获取 date-fns 中两个日期之间的持续时间（以天：小时：分钟：秒格式）

Question

我正在尝试使用 date-fns 库显示倒计时器并按以下方式执行操作，但无法在反应中找到解决方案。

预期输出：60 天：8 小时：9 分钟：剩余 5 秒

    const finishTime = new Date("01/01/2020");
    const currentTime = new Date();

将结果添加到数组以便稍后遍历：

    results.push(differenceInMonths(finishTime, currentTime));
    results.push(differenceInDays(finishTime, currentTime));
    results.push(differenceInHours(finishTime, currentTime));
    results.push(differenceInMinutes(finishTime, currentTime));
    results.push(differenceInSeconds(finishTime, currentTime));

添加手动逻辑以从秒获取时间。显然应该有一个更好的逻辑，使用/不使用该库，这是我所缺少的：

    const monthsRemaining = results[4] / (30 * 24 * 3600); // this will anyways fail as 30 days is not common for every month
    const daysRemaining = (monthsRemaining % 1) * 30;
    const hoursRemaining = (daysRemaining % 1) * 24;
    const minutesRemaining = (hoursRemaining % 1) * 60;
    const secondsRemaining = (minutesRemaining % 1) * 60;

return (
        <div>
            {Math.round(monthsRemaining)} Months : {Math.round(daysRemaining)}{" "}
            days : {Math.round(hoursRemaining)} hours :{" "}
            {Math.round(minutesRemaining)} minutes :{" "}
            {Math.round(secondsRemaining)} seconds
        </div>
    );

任何建议或指向正确方法的指针，因为我没有看到这样的直接实现，我只能看到只有一个单位的 formatDistance 方法

Answer 1

IntervalToDuration 是您正在寻找的。

// Get the duration between January 15, 1929 and April 4, 1968.
intervalToDuration({
  start: new Date(1929, 0, 15, 12, 0, 0),
   end: new Date(1968, 3, 4, 19, 5, 0)
})
// => { years: 39, months: 2, days: 20, hours: 7, minutes: 5, seconds: 0 }

https://date-fns.org/v2.16.1/docs/intervalToDuration