js datepicker-允许非连续日pcm

Question

I would like to limit users to selecting only the first and third Monday of each month. We have a volunteer intake only on these days, so I want to limit incorrect date selections as much as possible.

I'm not a js coder, but have managed to adapt some code I found online to allow the first or third Monday of each month, but I can't work out how to allow both of them.

Here's the code I have for the first Monday:

var firstMonday = new Date(date);
var mondays=0;
firstMonday.setDate(1);

while (mondays < 1) {
    firstMonday.setDate(firstMonday.getDate() + 1);
    if (firstMonday.getDay() == 1) {
        mondays++;
    }
}
var result = date.getDate() != firstMonday.getDate();

Answer 1

I think this is what you are asking. Credit to jabclab for the getMondays() function.

 // test: first monday of this month // result: true //var dates = [new Date(2017,8,4)]; // test: third monday of this month // result: true //var dates = [new Date(2017,8,18)]; // test: first and third monday of this month // result: true var dates = [new Date(2017,8,4), new Date(2017,8,18)]; // test: first monday, third monday, and random day from this month // result: false //var dates = [new Date(2017,8,4), new Date(2017,8,18), new Date(2017,8,22)]; alert(validate(dates)); function validate(dates) { var valid = true; var mondays = getMondays(); var firstMonday = mondays[0].setHours(0,0,0,0); var thirdMonday = mondays[2].setHours(0,0,0,0); if (dates && dates.length > 0) { for (var i = 0; i < dates.length; i++) { // Zero out time so only year, month, and day is compared var d = dates[i].setHours(0,0,0,0); if (d != firstMonday && d != thirdMonday) { return false; } } } else { valid = false; } return valid; } function getMondays() { var d = new Date(), month = d.getMonth(), mondays = []; d.setDate(1); // Get the first Monday in the month while (d.getDay() !== 1) { d.setDate(d.getDate() + 1); } // Get all the other Mondays in the month while (d.getMonth() === month) { mondays.push(new Date(d.getTime())); d.setDate(d.getDate() + 7); } return mondays; } 

Answer 2

Thanks, but I'm not sure if the above works or not as I was looking for a js code answer - I'll leave that for someone else to work out.

...which I've found in the meantime. Many thanks to Hugh at Fabrik for the following:

var thisdate = new Date(date);
thisdate.setHours(0,0,0,0);

var day = 1; // monday
var nth = 1; // first

var first = new Date(thisdate.getFullYear(), thisdate.getMonth(), 1),
  add = (day - first.getDay() + 7) % 7 + (nth - 1) * 7;
first.setDate(1 + add);

nth = 3; // third

var third = new Date(thisdate.getFullYear(), thisdate.getMonth(), 1),
  add = (day - third.getDay() + 7) % 7 + (nth - 1) * 7;
third.setDate(1 + add);

//console.log(thisdate + ', ' + first + ', ' + third);

var result = (first.getTime() !== thisdate.getTime()) && (third.getTime() !== thisdate.getTime());