如果您有一系列带有开始日期和结束日期的约会,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"}
}
因此,在这个示例中,我希望获得 1H 35M 22S 的唯一时间(事件)值。 有人知道这个的公式吗?
到目前为止,我已经做到了这一点,似乎有效,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);
最佳答案
不清楚你在问什么,但这演示了计算时间差
编辑哎呀javascript说这些是无效日期,它们来自哪里?
moment.js如果您必须使用这些作为输入,那么这是解析它们的一个不错的选择
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)
关于javascript - 计算给定多个开始和结束日期的唯一时间,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31870344/