CODEWITHSUNDEEP
javascripthtml
MacGyver_97
MacGyver_97

Reputation: 229

How to find every second Tuesday, starting from December 1st?

I've been able to show/hide a button if the time is within the range, however, how could I modify this to also check if the date is every second Tuesday, starting on December 1st?

<p id="newButton">LIVE</p>

window.addEventListener("load", function(){
var newButton = document.getElementById("newButton");
const start = 12 * 60 + 30;
const end =  13 * 60 + 30;
const date = new Date(); 
const now = date.getHours() * 60 + date.getMinutes();

if(start <= now && now <= end) {
 newButton.style.display = "block";
 alert("in time");
}
else {
 newButton.style.display = "none";
 alert("offline");
}
}, false);

Upvotes: 0

Views: 193

Answers (1)

Scott Sauyet
Scott Sauyet

Reputation: 50807

Here's what I suggested in the comments:

const msPerDay = 24 * 60 * 60 * 1000;

window.addEventListener("load", function(){
  var newButton = document.getElementById("newButton");
  const start = 12 * 60 + 30;
  const end =  13 * 60 + 30;
  const date = new Date(); 
  const now = date.getHours() * 60 + date.getMinutes();

  if(start <= now && now <= end 
     && Math.round((date - 1606798800000) / msPerDay) % 14 === 0) {
    newButton.style.display = "block";
    alert("in time");
  }
  else {
    newButton.style.display = "none";
    alert("offline");
  }
}, false);

Note the interesting ambiguity in the question. "Check if the date is every second Tuesday, starting on December 1st?" can be read -- as intended -- to represent alternating Tuesdays. But my initial misreading, "the second Tuesday of any month" is perfectly understandable, if, I suppose, slightly less likely given that the noted start date is itself a Tuesday.

1606798800000 is simply the result of new Date('2020-12-01').getTime(). It's probably a silly optimization not to simply include something like that in the code. So an alternative version might be

  if (start <= now && now <= end &&  
      Math.round((date - new Date(2020, 11, 1).getTime()) / msPerDay) % 14 === 0)

Upvotes: 1

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