我试图使用JS将日期对象转换为YYYYMMDD格式的字符串。有没有比连接Date.getYear(), Date.getMonth()和Date.getDay()更简单的方法?
当前回答
<pre>Date.prototype.getFromFormat = function(format) {
var yyyy = this.getFullYear().toString();
format = format.replace(/yyyy/g, yyyy)
var mm = (this.getMonth()+1).toString();
format = format.replace(/mm/g, (mm[1]?mm:"0"+mm[0]));
var dd = this.getDate().toString();
format = format.replace(/dd/g, (dd[1]?dd:"0"+dd[0]));
var hh = this.getHours().toString();
format = format.replace(/hh/g, (hh[1]?hh:"0"+hh[0]));
var ii = this.getMinutes().toString();
format = format.replace(/ii/g, (ii[1]?ii:"0"+ii[0]));
var ss = this.getSeconds().toString();
format = format.replace(/ss/g, (ss[1]?ss:"0"+ss[0]));
var ampm = (hh >= 12) ? "PM" : "AM";
format = format.replace(/ampm/g, (ampm[1]?ampm:"0"+ampm[0]));
return format;
};
var time_var = $('#899_TIME');
var myVar = setInterval(myTimer, 1000);
function myTimer() {
var d = new Date();
var date = d.getFromFormat('dd-mm-yyyy hh:ii:ss:ampm');
time_var.text(date);
} </pre>
use the code and get the output like **26-07-2017 12:29:34:PM**
check the below link for your reference
https://parthiban037.wordpress.com/2017/07/26/date-and-time-format-in-oracle-apex-using-javascript/
其他回答
当我需要这样做时,我通常使用下面的代码。
var date = new Date($.now());
var dateString = (date.getFullYear() + '-'
+ ('0' + (date.getMonth() + 1)).slice(-2)
+ '-' + ('0' + (date.getDate())).slice(-2));
console.log(dateString); //Will print "2015-09-18" when this comment was written
为了解释,.slice(-2)给出了字符串的最后两个字符。
所以无论如何,我们都可以在日期或月份后加上“0”,只要求最后两个,因为这两个总是我们想要的。
所以如果MyDate.getMonth()返回9,它将是:
("0" + "9") // Giving us "09"
加上。slice(-2)就得到了最后两个字符:
("0" + "9").slice(-2)
"09"
但是如果date.getMonth()返回10,它将是:
("0" + "10") // Giving us "010"
所以加上.slice(-2)会得到最后两个字符,或者:
("0" + "10").slice(-2)
"10"
const date = new Date()
console.log(date.toISOString().split('T')[0]) // 2022-12-27
纯JS (ES5)解决方案,没有任何可能的日期跳转问题,由date . toisostring()打印UTC:
var now = new Date();
var todayUTC = new Date(Date.UTC(now.getFullYear(), now.getMonth(), now.getDate()));
return todayUTC.toISOString().slice(0, 10).replace(/-/g, '');
这是为了回应@weberste对@Pierre Guilbert的回答的评论。
[day,,month,,year]= Intl.DateTimeFormat(undefined, { year: 'numeric', month: '2-digit', day: '2-digit' }).formatToParts(new Date()),year.value+month.value+day.value
or
new Date().toJSON().slice(0,10).replace(/\/|-/g,'')
您可以简单地使用这一行代码来获取日期
var date = new Date().getFullYear() + "-" + (parseInt(new Date().getMonth()) + 1) + "-" + new Date().getDate();