35

如何返回给定工作日的下一个日期（它可以是数字 0-6 或名称星期日至星期六）。

例如，如果今天，在2009 年 10 月 16 日星期五， 我通过了：

  • Friday，它将返回今天的日期2009 年 10 月 16 日
  • 星期六返回2009 年 10 月 17 日
  • 周四返回2009 年 10 月 22 日
4

6 回答 6

43

仅添加 7 并不能解决问题。

以下功能将为您提供一周的第二天。

function nextDay(x){
    var now = new Date();    
    now.setDate(now.getDate() + (x+(7-now.getDay())) % 7);
    return now;
}
于 2009-10-16T16:40:34.103 回答
20

这是蒂姆回答特定问题的稍微修改的版本——传入日期 d，以及所需的星期几（0-6 日），返回日期

function nextDay(d, dow){
    d.setDate(d.getDate() + (dow+(7-d.getDay())) % 7);
    return d;
}
于 2014-12-06T21:13:33.797 回答
9

这是另一个简单的解决方案

//takes dayIndex from sunday(0) to saturday(6)
function nextDate(dayIndex) {
    var today = new Date();
    today.setDate(today.getDate() + (dayIndex - 1 - today.getDay() + 7) % 7 + 1);
    return today;
}
document.write("Next Sunday is: "+nextDate(0).toLocaleString()+"<br/>");
document.write("Next Thursday is: "+nextDate(4).toLocaleString()+"<br/>");
document.write("Next Saturday is: "+nextDate(6).toLocaleString());

于 2017-04-26T03:44:11.343 回答
2

为了扩展用户 190106的答案，这段代码应该给你你想要的：

function getNextDay(day, resetTime){
  var days = {
    sunday: 0, monday: 1, tuesday: 2,
    wednesday: 3, thursday: 4, friday: 5, saturday: 6
  };

  var dayIndex = days[day.toLowerCase()];
  if (dayIndex !== undefined) {
    throw new Error('"' + day + '" is not a valid input.');
  }

  var returnDate = new Date();
  var returnDay = returnDate.getDay();
  if (dayIndex !== returnDay) {
    returnDate.setDate(returnDate.getDate() + (dayIndex + (7 - returnDay)) % 7);
  }

  if (resetTime) {
    returnDate.setHours(0);
    returnDate.setMinutes(0);
    returnDate.setSeconds(0);
    returnDate.setMilliseconds(0);
  }
  return returnDate;
}

alert(getNextDay('thursday', true));
于 2009-10-19T09:57:45.153 回答
1

如果您不想通过数字而是通过工作日名称（星期日 - 星期六）来查找某个工作日的未来日期，那么这也可以帮助您：

function getDateOfWeekday(refday){
    var days = {
        monday: 1,
        tuesday: 2,
        wednesday: 3,
        thursday: 4,
        friday: 5,
        saturday: 6,
        sunday: 0
    };
    if(!days.hasOwnProperty(refday))throw new Error(refday+" is not listed in "+JSON.stringify(days));
    var currDate = new Date();
    var currTimestamp = currDate.getTime();
    var triggerDay = days[refday];
    var dayMillDiff=0;
    var dayInMill = 1000*60*60*24;
    // add a day to dayMillDiff as long as the desired refday (sunday for instance) is not reached
    while(currDate.getDay()!=triggerDay){
        dayMillDiff += dayInMill;
        currDate = new Date(currDate.getTime()+dayInMill);
    }
    return new Date(currTimestamp + dayMillDiff);
}

var sunday = getDateOfWeekday("sunday");
document.write("Next Sunday is at: <strong>"+sunday.toLocaleString()+"</strong><br/>");

var thursday = getDateOfWeekday("thursday");
thursday.setHours(0,0,0,0); // set hours/minutes/seconds and millseconds to zero
document.write("Next Thursday is at: <strong>"+thursday.toLocaleString()+"</strong> on midnight<br/>");

var tuesday = getDateOfWeekday("tuesday");
document.write("Next Tuesday is at: <strong>"+tuesday.toLocaleString()+"</strong><br/>");

于 2017-12-19T14:36:22.717 回答
0

用于javascript DateTime 编程的瑞士刀工具。

http://www.datejs.com/

于 2009-10-19T10:00:44.377 回答