以下是计算年龄的简单方法:

//dob date dd/mm/yy 
var d = 01/01/1990


//today
//date today string format 
var today = new Date(); // i.e wed 04 may 2016 15:12:09 GMT
//todays year
var todayYear = today.getFullYear();
// today month
var todayMonth = today.getMonth();
//today date
var todayDate = today.getDate();

//dob
//dob parsed as date format   
var dob = new Date(d);
// dob year
var dobYear = dob.getFullYear();
// dob month
var dobMonth = dob.getMonth();
//dob date
var dobDate = dob.getDate();

var yearsDiff = todayYear - dobYear ;
var age;

if ( todayMonth < dobMonth ) 
 { 
  age = yearsDiff - 1; 
 }
else if ( todayMonth > dobMonth ) 
 {
  age = yearsDiff ; 
 }

else //if today month = dob month
 { if ( todayDate < dobDate ) 
  {
   age = yearsDiff - 1;
  }
    else 
    {
     age = yearsDiff;
    }
 }

为了测试生日是否已经过去，我定义了一个帮助函数Date.prototype。getDoY，它有效地返回一年中的天数。剩下的就不言自明了。

Date.prototype.getDoY = function() {
    var onejan = new Date(this.getFullYear(), 0, 1);
    return Math.floor(((this - onejan) / 86400000) + 1);
};

function getAge(birthDate) {
    function isLeap(year) {
        return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
    }

    var now = new Date(),
        age = now.getFullYear() - birthDate.getFullYear(),
        doyNow = now.getDoY(),
        doyBirth = birthDate.getDoY();

    // normalize day-of-year in leap years
    if (isLeap(now.getFullYear()) && doyNow > 58 && doyBirth > 59)
        doyNow--;

    if (isLeap(birthDate.getFullYear()) && doyNow > 58 && doyBirth > 59)
        doyBirth--;

    if (doyNow <= doyBirth)
        age--;  // birthday not yet passed this year, so -1

    return age;
};

var myBirth = new Date(2001, 6, 4);
console.log(getAge(myBirth));

不久前，我做了一个这样的函数:

function getAge(birthDate) {
  var now = new Date();

  function isLeap(year) {
    return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
  }

  // days since the birthdate    
  var days = Math.floor((now.getTime() - birthDate.getTime())/1000/60/60/24);
  var age = 0;
  // iterate the years
  for (var y = birthDate.getFullYear(); y <= now.getFullYear(); y++){
    var daysInYear = isLeap(y) ? 366 : 365;
    if (days >= daysInYear){
      days -= daysInYear;
      age++;
      // increment the age only if there are available enough days for the year.
    }
  }
  return age;
}

它接受一个Date对象作为输入，所以你需要解析'YYYYMMDD'格式的日期字符串:

var birthDateStr = '19840831',
    parts = birthDateStr.match(/(\d{4})(\d{2})(\d{2})/),
    dateObj = new Date(parts[1], parts[2]-1, parts[3]); // months 0-based!

getAge(dateObj); // 26

我相信在这种情况下，有时可读性更重要。除非我们要验证1000个字段，否则这应该足够准确和快速:

函数is18orOlder(dateString) { const dob = new Date(dateString); const dobPlus18 =新的日期(dob.getFullYear() + 18, dob.getMonth()， dob.getDate()); 返回dobPlus18 .valueOf() <= Date.now(); ｝ / /测试: console.log (is18orOlder (' 01/01/1910 '));/ /正确的 console.log (is18orOlder (' 01/01/2050 '));/ /错误 //当我在2020年10月2日发布这篇文章时，所以: console.log (is18orOlder (' 10/08/2002 '));/ /正确的 console.log(is18orOlder('10/19/2002')) // false

我喜欢这种方法，而不是用一个常数来表示一年有多少毫秒，然后再弄乱闰年等等。让内置的Date来做这个工作。

更新，发布这个片段，因为有人可能会发现它有用。因为我在输入字段上强制一个掩码，有mm/dd/yyyy的格式，并且已经验证了日期是否有效，在我的情况下，这也适用于验证18+年:

 function is18orOlder(dateString) {
   const [month, date, year] = value.split('/');
   return new Date(+year + 13, +month, +date).valueOf() <= Date.now();
}

这对我来说是最优雅的方式:

const getAge = (birthDateString) => { const today = new Date(); const birthDate = new Date(birthDateString); const yearsDifference = today.getFullYear() - birthDate.getFullYear(); 如果( 今天。getmonth () < birthDate.getMonth() || . getmonth ( (today.getMonth() === birthDate.getMonth() && today.getDate() < birthDate.getDate())) ) { 返回yearsDifference - 1; ｝ 返回yearsDifference; }; console.log (getAge (' 2018-03-12 '));

我认为可以简单地像这样:

function age(dateString){
    let birth = new Date(dateString);
    let now = new Date();
    let beforeBirth = ((() => {birth.setDate(now.getDate());birth.setMonth(now.getMonth()); return birth.getTime()})() < birth.getTime()) ? 0 : 1;
    return now.getFullYear() - birth.getFullYear() - beforeBirth;
}

age('09/20/1981');
//35

也适用于时间戳

age(403501000000)
//34