如何在javascript中获取两个日期之间的周六和周日的日期

how can i get dates on saturday and sunday between two dates in javascript

本文关键字：日期 之间 两个 周六 周日 javascript 获取      更新时间：2023-09-26

我对javascript之类的浏览器端脚本非常不熟悉，因为我是一名php开发人员。我需要知道周六和周日的日期。我已经找到了很多计算计数的答案，但还没有找到周六和周日的日期。我试过这些：

Date.prototype.endOfWeek = function(){
  return new Date( 
      this.getFullYear(), 
      this.getMonth(), 
      this.getDate() + 6 - this.getDay() 
  );
};
var now = new Date();
// returns next saturday; and returns saturday if it is saturday today.
alert(now.endOfWeek() );

它只在下个星期六给我回电话。请帮我拿这个

我也试过这个，但它会给我返回计数

function calcBusinessDays(dDate1, dDate2) { // input given as Date objects
        var iWeeks, iDateDiff, iAdjust = 0;
        if (dDate2 < dDate1) return -1; // error code if dates transposed
        var iWeekday1 = dDate1.getDay(); // day of week
        var iWeekday2 = dDate2.getDay();
        iWeekday1 = (iWeekday1 == 0) ? 7 : iWeekday1; // change Sunday from 0 to 7
        iWeekday2 = (iWeekday2 == 0) ? 7 : iWeekday2;
        if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1; // adjustment if both days on weekend
        iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays
        iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;
        // calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000)
        iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime()) / 604800000)
        if (iWeekday1 <= iWeekday2) {
          iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1)
                    alert(iDateDiff);
        } else {
          iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2)
        }
        iDateDiff -= iAdjust // take into account both days on weekend
        return (iDateDiff + 1); // add 1 because dates are inclusive
    }

alert(calcBusinessDays(new Date("August 11, 2010 11:13:00"),new Date("August 16, 2010 11:13:00")));

我怎样才能在星期六和星期日两次约会之间得到约会javascript

像这样的东西会在两天之间的所有周六和周日返回

function calcBusinessDays(dDate1, dDate2) {
    if (dDate1 > dDate2) return false;
    var date  = dDate1;
    var dates = [];
    while (date < dDate2) {
        if (date.getDay() === 0 || date.getDay() === 6) dates.push(new Date(date));
        date.setDate( date.getDate() + 1 );
    }
    return dates;
}


var d1 = new Date(2015, 3, 3);
var d2 = new Date(2015, 5, 3);
function calcBusinessDays(dDate1, dDate2) {
    if (dDate1 > dDate2) return false;
    var date  = dDate1;
    var dates = [];
    while (date < dDate2) {
        if (date.getDay() === 0 || date.getDay() === 6) dates.push(new Date(date));
        date.setDate( date.getDate() + 1 );
    }
    
    return dates;
}
document.body.innerHTML = '<pre>' + JSON.stringify(calcBusinessDays(d1,d2), null, 4) + '</pre>';

相关文章：