给定多个开始和结束日期计算唯一时间

Question

如果您有一组具有开始日期和结束日期的约会，您如何计算所有约会的唯一时间？

示例：

var appointments = {
    0:{"start":"2015-01-20 09:00:00","end":"2015-01-20 09:30:00"},
    1:{"start":"2015-01-20 09:15:00","end":"2015-01-20 09:42:22"},
    2:{"start":"2015-01-20 10:00:00","end":"2015-01-20 10:25:00"},
    3:{"start":"2015-01-20 10:10:00","end":"2015-01-20 10:53:00"}
}

所以在这个例子中我想得到一个唯一的时间（activity）值 1H 35M 22S。有人知道这方面的任何公式吗？

到目前为止我已经有了这个，似乎可以工作，但我认为日期必须按开始时间排序。这是最有效的计算方法吗？:

var totalElapsedAppointmentSeconds = 0;
var lastActiveTimestamp;

for (i in appointments) {
    if (totalElapsedAppointmentSeconds == 0) {
        totalElapsedAppointmentSeconds = new Date(appointments[i].end) - new Date(appointments[i].start);
        lastActiveTimestamp = new Date(appointments[i].end);
    } else {
        if (new Date(appointments[i].start) < lastActiveTimestamp) {
            if (new Date(appointments[i].end) > lastActiveTimestamp) {
                totalElapsedAppointmentSeconds += new Date(appointments[i].end) - lastActiveTimestamp;
                lastActiveTimestamp = new Date(appointments[i].end);
            } else {
                //nothing, already completely accounted for
            }
        } else {
            totalElapsedAppointmentSeconds += new Date(appointments[i].end) - new Date(appointments[i].start);
            lastActiveTimestamp = new Date(appointments[i].end);
        }
    }
}

totalElapsedAppointmentSeconds = totalElapsedAppointmentSeconds/1000;
var totalElapsedTime = Math.floor(totalElapsedAppointmentSeconds / 3600) + "H " + Math.floor((totalElapsedAppointmentSeconds % 3600)/60) + "M " + (totalElapsedAppointmentSeconds % 3600) % 60 + "S";
console.log("totalElapsedTime",totalElapsedTime);

Answer 1

不清楚你在问什么，但这证明了计算时差

EDIT 糟糕 javascript 说这些是无效日期，它们是从哪里来的？

moment.js 如果您必须将它们用作输入

，那么解析它们是一个不错的选择

var data = {
  "appointments": {
    0:{"start":"2015-01-20 09:00:00","end":"2015-01-20 09:30:00"},
    1:{"start":"20-01-2015 09:15:00","end":"20-01-2015 09:42:22"},
    2:{"start":"20-01-2015 10:00:00","end":"20-01-2015 10:25:00"},
    3:{"start":"20-01-2015 10:10:00","end":"20-01-2015 10:53:00"},
  }
}

function secondsDifference(ts1, ts2){
  startMs = new Date(ts1).valueOf();
  endMs = new Date(ts2).valueOf();
  deltaMs = endMs - startMs;
  deltaS = deltaMs /1000;
  deltaS = Math.floor(deltaS);
  return deltaS;
}

var a = data.appointments[0];
var result = secondsDifference(a.start, a.end);
console.log('first appointment length seconds:', result)

给定多个开始和结束日期计算唯一时间

Calculate unique time given multiple start and end dates

javascript

php

math

datetime

formula