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

function change(){
    setTimeout(function(){
        var dateObj  =      new Date();
                    var month    =      dateObj.getUTCMonth() + 1; //months from 1-12
                    var day      =      dateObj.getUTCDate();
                    var year     =      dateObj.getUTCFullYear();  
                    var newdate  =      year + "/" + month + "/" + day;
                    var entered_birthdate        =   document.getElementById('birth_dates').value;
                    var birthdate                =   new Date(entered_birthdate);
                    var birth_year               =   birthdate.getUTCFullYear();
                    var birth_month              =   birthdate.getUTCMonth() + 1;
                    var birth_date               =   birthdate.getUTCDate();
                    var age_year                =    (year-birth_year);
                    var age_month               =    (month-birth_month);
                    var age_date                =    ((day-birth_date) < 0)?(31+(day-birth_date)):(day-birth_date);
                    var test                    =    (birth_year>year)?true:((age_year===0)?((month<birth_month)?true:((month===birth_month)?(day < birth_date):false)):false) ;
                   if (test === true || (document.getElementById("birth_dates").value=== "")){
                        document.getElementById("ages").innerHTML = "";
                    }                    else{
                        var age                     =    (age_year > 1)?age_year:(   ((age_year=== 1 )&&(age_month >= 0))?age_year:((age_month < 0)?(age_month+12):((age_month > 1)?age_month:      (  ((age_month===1) && (day>birth_date) ) ? age_month:age_date)          )    )); 
                        var ages                    =    ((age===age_date)&&(age!==age_month)&&(age!==age_year))?(age_date+"days"):((((age===age_month+12)||(age===age_month)&&(age!==age_year))?(age+"months"):age_year+"years"));
                        document.getElementById("ages").innerHTML = ages;
                  }
                }, 30);

};

其他回答

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));

这是我的修改:

  function calculate_age(date) {
     var today = new Date();
     var today_month = today.getMonth() + 1; //STRANGE NUMBERING //January is 0!
     var age = today.getYear() - date.getYear();

     if ((today_month > date.getMonth() || ((today_month == date.getMonth()) && (today.getDate() < date.getDate())))) {
       age--;
     }

    return 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

使用ES6的干净的一行程序解决方案:

const getAge = birthDate => Math.floor((new Date() - new Date(birthDate).getTime()) / 3.15576e+10)

// today is 2018-06-13
getAge('1994-06-14') // 23
getAge('1994-06-13') // 24

我使用的是365.25天的一年(因为闰年是0.25天),分别是3.15576e+10毫秒(365.25 * 24 * 60 * 60 * 1000)。

它有几个小时的剩余时间,所以根据用例,它可能不是最佳选择。

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 '));