PRDV402: Introduction to JavaScript II

const today = new Date();
const birthday = new Date("December 17, 1995 03:24:00"); // DISCOURAGED: may not work in all runtimes
const birthday2 = new Date("1995-12-17T03:24:00"); // This is standardized and will work reliably
const birthday3 = new Date(1995, 11, 17); // the month is 0-indexed
const birthday4 = new Date(1995, 11, 17, 3, 24, 0);
const birthday5 = new Date(628021800000); // passing epoch timestamp

const date = new Date("2020-05-12T23:50:21.817Z");
date.toString(); // Tue May 12 2020 18:50:21 GMT-0500 (Central Daylight Time)
date.toDateString(); // Tue May 12 2020
date.toTimeString(); // 18:50:21 GMT-0500 (Central Daylight Time)
date[Symbol.toPrimitive]("string"); // Tue May 12 2020 18:50:21 GMT-0500 (Central Daylight Time)

date.toISOString(); // 2020-05-12T23:50:21.817Z
date.toJSON(); // 2020-05-12T23:50:21.817Z

date.toUTCString(); // Tue, 12 May 2020 23:50:21 GMT

date.toLocaleString(); // 5/12/2020, 6:50:21 PM
date.toLocaleDateString(); // 5/12/2020
date.toLocaleTimeString(); // 6:50:21 PM

const date = new Date("2000-01-17T16:45:30");
const [month, day, year] = [
  date.getMonth(),
  date.getDate(),
  date.getFullYear(),
];
// [0, 17, 2000] as month are 0-indexed
const [hour, minutes, seconds] = [
  date.getHours(),
  date.getMinutes(),
  date.getSeconds(),
];
// [16, 45, 30]

new Date() exhibits legacy undesirable, inconsistent behavior with two-digit year values; specifically, when a new Date() call is given a two-digit year value, that year value does not get treated as a literal year and used as-is but instead gets interpreted as a relative offset - in some cases as an offset from the year 1900, but in other cases, as an offset from the year 2000.

let date = new Date(98, 1); // Sun Feb 01 1998 00:00:00 GMT+0000 (GMT)
date = new Date(22, 1); // Wed Feb 01 1922 00:00:00 GMT+0000 (GMT)
date = new Date("2/1/22"); // Tue Feb 01 2022 00:00:00 GMT+0000 (GMT)

// Legacy method; always interprets two-digit year values as relative to 1900
date.setYear(98);
date.toString(); // Sun Feb 01 1998 00:00:00 GMT+0000 (GMT)
date.setYear(22);
date.toString(); // Wed Feb 01 1922 00:00:00 GMT+0000 (GMT)

So, to create and get dates between the years 0 and 99, instead use the preferred setFullYear() and getFullYear() methods:.

// Preferred method; never interprets any value as being a relative offset,
// but instead uses the year value as-is
date.setFullYear(98);
date.getFullYear(); // 98 (not 1998)
date.setFullYear(22);
date.getFullYear(); // 22 (not 1922, not 2022)

Calculating elapsed time

The following examples show how to determine the elapsed time between two JavaScript dates in milliseconds.

Due to the differing lengths of days (due to daylight saving changeover), months, and years, expressing elapsed time in units greater than hours, minutes, and seconds requires addressing a number of issues, and should be thoroughly researched before being attempted.

// Using Date objects
const start = Date.now();

// The event to time goes here:
doSomethingForALongTime();
const end = Date.now();
const elapsed = end - start; // elapsed time in milliseconds

// Using built-in methods
const start = new Date();

// The event to time goes here:
doSomethingForALongTime();
const end = new Date();
const elapsed = end.getTime() - start.getTime(); // elapsed time in milliseconds

// To test a function and get back its return
function printElapsedTime(testFn) {
  const startTime = Date.now();
  const result = testFn();
  const endTime = Date.now();

  console.log(`Elapsed time: ${String(endTime - startTime)} milliseconds`);
  return result;
}

const yourFunctionReturn = printElapsedTime(yourFunction);

Note: In browsers that support the Web Performance API's high-resolution time feature, Performance.now() can provide more reliable and precise measurements of elapsed time than Date.now().

Get the number of seconds since the ECMAScript Epoch

const seconds = Math.floor(Date.now() / 1000);

In this case, it's important to return only an integer-so a simple division won't do. It's also important to only return actually elapsed seconds. (That's why this code uses Math.floor(), and not Math.round()).