Question

如果您有一系列约会的开始和结束日期，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，我希望获得1H 35M 22S的唯一时间（活动）值。有人知道任何公式吗？

到目前为止，我有这个，似乎工作，但我认为日期必须按开始时间排序。这是计算这个的最有效方法吗？：

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

编辑哎呀javascript说这些是无效日期，它们来自哪里？

如果必须将这些作为输入

，

moment.js是一个很好的解析它们的选项

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

1 个答案: