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我正在为学校制作一个网站页面，一个人可以在其中输入两个日期：

  1. 一天开始为公司工作
  2. 公司停止工作的日子（合同结束时）

这些天是通过格式填写的MM-DD-YYYY

当一个人填写开始日期时，它将计算以下公式： “工作天数 = 今天的日期 - 该人开始工作的日期” 之后它将计算为天而不是毫秒（工作天数/1000/ 60/60/24）。

现在我必须摆脱一个人工作的每一周Saturday。Sunday

编辑：已修复，谢谢大家

Javascript代码

    function days_of_a_year(year) { return isLeapYear(year) ? 366 : 365; }
    function isLeapYear(year) { return year % 400 === 0 || (year % 100 !== 0 && year % 4 === 0); }
    var year    = moment().year();
    var days_year = days_of_a_year(moment().year());
    $(document).ready(function(){


    $(".form-control").keyup(function(){
        //get
        var leave_days = $('#leave_days').val(),
         leave_hours = $('#leave_hours').val(),
         hours_employee_week = $('#hours_employee_week').val(),
         hours_week = $('#hours_week').val(),
         date_employed = $('#date_employed').val(),
         date_unemployed = $('#date_unemployed').val(),

         start = new Date(date_employed),
         end   = new Date(date_unemployed),
         diff  = new Date(end - start),
         days  = Math.round(diff/1000/60/60/24),
            now = new Date(),
            days_worked = new Date(now - start),
         year = moment().year();


        var leave_hours_full = leave_days * leave_hours;
        var perc_employment =  hours_employee_week / hours_week * 100;
        var leave_hours_year = leave_hours_full * (hours_employee_week / hours_week);

        var days_worked_year = Math.round(days_worked/1000/60/60/24);

        console.log(parseInt(days_worked_year));


        $('#days_worked_year').val(days);
        $('#days_full_year').val(days_year);
        $('#perc_worked_year').val(days_worked_year);
4

2 回答 2

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我相信这个问题和你的类似。该问题中接受的答案提供了此功能：

// Expects start date to be before end date
// start and end are Date objects
function dateDifference(start, end) {

  // Copy date objects so don't modify originals
  var s = new Date(+start);
  var e = new Date(+end);

  // Set time to midday to avoid dalight saving and browser quirks
  s.setHours(12,0,0,0);
  e.setHours(12,0,0,0);

  // Get the difference in whole days
  var totalDays = Math.round((e - s) / 8.64e7);

  // Get the difference in whole weeks
  var wholeWeeks = totalDays / 7 | 0;

  // Estimate business days as number of whole weeks * 5
  var days = wholeWeeks * 5;

  // If not even number of weeks, calc remaining weekend days
  if (totalDays % 7) {
    s.setDate(s.getDate() + wholeWeeks * 7);

    while (s < e) {
      s.setDate(s.getDate() + 1);

      // If day isn't a Sunday or Saturday, add to business days
      if (s.getDay() != 0 && s.getDay() != 6) {
        ++days;
      }
    }
  }
  return days;
}
于 2019-09-02T12:17:32.807 回答
0

这是我的做法：

function daysWorked(firstDay, lastDay) {
  let daysWorked = 0;

  for (
    let cursor = new Date(+firstDay);
    cursor.getTime() <= lastDay.getTime();
    cursor.setDate(cursor.getDate() + 1)
  ) {
    let day = cursor.getDay();
    // skip Saturdays and Sundays
    if (day === 0 || day === 6) {
      continue;
    }
    daysWorked++;
  }
  return daysWorked;
}

我觉得这可能比以前的答案效率低一点，这取决于 JavaScript 处理日期的好坏。

让我尝试一个小基准...

2019-06-16 至 2019-09-02

  • 上一个答案的100 万次dateDifference()：3698 毫秒
  • 100 万次daysWorked()来自我的回答：47979 毫秒

慢了12倍...

2015-06-16 至 2019-09-02

  • 上一个答案的10k 次dateDifference()：95 毫秒
  • 我的回答是10k 次daysWorked()：9054 毫秒

约慢 100 倍...

好吧忘记我的答案xD

于 2019-09-02T12:46:07.680 回答