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

从naveen和原始OP的帖子中采用,我最终得到了一个可重用的方法存根,它接受字符串和/或JS Date对象。

我将其命名为gregorianAge(),因为这个计算准确地给出了我们如何使用公历表示年龄。即,如果月和日在出生年份的月和日之前,则不计算结束年。

/** * Calculates human age in years given a birth day. Optionally ageAtDate * can be provided to calculate age at a specific date * * @param string|Date Object birthDate * @param string|Date Object ageAtDate optional * @returns integer Age between birthday and a given date or today */ function gregorianAge(birthDate, ageAtDate) { // convert birthDate to date object if already not if (Object.prototype.toString.call(birthDate) !== '[object Date]') birthDate = new Date(birthDate); // use today's date if ageAtDate is not provided if (typeof ageAtDate == "undefined") ageAtDate = new Date(); // convert ageAtDate to date object if already not else if (Object.prototype.toString.call(ageAtDate) !== '[object Date]') ageAtDate = new Date(ageAtDate); // if conversion to date object fails return null if (ageAtDate == null || birthDate == null) return null; var _m = ageAtDate.getMonth() - birthDate.getMonth(); // answer: ageAt year minus birth year less one (1) if month and day of // ageAt year is before month and day of birth year return (ageAtDate.getFullYear()) - birthDate.getFullYear() - ((_m < 0 || (_m === 0 && ageAtDate.getDate() < birthDate.getDate())) ? 1 : 0) } // Below is for the attached snippet function showAge() { $('#age').text(gregorianAge($('#dob').val())) } $(function() { $(".datepicker").datepicker(); showAge(); }); <link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css"> <script src="//code.jquery.com/jquery-1.10.2.js"></script> <script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script> DOB: <input name="dob" value="12/31/1970" id="dob" class="datepicker" onChange="showAge()" /> AGE: <span id="age"><span>

其他回答

我有点晚了,但我发现这是计算出生日期的最简单的方法。

希望这能有所帮助。

function init() { writeYears("myage", 0, Age()); } function Age() { var birthday = new Date(1997, 02, 01), //Year, month-1 , day. today = new Date(), one_year = 1000 * 60 * 60 * 24 * 365; return Math.floor((today.getTime() - birthday.getTime()) / one_year); } function writeYears(id, current, maximum) { document.getElementById(id).innerHTML = current; if (current < maximum) { setTimeout(function() { writeYears(id, ++current, maximum); }, Math.sin(current / maximum) * 200); } } init() <span id="myage"></span>

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。

这是我修改的尝试(用一个字符串传递给函数而不是一个日期对象):

function calculateAge(dobString) {
    var dob = new Date(dobString);
    var currentDate = new Date();
    var currentYear = currentDate.getFullYear();
    var birthdayThisYear = new Date(currentYear, dob.getMonth(), dob.getDate());
    var age = currentYear - dob.getFullYear();

    if(birthdayThisYear > currentDate) {
        age--;
    }

    return age;
}

和用法:

console.log(calculateAge('1980-01-01'));

这是我的解决方案,只需要传入一个可解析的日期:

function getAge(birth) {
  ageMS = Date.parse(Date()) - Date.parse(birth);
  age = new Date();
  age.setTime(ageMS);
  ageYear = age.getFullYear() - 1970;

  return ageYear;

  // ageMonth = age.getMonth(); // Accurate calculation of the month part of the age
  // ageDay = age.getDate();    // Approximate calculation of the day part of the age
}

我对之前的答案做了一些更新。

var calculateAge = function(dob) {
    var days = function(date) {
            return 31*date.getMonth() + date.getDate();
        },
        d = new Date(dob*1000),
        now = new Date();

    return now.getFullYear() - d.getFullYear() - ( measureDays(now) < measureDays(d));
}

我希望这对你有所帮助