Question

我想使用JavaScript生成下一个工作日。

这是我现在的代码

var today = new Date();
    today.setDate(today.getDate());
    var tdd = today.getDate();
    var tmm = today.getMonth()+1;
    var tyyyy = today.getYear();

    var date = new Date();

    date.setDate(date.getDate()+3);

问题是，星期五它会返回星期六的日期，而我希望它是星期一

Answer 1

这将选择下一个工作日。

＆＃13;

var today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
console.log("today, Monday",today,"day #"+today.getDay());
var next = new Date(today.getTime());
next.setDate(next.getDate()+1); // tomorrow
while (next.getDay() == 6 || next.getDay() == 0) next.setDate(next.getDate() + 1);
console.log("no change    ",next,"day #"+next.getDay());
console.log("-------");
// or without a loop:

function getNextWork(d) {
  d.setDate(d.getDate()+1); // tomorrow
  if (d.getDay()==0) d.setDate(d.getDate()+1);
  else if (d.getDay()==6) d.setDate(d.getDate()+2);
  return d;
}
next = getNextWork(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 29,12,0,0,0); // Monday at noone
next = getNextWork(today); // Still Monday at noon
console.log("today, Monday",today);
console.log("no change    ",next);
console.log("-------");

// Implementing Rob's comment

function getNextWork1(d) {
  var day = d.getDay(),add=1;
  if (day===5) add=3;
  else if (day===6) add=2;
  d.setDate(d.getDate()+add);  
  return d;
}
today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
next = getNextWork1(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 26,12,0,0,0,0); // Monday at noon
next = getNextWork1(today); // Monday
console.log("today, Monday",today);
console.log("no change    ",next);

＆＃13;

Answer 2

您可以按时加1天，直到您到达周六或周日的那一天：

＆＃13;

function getNextBusinessDay(date) {
  // Copy date so don't affect original
  date = new Date(+date);
  // Add days until get not Sat or Sun
  do {
    date.setDate(date.getDate() + 1);
  } while (!(date.getDay() % 6))
  return date;
}

// today,    Friday 26 Aug 2016
[new Date(), new Date(2016,7,26)].forEach(function(d) {
  console.log(d.toLocaleString() + ' : ' + getNextBusinessDay(d).toLocaleString());
});

＆＃13;

Answer 3

检查出来：https://jsfiddle.net/e9a4066r/

function get_next_weekday (date) {
    var tomorrow = new Date(date.setDate(date.getDate() + 1))
    return tomorrow.getDay() % 6
        ? tomorrow
        : get_next_weekday(tomorrow)
}

Answer 4

接受的答案将一次跳过一天，这会回答OP的问题，但是对于那些希望在周末跳过时添加可变天数的人，下面的功能可能会有所帮助：

function addWorkDays(date, days) {   
  while (days > 0) {
    date.setDate(date.getDate() + 1);
    if (date.getDay() != 0 && date.getDay() != 6) {
      days -= 1;
    }
  }          
  return date;
}

第二天，跳过周末

4 个答案: