Question

我有这段代码可以正确找到格式hh：mm的总和我很难将其转换为格式hh：mm：ss 显然，所有内容都应正确舍入。

这是我要使用的数组格式，但是代码显然仅适用于[“ 00”，“ 10”]为例

const array = [
  ["00", "00", "09"],
  ["00", "00", "09"],
  ["00", "00", "04"],
  ["00", "00", "02"],
  ["00", "00", "21"],
  ["00", "00", "14"],
  ["00", "00", "26"],
  ["00", "00", "02"],
  ["00", "00", "14"],
  ["00", "00", "10"],
  ["00", "00", "48"],
  ["00", "00", "12"],
  ["00", "00", "09"],
  ["00", "00", "09"]
]

var mins = 0;
var hrs = 0;

if (array == null) {
  mins = Number(t1[1]) + Number(t2[1]);
  hrs = Math.floor(parseInt(mins / 60));
  hrs = Number(t1[0]) + Number(t2[0]) + hrs;
} else {
  $.each(array, function(key) {
    mins += Number(array[key][1]);
    hrs = Math.floor(parseInt(mins / 60));
    hrs += Number(array[key][0]) + hrs;
  });
}

console.log(mins, hrs)

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

在这种情况下，

t1和t2只是两个单独的值要加在一起，然后是一堆这样的数组。两者都应该正常工作。

也请帮助我在此处添加适当的秒数。

Answer 1

您可以花几秒钟的时间将所有值相加并构建一个新的总和字符串。

function sumTime(t1, t2, array = []) {
    var times = [3600, 60, 1],
        sum = [t1, t2, ...array]
            .map(s => s.split(':').reduce((s, v, i) => s + times[i] * v, 0))
            .reduce((a, b) => a + b, 0);

    return times
        .map(t => [Math.floor(sum / t), sum %= t][0])
        .map(v => v.toString().padStart(2, 0))
        .join(':');
}

console.log(sumTime("01:02:03", "04:57:57"));
console.log(sumTime("00:00:09", "00:00:09", ["00:00:04", "00:00:02", "00:00:21", "00:00:14", "00:00:26", "00:00:02", "00:00:14", "00:00:10", "00:00:48", "00:00:12", "00:00:09", "00:00:09"]));

ES5

function sumTime(t1, t2, array) {
    var times = [3600, 60, 1],
        sum = [t1, t2].filter(Boolean).concat(array || [])
            .map(function (s) {
                return s
                    .split(':') // omit this line for splitted values ['02', '04', '45']
                    .reduce(function (s, v, i) {
                        return s + times[i] * v;
                    }, 0);
            })
            .reduce(function (a, b) { return a + b; }, 0);

    return times
        .map(function (t) {
            var value = Math.floor(sum / t);
            sum %= t;
            return value;
        })
        .map(function (v) {
            return v.toString().padStart(2, 0);
         })
        .join(':');
}


console.log(sumTime("01:02:03", null));
console.log(sumTime(null, "04:57:57"));
console.log(sumTime(null, null, ["01:02:03", "04:57:57"]));
console.log(sumTime("00:00:09", "00:00:09", ["00:00:04", "00:00:02", "00:00:21", "00:00:14", "00:00:26", "00:00:02", "00:00:14", "00:00:10", "00:00:48", "00:00:12", "00:00:09", "00:00:09"]));

具有拆分的数组。

function sumTime(t1, t2, array) {
    var times = [3600, 60, 1],
        sum = [t1, t2].filter(Boolean).concat(array || [])
            .map(function (s) {
                return s
                    //.split(':') // omit this line for splitted values ['02', '04', '45']
                    .reduce(function (s, v, i) {
                        return s + times[i] * v;
                    }, 0);
            })
            .reduce(function (a, b) { return a + b; }, 0);

    return times
        .map(function (t) {
            var value = Math.floor(sum / t);
            sum %= t;
            return value;
        })
        .map(function (v) {
            return v.toString().padStart(2, 0);
         })
        .join(':');
}

console.log(sumTime(["01", "02", "03"], null));
console.log(sumTime(null, ["04", "57", "57"]));
console.log(sumTime(null, null, [["01", "02", "03"], ["04", "57", "57"]]));
console.log(sumTime(["00", "00", "09"], ["00", "00", "09"], [["00", "00", "04"], ["00", "00", "02"], ["00", "00", "21"], ["00", "00", "14"], ["00", "00", "26"], ["00", "00", "02"], ["00", "00", "14"], ["00", "00", "10"], ["00", "00", "48"], ["00", "00", "12"], ["00", "00", "09"], ["00", "00", "09"]]));

Answer 2

就像提到的评论之一一样，moment是一个流行的npm软件包，用于处理时间。您可以使用时刻的formatting options来解析数组数据结构中的时间：

const array = [
  [0, 0, 9],
  [0, 2, 3],
  [1, 2, 3],
   ...
];

const dates = array.map((date) => {
  return moment(`${date[0]}-${date[1]}-${date[2]}`, 'hh-mm-ss');
  // make sure you account for 12 or 24 hour clocks, not sure which you are using
});

const sum = dates.reduce((date, total) => { return total.add(date); }, moment.duration(0));

Answer 3

这几乎是相同的数学，只是又花了几秒钟时间->分钟。

seconds = t1[2] + t2[2];
mins = parseInt(seconds / 60);
seconds = seconds % 60;
mins += t1[1] + t2[1];
hrs = parseInt(minutes / 60);
mins = mins % 60;
hrs += t1[0] + t2[0];

我假设您已经知道如何通过简单地使用hh:mm:ss和split将Number转换为适当的数组，但是以防万一：

t1 = time1.split(':').map(part => Number(part));
t2 = time2.split(':').map(part => Number(part));

Answer 4

您可以使用 reducer 函数汇总相应的数组元素，然后修改结果。

var timeArrays = [
                   ['00', '52', '00'], ['00', '08', '00'], ['00', '00', '40'],
                   ['00', '00', '01'], ['00', '00', '20']
                 ];

function totalTimeAsString(timeArray) {
    var result = ['00', '00', '00'];

    var reducer = function (acc, current) {
        return [
                 Number(acc[0]) + Number(current[0]),
                 Number(acc[1]) + Number(current[1]),
                 Number(acc[2]) + Number(current[2])
               ];
    };

    var [hh, mm, ss] = timeArray.reduce(reducer, result);

    // seconds
    if (ss > 59) {
        mm += Math.floor(ss / 60);
    }
    ss = ss % 60;
    result[2] = ((ss < 10) ? '0' : '') + ss;

    // minutes
    if (mm > 59) {
        hh += Math.floor(mm / 60);
    }
    mm = mm % 60;
    result[1] = ((mm < 10) ? '0' : '') + mm;

    // hours
    result[0] = ((hh < 10) ? '0' : '') + hh;

    return result.join(':');
}

console.log(totalTimeAsString(timeArrays));
// "01:01:01"

以时间格式hh：mm：ss

4 个答案: