如何添加10秒到一个JavaScript日期对象?
就像这样:
var timeObject = new Date()
var seconds = timeObject.getSeconds() + 10;
timeObject = timeObject + seconds;
当前回答
如果你需要在当前时间上增加10秒，你可以使用:
const date = new Date();
date.setUTCSeconds(date.getUTCSeconds() + 10); // Change 10 to any number of seconds
其他回答
还有一个setSeconds方法:
var t = new Date();
t.setSeconds(t.getSeconds() + 10);
有关其他Date函数的列表，您应该查看MDN
setSeconds将正确处理换行的情况:
var d; d =新日期('2014-01-01 10:11:55'); alert(d.getMinutes() + ':' + d.getSeconds());/ / 34) d.setSeconds(d.getSeconds() + 10); alert(d.getMinutes() + ':0' + d.getSeconds());/ / 12:05)
几年前，我写了一个“通用的”日期函数:
function addToDate({time_unit, operator, offset_value }) {
const date = new Date();
operator = operator == "after" ? "+" : "-";
switch (time_unit) {
case "seconds":
date.setSeconds(eval(`${date.getSeconds()} ${operator} ${offset_value}`));
break;
case "hours":
date.setHours(eval(`${date.getHours()} ${operator} ${offset_value}`));
break;
case "minutes":
date.setMinutes(eval(`${date.getMinutes()} ${operator} ${offset_value}`));
break;
case "days":
date.setDate(eval(`${date.getDate()} ${operator} ${offset_value}`));
break;
case "months":
date.setMonth(eval(`${date.getMonth()} ${operator} ${offset_value}`));
break;
case "years":
date.setFullYear(eval(`${date.getFullYear()} ${operator} ${offset_value}`));
break;
default:
break;
}
return date;
}
const new_date = addToDate({time_unit:'seconds','operator':'after','offset_value':10});
console.log(new_date.toISOString());
timeObject.setSeconds(timeObject.getSeconds() + 10)
如果你需要在当前时间上增加10秒，你可以使用:
const date = new Date();
date.setUTCSeconds(date.getUTCSeconds() + 10); // Change 10 to any number of seconds
你可以使用setSeconds方法获取今天的秒数，并在其中添加10秒 var today = new Date(); today.setSeconds(today.getSeconds() + 10); 您可以添加10 *1000毫秒的新日期: var today = new Date(); today = new Date(today. gettime () + 1000*10); 你可以使用setTime: settime (now.getTime() + 10000)
