blocks|key|410240|text|我发现，将两个日期(提供的日期和当前日期)的月份都重置为1月份是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|410241|>+moment("02/26/1978",+"MM/DD/YYYY").month(0).from(moment().month(0))
"34+years+ago"|code-block|syntax|javascript|410242|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I found that it would work to reset the month to January for both dates (the provided date and the present):

<pre><code>&gt; moment("02/26/1978", "MM/DD/YYYY").month(0).from(moment().month(0))
"34 years ago"
</code></pre>

blocks|key|410328|text|使用moment.js非常简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|410329|var+years+=+moment().diff('1981-01-01',+'years');
var+days+=+moment().diff('1981-01-01',+'days');|code-block|syntax|javascript|410330|有关其他参考，请阅读moment.js+official+documentation。|offset|length|410331|entityMap|0|LINK|mutability|MUTABLE|url|https://momentjs.com/docs/#/displaying/difference/^0|0|0|K|M|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Using moment.js is as easy as:

<pre><code>var years = moment().diff('1981-01-01', 'years');
var days = moment().diff('1981-01-01', 'days');
</code></pre>

For additional reference, you can read moment.js <a href="https://momentjs.com/docs/#/displaying/difference/" rel="noreferrer">official documentation</a>.

blocks|key|192299|text|http://jsfiddle.net/xR8t5/27/|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|192300|如果您不需要分数值：|192301|var+years+=+moment().diff('1981-01-01',+'years',false);
alert(+years);|code-block|syntax|javascript|192302|如果您想要分数值：|192303|var+years+=+moment().diff('1981-01-01',+'years',true);
alert(+years);|192304|单位可以是秒、分钟、小时、天、周、月、年|192305|entityMap|0|LINK|mutability|MUTABLE|url^0|0|T|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|6|7|11|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|14|8|@]|9|@]|C|$I|J]]|$1|O|3|P|5|6|7|15|8|@]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|4]]]]

<a href="http://jsfiddle.net/xR8t5/27/" rel="noreferrer">http://jsfiddle.net/xR8t5/27/</a>

if you do not want fraction values:

<pre><code>var years = moment().diff('1981-01-01', 'years',false);
alert( years);
</code></pre>

if you want fraction values:

<pre><code>var years = moment().diff('1981-01-01', 'years',true);
alert( years);
</code></pre>

Units can be [seconds, minutes, hours, days, weeks, months, years]

blocks|key|410525|text|这种方法简单且功能强大。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|410526|Value是日期，"DD-MM-YYYY“是日期的掩码。|410527|moment().diff(moment(value,+"DD-MM-YYYY"),+'years');|code-block|syntax|javascript|410528|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This method is easy and powerful.

Value is a date and "DD-MM-YYYY" is the mask of the date.

<pre><code>moment().diff(moment(value, "DD-MM-YYYY"), 'years');
</code></pre>

blocks|key|192208|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|192209|+moment("02/26/1978",+"MM/DD/YYYY").fromNow().split("+")[0];|code-block|syntax|javascript|192210|解释：|192211|我们收到类似这样的字符串：'23天前‘。将其拆分成数组：'23'，'days'，‘array’，然后取第一项'23‘。|192212|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try this: 

<pre><code> moment("02/26/1978", "MM/DD/YYYY").fromNow().split(" ")[0];
</code></pre>

Explanation:

We receive string something like this: '23 days ago'. Split it to array: ['23', 'days', 'ago'] and then take first item '23'.

blocks|key|410282|text|这个方法对我很有效。它检查这个人今年是否过了生日，否则减去一年。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|410283|//+date+is+the+moment+you're+calculating+the+age+of
var+now+=+moment().unix();
var+then+=+date.unix();
var+diff+=+(now+-+then)+/+(60+*+60+*+24+*+365);
var+years+=+Math.floor(diff);|code-block|syntax|javascript|410284|编辑:第一个版本不是很完美。更新后的版本应该|410285|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This method works for me. It's checking if the person has had their birthday this year and subtracts one year otherwise.

<pre><code>// date is the moment you're calculating the age of
var now = moment().unix();
var then = date.unix();
var diff = (now - then) / (60 * 60 * 24 * 365);
var years = Math.floor(diff);
</code></pre>

Edit: First version didn't quite work perfectly. The updated one should

blocks|key|192263|text|如果您不想使用任何模块来计算年龄|type|unstyled|depth|inlineStyleRanges|entityRanges|data|192264|var+age+=+Math.floor((new+Date()+-+new+Date(date_of_birth))+/+1000+/+60+/+60+/+24+/+365.25)|code-block|syntax|javascript|192265|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you dont want to use any module for age calculation

<pre><code>var age = Math.floor((new Date() - new Date(date_of_birth)) / 1000 / 60 / 60 / 24 / 365.25)
</code></pre>

blocks|key|410576|text|当您想要显示年份和剩余天数时：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|410577|var+m+=+moment(d.birthday.date,+"DD.MM.YYYY");
var+years+=+moment().diff(m,+'years',+false);
var+days+=+moment().diff(m.add(years,+'years'),+'days',+false);
alert(years+%2B+'+years,+'+%2B+days+%2B+'+days');|code-block|syntax|javascript|410578|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

When you want to show years and the remaining days:

<pre><code>var m = moment(d.birthday.date, "DD.MM.YYYY");
var years = moment().diff(m, 'years', false);
var days = moment().diff(m.add(years, 'years'), 'days', false);
alert(years + ' years, ' + days + ' days');
</code></pre>

blocks|key|192450|text|我更喜欢这个小方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|192451|function+getAgeFromBirthday(birthday)+{
++++if(birthday){
++++++var+totalMonths+=+moment().diff(birthday,+'months');
++++++var+years+=+parseInt(totalMonths+/+12);
++++++var+months+=+totalMonths+%25+12;
++++++++if(months+!==+0){
+++++++++++return+parseFloat(years+%2B+'.'+%2B+months);
+++++++++}
++++return+years;
++++++}
++++return+null;
}|code-block|syntax|javascript|192452|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I prefer this small method.

<pre><code>function getAgeFromBirthday(birthday) {
 if(birthday){
 var totalMonths = moment().diff(birthday, 'months');
 var years = parseInt(totalMonths / 12);
 var months = totalMonths % 12;
 if(months !== 0){
 return parseFloat(years + '.' + months);
 }
 return years;
 }
 return null;
}
</code></pre>

I'm trying to calculate a person's age using <a href="http://momentjs.com/" rel="noreferrer">Moment.js</a>, but I'm finding that the otherwise useful <a href="http://momentjs.com/docs/#/displaying/fromnow/" rel="noreferrer">fromNow</a> method rounds up the years. For instance, if today is 12/27/2012 and the person's birth date is 02/26/1978, <code>moment("02/26/1978", "MM/DD/YYYY").fromNow()</code> returns '35 years ago'. How can I make Moment.js ignore the number of months, and simply return the number of years (i.e. 34) since the date?

Moment.js - how do I get the number of years since a date, not rounded up?

我正在尝试使用计算一个人的年龄，但我发现原本有用的方法会将年份四舍五入。例如，如果今天是2012年12月27日，而此人的出生日期是1978年2月26日，则moment("02/26/1978", "MM/DD/YYYY").fromNow()返回'35年前‘。如何才能让Moment.js忽略月数，而只是返回自该日期以来的年数(即34年)？

问Moment.js -如何获得从日期开始的年数，而不是四舍五入？
EN

回答 9

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Moment.js -如何获得从日期开始的年数，而不是四舍五入？EN

回答 9

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Moment.js -如何获得从日期开始的年数，而不是四舍五入？
EN