我有一个从输入值定义的 javascript 变量。
$d = $('#date').val();
$myDateParts = $d.split("-");
$dflip = new Date($myDateParts[2], ($myDateParts[1]-1), $myDateParts[0]);
console.log($dflip);
$dflip = Wed Sep 19 00:00:00 UTC+0100 2012
我如何将输出格式化为:
Wed Sep 19
我有一个从输入值定义的 javascript 变量。
$d = $('#date').val();
$myDateParts = $d.split("-");
$dflip = new Date($myDateParts[2], ($myDateParts[1]-1), $myDateParts[0]);
console.log($dflip);
$dflip = Wed Sep 19 00:00:00 UTC+0100 2012
我如何将输出格式化为:
Wed Sep 19
substring
您可以使用或使用toDateString
两者来做某事
例如:
var dateString = new Date(2012, 0, 31).toDateString();
var noYear = dateString.substring(0, dateString.length-5);
console.log(noYear);
这可能不是完成您正在寻找的最干净、最有效或最好的方法,但我创建了一个函数来返回没有时区的日期。您可以调整“theDate”变量以仅返回您想要的日期部分。
function properDate(){
var d = new Date();
var DayOfMonth = d.getDate();
var DayOfWeek = d.getDay();
var Month = d.getMonth();
var Year = d.getFullYear();
var Hours = d.getHours();
var Minutes = d.getMinutes();
var Seconds = d.getSeconds();
switch (DayOfWeek) {
case 0:
day = "Sun";
break;
case 1:
day = "Mon";
break;
case 2:
day = "Tue";
break;
case 3:
day = "Wed";
break;
case 4:
day = "Thu";
break;
case 5:
day = "Fri";
break;
case 6:
day = "Sat";
break;
}
switch (Month) {
case 0:
month = "Jan";
break;
case 1:
month = "Feb";
break;
case 2:
month = "Mar";
break;
case 3:
month = "Apr";
break;
case 4:
month = "May";
break;
case 5:
month = "Jun";
break;
case 6:
month = "Jul";
break;
case 7:
month = "Aug";
break;
case 8:
month = "Sep";
break;
case 9:
month = "Oct";
break;
case 10:
month = "Nov";
break;
case 11:
month = "Dec";
break;
}
var theDate = day + " " + month + " " + DayOfMonth + " " + Year + " " + Hours + ":" + Minutes + ":" + Seconds;
return theDate;
}
尝试使用以下代码。
<script src="../../ui/jquery.ui.datepicker.js"></script>
$( "#datepicker" ).datepicker( "D M yy", dflip.val());
我的 DateExtentions 库会这样做——尽管如果你所做的只是一种简单的格式,它可能会有点过分。
http://depressedpress.com/javascript-extensions/dp_dateextensions/
我可以根据传递的掩码解析日期并根据需要格式化输出(它还会执行各种日期数学和实用程序,因此它可能比您需要的要重)。