这个脚本有什么问题？
当我将时钟设置为 29/04/2011 时，它会在周输入中增加36/4/2011！但正确的日期应该是6/5/2011
var d = new Date();
var curr_date = d.getDate();
var tomo_date = d.getDate()+1;
var seven_date = d.getDate()+7;
var curr_month = d.getMonth();
curr_month++;
var curr_year = d.getFullYear();
var tomorrowsDate =(tomo_date + "/" + curr_month + "/" + curr_year);
var weekDate =(seven_date + "/" + curr_month + "/" + curr_year);
{
jQuery("input[id*='tomorrow']").val(tomorrowsDate);
jQuery("input[id*='week']").val(weekDate);
}
无需声明
返回时间戳
new Date().setDate(new Date().getDate() + 7)
返回日期
new Date(new Date().setDate(new Date().getDate() + 7))
像这样的东西？
var days = 7;
var date = new Date();
var res = date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
alert(res);
再次转换为日期：
date = new Date(res);
alert(date)
或者：
date = new Date(res);
// hours part from the timestamp
var hours = date.getHours();
// minutes part from the timestamp
var minutes = date.getMinutes();
// seconds part from the timestamp
var seconds = date.getSeconds();
// will display time in 10:30:23 format
var formattedTime = date + '-' + hours + ':' + minutes + ':' + seconds;
alert(formattedTime)
获取未来 x 天日期的简单方法是增加日期：
function addDays(dateObj, numDays) {
return dateObj.setDate(dateObj.getDate() + numDays);
}
请注意，这会修改提供的日期对象，例如
function addDays(dateObj, numDays) {
dateObj.setDate(dateObj.getDate() + numDays);
return dateObj;
}
var now = new Date();
var tomorrow = addDays(new Date(), 1);
var nextWeek = addDays(new Date(), 7);
alert(
'Today: ' + now +
'\nTomorrow: ' + tomorrow +
'\nNext week: ' + nextWeek
);
在一行中：
new Date(Date.now() + 7 * 24 * 60 * 60 * 1000)