我在数组[0]中有以下对象:
pv
end_datatime格式为yyyymmddMhhmmss(前4位数字为年份,后跟月份,天数,分隔符“M”,小时,分钟和秒。
我想计算所有活动的总持续时间? (总共11小时,即2小时的睡眠+ 4小时的早餐+ 1小时的课程+ 4小时的滑雪)
答案 0 :(得分:1)
您需要一个累加器来跟踪到目前为止的时间。我假设数字指的是分钟或其他东西。如果它实际上是时间(如在1700 = 5PM),那么你显然需要做更多的逻辑(因为事件可以从1159开始并在0100结束,在这种情况下,startTime - endTime将不起作用) ,但是从你的尝试来看,你似乎只是在寻找两个数字之间的差异。这是我得到的:
arr = [
{
"startTime": "1300",
"endTime": "1700",
"eventName": "Tea"
},
{
"startTime": "0800",
"endTime": "1200",
"eventName": "Breakfast"
},
{
"startTime": "1300",
"endTime": "1400",
"eventName": "Lesson"
},
{
"startTime": "1300",
"endTime": "1700",
"eventName": "Ski"
}
]
arr.reduce((acc, val) => {
return acc + (parseInt(val.endTime) - parseInt(val.startTime))
}, 0)
// returns 1300
答案 1 :(得分:1)
这是一种可能的实施方式,考虑到跨越多天的时间:
const input = [{
"startTime": "1300",
"endTime": "1700",
"eventName": "Sleep",
"end_datetime": "20180510M0100",
"start_datetime": "20180509M2300",
},
{
"startTime": "0800",
"endTime": "1200",
"eventName": "Breakfast",
"end_datetime": "20180507M1200",
"start_datetime": "20180507M0800",
},
{
"startTime": "1300",
"endTime": "1400",
"eventName": "Lesson",
"end_datetime": "20180507M1400",
"start_datetime": "20180507M1300",
},
{
"startTime": "1300",
"endTime": "1700",
"eventName": "Ski",
"end_datetime": "20180511M170000",
"start_datetime": "20180511M130000",
}
];
function getDiff(datestr1, datestr2) {
const m1 = moment(datestr1, 'YYYYMMDD*hhmm');
const m2 = moment(datestr2, 'YYYYMMDD*hhmm');
const minuteDifference = m2.diff(m1, 'minutes');
return minuteDifference;
}
const totalMinutes = input.reduce(
(accum, {
end_datetime,
start_datetime
}) => accum + getDiff(start_datetime, end_datetime),
0
);
const totalHours = totalMinutes / 60;
console.log(totalMinutes + ' minutes = ' + totalHours + ' hours');

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