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如何创建一个函数来判断当前时间是否在指定时间范围内

来源:stackoverflow

时间:2024-02-19 12:12:25 282浏览 收藏

Golang不知道大家是否熟悉?今天我将给大家介绍《如何创建一个函数来判断当前时间是否在指定时间范围内》,这篇文章主要会讲到等等知识点,如果你在看完本篇文章后,有更好的建议或者发现哪里有问题,希望大家都能积极评论指出,谢谢!希望我们能一起加油进步!

问题内容

我正在尝试找到一种方法来检查当前时间是否在时间窗口内。

输入是:

upgradeday []string - 一段日期(例如 ["sunday", "tuesday"]

upgradetime string - 小时:分钟(例如 "22:04"

upgradeduration int64 - 时间量,来自时间窗口有效的 upgradetime。最长可达 12 小时。

完整示例:

upgradeday = ["sunday", tuesday"] , upgradetime = "10:00", upgradeduration = 2 -> 时间窗口为每周日和周二,从 10:00 到 12:00 点。 我尝试编写以下函数,但它在天/月/年之间的转换中不起作用:

func isInsideTimeWindow(upgradeDay []string, upgradeTime string, upgradeDuration int64) bool {
    now := time.Now()

    ut := strings.Split(upgradeTime, ":")
    hour, _ := strconv.Atoi(ut[0])
    min, _ := strconv.Atoi(ut[1])

    // !! not working when now it's Monday 00:01 and got: upgradeDay = ["Sunday"], upgradeTime = 23:59, upgradeDuration = 2
    twStart := time.Date(now.Year(), now.Month(), now.Day(), hour, min, 0, 0, now.Location())
    twEnd := twStart.Add(time.Hour * time.Duration(upgradeDuration))


    if !(now.After(twStart) && now.Before(twEnd)) {
        return false
    }
    
    wd := now.Weekday().String()
    
    for i := range upgradeDay {
      if upgradeDay[i] == wd  {
         return true
      }
    }
    
    return false
}

有人知道如何在 go 中解决这个问题吗?


解决方案


时代有很多复杂性。例如:

  • 如果升级日是“søndag”(丹麦语)而不是“星期日”怎么办?
  • 我们应该使用当地时间还是 utc 时间?如果是本地的,谁的位置重要?如果服务器在伦敦而我在旧金山,我们使用服务器的时间还是我的时间?
  • 如果升级间隔包括凌晨 2 点,那么是否也计算太平洋夏令时上午 2 点和太平洋标准时间上午 2 点?我住的地方这些时间相差一小时。如果该时间间隔从凌晨 2 点开始,到 2:59:59 结束,则在许多夏令时转变为一小时的地区,一年中的某一天不存在该时间。

如果您忽略所有这些复杂性 - 国际化 (i18n)、本地化 (l10n)、dst 等等 - 某人可以设置日期和时间或升级这一事实仍然存在一些问题其本身可能需要一些时间,但通常我们也会忽略这些。

请注意,go 的 time.now() 返回本地时间,但是,谁的位置?由于我们尚未回答使用谁的时区问题,因此我们可能希望避免担心这一点。考虑到其余的输入约束,让我们编写一个函数来确定提供的时间是否满足输入约束,而不是 time.now() 是否满足输入约束。然后,调用者可以提供 utc 时间用户所在位置的挂钟时间:

somenow = time.time()
localnow = somenow.in(location) // from time.loadlocation() or similar

我们还有一些与您的类型似乎不一致的东西:

upgradeduration int64 - 从升级时间算起的时间量,在该时间窗口有效。最长可达 12 小时

0 到 12 之间的小时值很容易适合普通 int。这是否已经是一个以纳秒表示的 time.duration 值?如果是这样,为什么是 int64 而不是 time.duration?或者它是以秒为单位的值,因此可以在 0 到 43200 之间?如果是这样,它仍然适合 int

我做了很多假设并得出以下结论,您可以在 Go Playground 上尝试一下。

package main

import (
    "fmt"
    "strconv"
    "strings"
    "time"
)

