从 'current time' 计算具有最近几天、几周和几个月的时间戳的对象（在数组中）

Question

我有一个对象数组，时间戳为 属性:

示例输入：

const data = [
  {
    _id: "602102db3acc4515d4b2f687",
    createdDt: "2021-02-08T09:22:35.000Z",
  },
  {
    _id: "6021024da706a260d8932da2",
    createdDt: "2021-02-08T09:20:13.000Z",
  },
  // ...
  // ...
  {
    _id: "6020fd863acc4515d4b2f685",
    createdDt: "2021-02-08T08:59:50.000Z",
  },
];

现在从当前日期开始，我需要 counts 个时间戳为今天[=的对象总数42=]、前一天等（如果前一天没有进入，计数可以为零）

同理，我需要计数这一天，前一天等等，周和月也是如此。

例如，我期望的输出可能是这样的：

示例输出：

const result = {
  days: [0, 0, 5, 10, ...],
  weeks: [15, 5, 8, 0, ...],
  months: [30, 42, 33, 23, ...]
}

我正在使用 ES6、Lodash 和 moment.js。这是用于基本图形表示。

---

更新：

这是我写的代码，有人可以提出更简单的解决方案吗？

我目前的解决方案：

for (var i = 0, k = 1; i < 365; i++, k++) {
  let dt = moment().subtract(i, "days");
  let td = moment().subtract(k, "days");
  builddays.push(0);
  for (var j = 0; j < drivers.length; j++) {
    let ddt = new Date(drivers[j].createdDt);
    if (moment(ddt).isBetween(td, dt)) builddays[i] = drivers[j].count;
  }
}

var weeksbifurcate = builddays.reduce((resultArray, item, index) => {
  const chunkIndex = Math.floor(index / 7);

  if (!resultArray[chunkIndex]) {
    resultArray[chunkIndex] = [];
  }

  resultArray[chunkIndex].push(item);

  return resultArray;
}, []);

var monthsbifurcate = builddays.reduce((resultArray, item, index) => {
  const chunkIndex = Math.floor(index / 30);

  if (!resultArray[chunkIndex]) {
    resultArray[chunkIndex] = [];
  }

  resultArray[chunkIndex].push(item);

  return resultArray;
}, []);

for (i = 0; i < 12; i++) {
  days.push(builddays[i]);
  weeks.push(weeksbifurcate[i].reduce(getSum, 0));
  months.push(monthsbifurcate[i].reduce(getSum, 0));
}

Answer 1

使用moment # diff查找索引，即dayIndex、weekIndex、monthIndex并在相应的[=16=中放入或增加右边的index ] 在 result 对象中：

const solve = (data) => {
  const result = {
    days: [],
    weeks: [],
    months: [],
  };
  data.forEach((item) => {
    const today = moment();
    const createdDt = moment(item.createdDt);

    const dayIndex = today.diff(createdDt, "days");
    const weekIndex = today.diff(createdDt, "weeks");
    const monthIndex = today.diff(createdDt, "months");

    result.days[dayIndex] = (result.days[dayIndex] || 0) + 1;
    result.weeks[weekIndex] = (result.weeks[weekIndex] || 0) + 1;
    result.months[monthIndex] = (result.months[monthIndex] || 0) + 1;
  });
  return result;
};

const data = [
  {createdDt: "2021-03-03T00:00:00.000Z",},
  {createdDt: "2021-03-03T00:00:00.000Z",},
  {createdDt: "2021-03-01T00:00:00.000Z",},
  {createdDt: "2021-02-28T00:00:00.000Z",},
  {createdDt: "2021-02-27T00:00:00.000Z",},
  {createdDt: "2021-02-27T00:00:00.000Z",},
  {createdDt: "2021-02-27T00:00:00.000Z",},
  {createdDt: "2021-02-24T00:00:00.000Z",},
  {createdDt: "2021-02-23T00:00:00.000Z",},
];

console.log(solve(data));

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.29.1/moment.min.js"></script>

结果中的undefined值可能很少，但是没有必要设置为0，因为无论你去哪里都可以认为是0show/use它。