我将某个日期(以毫秒为单位)转换为日期格式,如下所示:
var dateFrom = new Date(1312883657720); //dateFrom = 2011-08-09T12:57:01
现在我需要以这种格式使用dateFrom(2011-08-09T12:57:01)的值:2011/08/09。我似乎找不到删除时间字符串的日期方法,是否需要解决方法?
答案 0 :(得分:1)
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
Date date = sdf.parse(sdf.format(new Date()));
或其他方法使用日历,
Calendar cal = Calendar.getInstance();
cal.set(Calendar.HOUR_OF_DAY, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
cal.set(Calendar.MILLISECOND, 0);
dateWithoutTime = cal.getTime();
答案 1 :(得分:0)
您需要使用对象
形成一个字符串var dateFrom = new Date(1312883657720); //dateFrom = 2011-08-09T12:57:01
var humanDate = dateFrom.getFullYear()
+ "/" + ("0" + (dateFrom.getMonth()+1)).slice(-2)
+ "/" + ("0" + dateFrom.getDate()).slice(-2);
// ("0" + dateFrom.method()).slice(-2) is used to add the leading zero
console.log(humanDate); // 2011/08/09
答案 2 :(得分:0)
使用此
<div id="res">res</div>
function formatDate(date) {
var d = new Date(date),
month = '' + (d.getMonth() + 1),
day = '' + d.getDate(),
year = d.getFullYear();
if (month.length < 2) month = '0' + month;
if (day.length < 2) day = '0' + day;
return [year, month, day].join('/');
}
document.getElementById('res').innerHTML = formatDate('2011-08-09T12:57:01') ;
产出:2011/08/09
答案 3 :(得分:0)
尝试以下两种方法中的任何一种
var newDate = moment(new Date(1312883657720)).format(YYYY/MM/DD);
或
var newDate = new Date(1312883657720);
newDate = newDate.getFullYear()+'/'+newDate.getMonth()+'/'+newDate getDate()