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();
}

还有两个选择:

// Int Age to Date as string YYY-mm-dd
function age_to_date(age)
{
    try {
        var d = new Date();
        var new_d = '';
        d.setFullYear(d.getFullYear() - Math.abs(age));
        new_d = d.getFullYear() + '-' + d.getMonth() + '-' + d.getDate();

        return new_d;
    } catch(err) {
        console.log(err.message);
    }
}
// Date string (YYY-mm-dd) to Int age (years old)
function date_to_age(date)
{
    try {
        var today = new Date();
        var d = new Date(date);

        var year = today.getFullYear() - d.getFullYear();
        var month = today.getMonth() - d.getMonth();
        var day = today.getDate() - d.getDate();
        var carry = 0;

        if (year < 0)
            return 0;
        if (month <= 0 && day <= 0)
            carry -= 1;

        var age = parseInt(year);
        age += carry;

        return Math.abs(age);
    } catch(err) {
        console.log(err.message);
    }
}

看这个例子，你从这里得到完整的年、月、日信息

function getAge(dateString) {
    var today = new Date();
    var birthDate = new Date(dateString);
    var age = today.getFullYear() - birthDate.getFullYear();
    var m = today.getMonth() - birthDate.getMonth();
    var da = today.getDate() - birthDate.getDate();
    if (m < 0 || (m === 0 && today.getDate() < birthDate.getDate())) {
        age--;
    }
    if(m<0){
        m +=12;
    }
    if(da<0){
        da +=30;
    }
    return age+" years "+ Math.abs(m) + "months"+ Math.abs(da) + " days";
}
alert('age: ' + getAge("1987/08/31"));    
[http://jsfiddle.net/tapos00/2g70ue5y/][1]

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

moment.js还有一个可能的解决方案:

var moment = require('moment');
var startDate = new Date();
var endDate = new Date();
endDate.setDate(endDate.getFullYear() + 5); // Add 5 years to second date
console.log(moment.duration(endDate - startDate).years()); // This should returns 5

$("#birthday").change(function (){


var val=this.value;

var current_year=new Date().getFullYear();
if(val!=null){
    var Split = val.split("-");
    var birth_year=parseInt(Split[2]);

    if(parseInt(current_year)-parseInt(birth_year)<parseInt(18)){

  $("#maritial_status").attr('disabled', 'disabled');
        var val2= document.getElementById("maritial_status");
        val2.value = "Not Married";
        $("#anniversary").attr('disabled', 'disabled');
        var val1= document.getElementById("anniversary");
        val1.value = "NA";

    }else{
        $("#maritial_status").attr('disabled', false);
        $("#anniversary").attr('disabled', false);

    }
}
});