从一系列繁忙的时间范围中获取可用的时间范围 [英] Get available time ranges from an array of busy time ranges

问题描述

假设您有一个会议的 BUSY 时间范围数组

Lets say you have an array of BUSY time ranges for a meeting

[{'start':'9:00 AM', 'end':'10:00 AM'},
{'start':'12:00 PM', 'end':'2:00 PM'},
{'start':'5:00 AM', 'end':'7:00 PM'}]

我希望在24小时内返回一组 AVAILABLE 次，与上述时间相反.喜欢...

I would like to get in return an array of AVAILABLE times in a 24 hour frame, that are the opposite of the above times. Like...

[{'start':'00:00 AM', 'end':'9:00 AM'},
 {'start':'10:00 AM', 'end':'12:00 PM'},
 {'start':'2:00 PM', 'end':'5:00 PM'},
 {'start':'7:00 PM', 'end':'11:59 PM'}]

我尝试过使用moment.js以及 https://www.npmjs.com /package/moment-range ，特别是.subtract()方法.

I have tried using moment.js as well as https://www.npmjs.com/package/moment-range, specifically the .subtract() method.

我知道类似的stackoverflow问题，但是无法找到适用于此格式的问题(使用javascript)，momentJS和优雅的ES6数组方法解决方案.

I am aware of similar stackoverflow questions but couldn't find ones that were applicable to this format, in javascript, with momentJS, and elegant ES6 array method solution.

推荐答案

function giveUtc(start) {
  var t = moment().format("YYYY-MM-DD")
  var t1 = t + " " + start
  return moment(t1, "YYYY-MM-DD h:mm A").format()

}


const timeRange = [{
    'start': '9:00 AM',
    'end': '10:00 AM'
},
{
    'start': '12:00 PM',
    'end': '2:00 PM'
},
{
    'start': '5:00 PM',
    'end': '7:00 PM'
},
{
    "start": "11:00 AM",
    "end": "3:00 PM",
},
{
    "start": "6:00 PM",
    "end": "9:00 PM",
}]


timeRange.sort((a, b) => {
  var utcA = giveUtc(a.start)
  var utcB = giveUtc(b.start)
  if (utcA < utcB) {
return -1

  }
  if (utcA > utcB) {
return 1


  }
  return 0
})
const availableTimeArray = []

let endTimeFarthest = moment(giveUtc("0.00 AM"))
let startTimeMinimum = moment(giveUtc("12.59 PM"))
timeRange.forEach((element, index) => {
  let currentEndTime = moment(giveUtc(element.end))
  const currentStartTime = moment(giveUtc(element.start))
  if (currentStartTime.isBefore(startTimeMinimum)) {
startTimeMinimum = currentStartTime
  }
  if (currentEndTime.isAfter(endTimeFarthest)) {
endTimeFarthest = currentEndTime
  }
  /* console.log(startTimeMinimum.format("h:mm A"), endTimeFarthest.format("h:mm A")) */
  if (index === timeRange.length - 1) {
if (timeRange.length === 1) {
  availableTimeArray.push({
    start: "00:00 AM",
    end: currentStartTime.format("h:mm A")
  })
}
availableTimeArray.push({
  //start: currentEndTime.format("h:mm A"),
  start: endTimeFarthest.format("h:mm A"),
  end: "11.59 PM"
})

  } else {
const nextBusyTime = timeRange[index + 1]
const nextStartTime = moment(giveUtc(nextBusyTime.start))
if (index === 0) {
  availableTimeArray.push({
    start: "00:00 AM",
    end: currentStartTime.format("h:mm A")
  })
}
let endTimeToCompare = currentEndTime.isBefore(endTimeFarthest) ?
  endTimeFarthest :
  currentEndTime
if (endTimeToCompare.isBefore(nextStartTime)) {
  availableTimeArray.push({
    start: endTimeToCompare.format("h:mm A"),
    end: nextStartTime.format("h:mm A")
  })
}

  }

})
console.log(availableTimeArray)

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.5.1/moment.min.js"></script>

我已使用utc时间戳比较时间，并假设所有间隔都属于一天.某些情况可能会丢失，但是您可以采用这个想法.我已经利用了贪婪算法.首先根据开始时间对所有间隔进行排序.然后遍历排序后的数组以选择正确的时间间隔

I have used utc timestamp to compare between timings and assumed that all the interval belongs to a single day. Some edge case might be missing but you can take the idea. I have made use of greedy algorithm. First sorting all the interval based on the start time. Then iterating through the sorted array to pick correct interval

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