 1440..2879: return "a day"
        case 2880..43199: return "${Math.round(distance_in_mins / 1440)} days"
        case 43200..86399: return "a month"
        case 86400..525599: return "${Math.round(distance_in_mins / 43200)} months"
        case 525600..1051199: return "a year"
        default: return "over ${Math.round(distance_in_mins / 525600)} years"
    }

}

def d = new Date().time
println distance_of_time_in_words(d, d + 10, true, [])
println distance_of_time_in_words(d, d + 60, true, [])
println distance_of_time_in_words(d, d + 35, true, [])
println distance_of_time_in_words(d, d + 123, true, [])
println distance_of_time_in_words(d, d + 400, true, [])
println distance_of_time_in_words(d, d + 1300*60, true, [])
println distance_of_time_in_words(d, d + 2000*60, true, [])
println distance_of_time_in_words(d, d + 3.5*60*60*24, true, [])
println distance_of_time_in_words(d, d + 32*60*60*24, true, [])
println distance_of_time_in_words(d, d + 65*60*60*24, true, [])
println distance_of_time_in_words(d, d + 1.5*12*30*60*60*24, true, [])
println distance_of_time_in_words(d, d + 7*12*30*60*60*24, true, [])

Answer 2

查看this library，它称为人类时间，它允许解析和渲染人类更容易消化的时间。

Answer 3

我不知道Groovy，但是移植the one from rails非常简单。您可以通过单击“show source”来查看Ruby源代码。

Answer 4

您可以使用此PrettyDate插件 - http://www.grails.org/plugin/pretty-date

或此PrettyTime插件 - http://www.grails.org/plugin/pretty-time

Answer 5

您还可以使用timeago结帐，这是在javascript端完成的。编写一个taglib非常容易，您可以使用它来在标记中使用适当的类创建日期。

Answer 6

我使用joda-time插件，Period和PeriodFormatter，例如here。

如何使用Groovy显示一段时间？

6 个答案: