[SOLVED] Get next day, skip weekends

Get next day, skip weekends

I want to generate next working day using JavaScript.

This is my code as of now

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);

Problem is, on Fridays it returns Saturday's date whereas I want it to be Monday

Solution

This will choose the next working day when a date is passed to it.

I suggest you normalise the date you pass, so you will not be surprised around summertime/wintertime change

Updated in 2023

// update and return passed date object to next working day
const getNextWork = date => {
  let day = date.getDay(), add = 1;
  if (day === 6)           add = 2; else 
  if (day === 5)           add = 3;
  date.setDate(date.getDate() + add); // will correctly handle 31+1 > 32 > 1st next month
  return date;
};


// tests:
const dt = new Intl.DateTimeFormat("en-US", {
  weekday: "short",
  year: "numeric",
  month: "long",
  day: "numeric",
  timeZone: "UTC",
  timeZoneName: "short",
  hour: "numeric",
  minute: "numeric",
});
const aDay = 24 * 60 * 60 * 1000;
// 26th of March 2023 is daylight savings date in my country
let date = new Date(2023, 2, 24, 15, 0, 0, 0).getTime();

for (let i = 0; i < 7; i++) {
  const d = new Date(date + i * aDay);
  console.log(dt.format(d), "-->", dt.format(getNextWork(d)));
}

If you do not want to update the passed date object have this instead

// find next working day and return a new date for that date
const getNewNextWork = date => {
  const dateCopy = new Date(date);
  let day = dateCopy.getDay(), add = 1;
  if (day === 6)           add = 2; else 
  if (day === 5)           add = 3;
  dateCopy.setDate(dateCopy.getDate() + add); // Modifying the copy, not the original date
  return dateCopy;
};

Older code:

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 noon
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);