在JavaScript中从工作日获取下一个日期

Question

如何返回给定工作日的下一个日期（可以是数字0-6或名称周日到周六）。

例如，如果今天，在周五16月 - 2009年我传入：

• 周五，它将返回今天的日期16月 - 2009年
• 周六返回17-Oct-2009
• 星期四返回22-Oct-2009

Answer 1

只要加上7并不能解决问题。

下面的函数会给你一周中的第二天。

function nextDay(x){ 
    var now = new Date();  
    now.setDate(now.getDate() + (x+(7-now.getDay())) % 7); 
    return now; 
} 

Answer 2

要在user 190106的answer扩大，此代码应该给你你想要的东西：

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) { 
    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)); 

Answer 3

瑞士刀工具javascript DateTime programming。

http://www.datejs.com/

Answer 4

这里有一个稍作修改的版本蒂姆回答解决具体question--传递日期d，和，和上周的希望日期（DOW 0-6），返回日期

function nextDay(d, dow){ 
    d.setDate(d.getDate() + (dow+(7-d.getDay())) % 7); 
    return d; 
} 

Answer 5

这里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());

Answer 6

如果你不想做通号码，但平日-名（周日 - 周六）找到某个工作日的将来的日期，那么这可以帮助您，以及：

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/>");