// startok determines whether the given starting-time is within a
// time window.
//
// the window starts at a time given as two integers,
// h and m, representing hours and minutes, and extends for
// the given duration d in hours, which in general should not
// extend into another day.  if it does extend past the end of
// the day into the next day, we ignore the extension.
//
// the days on which the time *is* in that window are further
// limited by the days[] slice of weekday values.
//
// note: it would probably be sensible to return a time.duration
// value that is how long it will be until the next ok time, but
// we leave that as an exercise.
//
// it would also be sensible to allow the duration d to extend
// into the next day, which is also left as an exercise.
func startok(when time.time, days []time.weekday, h, m, d int) bool {
    // find ok-to-start time, and end-time.  if end exceeds
    // 24*60, we ignore the extra end time, rather than
    // allowing some minutes into the next day.
    start := h*60 + m
    end := start + d*60

    // convert when to hour-and-minute and see if we are
    // in the allowed range.
    wh, wm, _ := when.clock()
    now := wh*60 + wm
    if now < start || now >= end {
        // not in hh:mm through hh+d:mm; say no.
        return false
    }

    // the time-of-day is ok; check the day-of-week.
    // we could do this earlier but by positioning it
    // here, we leave room to check to see if it's
    // the *next* day, if needed.
    if !func(wd time.weekday) bool {
        for _, allowed := range days {
            if wd == allowed {
                return true
            }
        }
        return false
    }(when.weekday()) {
        return false // when.weekday() not in days[]
    }

    // time is ok, day is ok
    return true
}

// startokstr is like startok but the window starts at a time
// given as a string encoded as hh:mm, with the days being a
// slice of strings instead of weekday.  because of these strings,
// parsing can produce an error, so this function has an error
// return.
func startokstr(when time.time, days []string, hhmm string, d int) (bool, error) {
    parts := strings.split(hhmm, ":")
    // optional: be strict about two-digit values
    if len(parts) != 2 {
        return false, fmt.errorf("invalid time string %q", hhmm)
    }
    h, err := strconv.atoi(parts[0])
    if err != nil {
        return false, err
    }
    if h < 0 || h >= 60 {
        return false, fmt.errorf("invalid hour value %s", parts[0])
    }
    m, err := strconv.atoi(parts[1])
    if err != nil {
        return false, err
    }
    if m < 0 || m >= 60 {
        return false, fmt.errorf("invalid minute value %s", parts[1])
    }
    var wd []time.weekday
    for _, s := range days {
        w, err := parseweekday(s)
        if err != nil {
            return false, err
        }
        wd = append(wd, w)
    }
    ok := startok(when, wd, h, m, d)
    return ok, nil
}

// parseweekday handles weekday strings.
//
// ideally we'd use time.parse for this, as it already has
// these in it, but they are not exported in usable form.
func parseweekday(s string) (time.weekday, error) {
    strtoweekday := map[string]time.weekday{
        "sunday":    time.sunday,
        "monday":    time.monday,
        "tuesday":   time.tuesday,
        "wednesday": time.wednesday,
        "thursday":  time.thursday,
        "friday":    time.friday,
        "saturday":  time.saturday,
    }
    if v, ok := strtoweekday[s]; ok {
        return v, nil
    }
    return time.sunday, fmt.errorf("invalid day-of-week %q", s)
}

// tests should be converted to real tests and put in
// a separate file.
func tests() {
    okdays := []string{"sunday", "wednesday"}
    okstart := "04:00"
    okduration := 2 // hours

    tfmt := "mon jan 2 15:04:05 2006"
    t1 := "sat sep 5 04:30:00 2020" // time ok, day not
    t2 := "sun sep 6 04:30:00 2020" // time ok, day ok
    check := func(s string, expect bool) {
        when, err := time.parse(tfmt, s)
        if err != nil {
            panic(err)
        }
        result, err := startokstr(when, okdays, okstart, okduration)
        if err != nil {
            panic(err)
        }
        if result != expect {
            fmt.printf("fail: expected %v for %q\n", expect, s)
        }
    }
    check(t1, false)
    check(t2, true)
    fmt.println("2 tests run")
}

func main() {
    tests()
}

这是解决该问题的一种方法:

package main
import "time"
type window struct { time.Time }

func (w window) isDay(s string) bool {
   return w.Weekday().String() == s
}

func (w window) isHourRange(begin, end int) bool {
   return w.Hour() >= begin && w.Hour() <= end
}

func main() {
   w := window{
      time.Now(),
   }
   {
      b := w.isDay("Friday")
      println(b)
   }
   {
      b := w.isHourRange(20, 23)
      println(b)
   }
}

这假设只有一天有效,因此您需要修改它来处理 多日。不过,这应该可以帮助您入门。

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