如果给出格式为YYYYMMDD的出生日期,如何以年计算年龄?是否可以使用Date()函数?

我正在寻找一个比我现在使用的更好的解决方案:

Var dob = '19800810'; var年=数字(dob.)substr (0, 4)); var月=数字(dob.)Substr (4, 2)) - 1; var day =数字(dob.)2) substr(6日); var today = new Date(); var age = today.getFullYear() -年份; if (today.getMonth() < month || (today.getMonth() == month && today.getDate() < day)) { 年龄——; } 警报(年龄);


当前回答

试试这个。

function getAge(dateString) {
    var today = new Date();
    var birthDate = new Date(dateString);
    var age = today.getFullYear() - birthDate.getFullYear();
    var m = today.getMonth() - birthDate.getMonth();
    if (m < 0 || (m === 0 && today.getDate() < birthDate.getDate())) {
        age--;
    }
    return age;
}

我相信你的代码中唯一看起来粗糙的是substr部分。

小提琴:http://jsfiddle.net/codeandcloud/n33RJ/

其他回答

你可以在你的表格中使用这个年龄限制

function dobvalidator(birthDateString){
    strs = birthDateString.split("-");
    var dd = strs[0];
    var mm = strs[1];
    var yy = strs[2];

    var d = new Date();
    var ds = d.getDate();
    var ms = d.getMonth();
    var ys = d.getFullYear();
    var accepted_age = 18;

    var days = ((accepted_age * 12) * 30) + (ms * 30) + ds;
    var age = (((ys - yy) * 12) * 30) + ((12 - mm) * 30) + parseInt(30 - dd);

    if((days - age) <= '0'){
        console.log((days - age));
        alert('You are at-least ' + accepted_age);
    }else{
        console.log((days - age));
        alert('You are not at-least ' + accepted_age);
    }
}

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

试试这个:

$('#Datepicker').change(function(){

var $bef = $('#Datepicker').val();
var $today = new Date();
var $before = new Date($bef);
var $befores = $before.getFullYear();
var $todays = $today.getFullYear();
var $bmonth = $before.getMonth();
var $tmonth = $today.getMonth();
var $bday = $before.getDate();
var $tday = $today.getDate();

if ($bmonth>$tmonth)
{$('#age').val($todays-$befores);}

if ($bmonth==$tmonth)
{   
if ($tday > $bday) {$('#age').val($todays-$befores-1);}
else if ($tday <= $bday) {$('#age').val($todays-$befores);}
}
else if ($bmonth<$tmonth)
{ $('#age').val($todays-$befores-1);} 
})
$("#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);

    }
}
});

还有另一个解决方案:

/**
 * Calculate age by birth date.
 *
 * @param int birthYear Year as YYYY.
 * @param int birthMonth Month as number from 1 to 12.
 * @param int birthDay Day as number from 1 to 31.
 * @return int
 */
function getAge(birthYear, birthMonth, birthDay) {
  var today = new Date();
  var birthDate = new Date(birthYear, birthMonth-1, birthDay);
  var age = today.getFullYear() - birthDate.getFullYear();
  var m = today.getMonth() - birthDate.getMonth();
  if (m < 0 || (m === 0 && today.getDate() < birthDate.getDate())) {
    age--;
  }
  return age;
}