更好的捷径:

+new Date # Milliseconds since Linux epoch
+new Date / 1000 # Seconds since Linux epoch
Math.round(+new Date / 1000) #Seconds without decimals since Linux epoch

更好的捷径:

+new Date # Milliseconds since Linux epoch
+new Date / 1000 # Seconds since Linux epoch
Math.round(+new Date / 1000) #Seconds without decimals since Linux epoch

一种简单快速的方法，不需要创建日期对象并给出一个整数作为结果(如果发送小数，接受秒的api将会出错)

var nowincos = ~~(日期。 游戏机。log (nowInSeconds);

//当前Unix时间戳 // 1443535752秒自1970年1月1日。(UTC) //当前时间，单位为秒 console.log(数学。floor(new Date().valueOf() / 1000));/ / 1443535752 console.log(Math.floor(Date.now() / 1000));/ / 1443535752 console.log(数学。floor(new Date().getTime() / 1000));/ / 1443535752 < script src = " https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js " > < /脚本>

jQuery

console.log(Math.floor($.now() / 1000));/ / 1443535752 < script src = " https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js " > < /脚本>

在2020年的某一天，在Chrome 80.0.3987.132中，这给出了1584533105

~~(new Date()/1000) // 1584533105
Number.isInteger(~~(new Date()/1000)) // true

这些JavaScript解决方案提供了从1970年1月1日午夜开始的毫秒或秒。

IE 9+解决方案(IE 8或旧版本不支持):

var timestampInMilliseconds = Date.now();
var timestampInSeconds = Date.now() / 1000; // A float value; not an integer.
    timestampInSeconds = Math.floor(Date.now() / 1000); // Floor it to get the seconds.
    timestampInSeconds = Date.now() / 1000 | 0; // Also you can do floor it like this.
    timestampInSeconds = Math.round(Date.now() / 1000); // Round it to get the seconds.

要获得有关Date.now()的更多信息，请访问https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date/now

通用解决方案:

// ‘+’ operator makes the operand numeric.
// And ‘new’ operator can be used without the arguments ‘(……)’.
var timestampInMilliseconds = +new Date;
var timestampInSeconds = +new Date / 1000; // A float value; not an intger.
    timestampInSeconds = Math.floor(+new Date / 1000); // Floor it to get the seconds.
    timestampInSeconds = +new Date / 1000 | 0; // Also you can do floor it like this.
    timestampInSeconds = Math.round(+new Date / 1000); // Round it to get the seconds.

如果你不想要这种情况，请小心使用。

if(1000000 < Math.round(1000000.2)) // false.