除了o-o的答案之外，我还建议将逻辑操作与返回值分离，并将它们作为三元放入变量中。

另外，使用concat()来确保变量的安全连接

Date.prototype.yyyymmdd = function() { var yyyy = this.getFullYear(); var mm = this.getMonth() < 9 ? "0" + (this.getMonth() + 1) : (this.getMonth() + 1); // getMonth() is zero-based var dd = this.getDate() < 10 ? "0" + this.getDate() : this.getDate(); return "".concat(yyyy).concat(mm).concat(dd); }; Date.prototype.yyyymmddhhmm = function() { var yyyymmdd = this.yyyymmdd(); var hh = this.getHours() < 10 ? "0" + this.getHours() : this.getHours(); var min = this.getMinutes() < 10 ? "0" + this.getMinutes() : this.getMinutes(); return "".concat(yyyymmdd).concat(hh).concat(min); }; Date.prototype.yyyymmddhhmmss = function() { var yyyymmddhhmm = this.yyyymmddhhmm(); var ss = this.getSeconds() < 10 ? "0" + this.getSeconds() : this.getSeconds(); return "".concat(yyyymmddhhmm).concat(ss); }; var d = new Date(); document.getElementById("a").innerHTML = d.yyyymmdd(); document.getElementById("b").innerHTML = d.yyyymmddhhmm(); document.getElementById("c").innerHTML = d.yyyymmddhhmmss(); <div> yyyymmdd: <span id="a"></span> </div> <div> yyyymmddhhmm: <span id="b"></span> </div> <div> yyyymmddhhmmss: <span id="c"></span> </div>

以下是对https://stackoverflow.com/users/318563/o-o的回答的一点改进

Date.prototype.ddmmyyyy = function(delimiter) {
    var yyyy = this.getFullYear().toString();
    var mm = (this.getMonth()+1).toString(); // getMonth() is zero-based
    var dd  = this.getDate().toString();
    return (dd[1]?dd:"0"+dd[0]) + delimiter + (mm[1]?mm:"0"+mm[0]) + delimiter +yyyy  ; // padding
};

希望对大家有所帮助!

:)

我经常使用的一段修改代码:

Date.prototype.yyyymmdd = function() {
  var mm = this.getMonth() + 1; // getMonth() is zero-based
  var dd = this.getDate();

  return [this.getFullYear(),
          (mm>9 ? '' : '0') + mm,
          (dd>9 ? '' : '0') + dd
         ].join('');
};

var date = new Date();
date.yyyymmdd();

最短的

.toJSON().slice(0,10).split`-`.join``;

let d = new Date(); let s = d.toJSON().slice(0,10).split ' - ' .join '; console.log(年代);

对公认答案的一点变化:

函数getDate_yyyymmdd() { const date = new date (); const yyyy = date.getFullYear(); const mm = String(date.getMonth() + 1).padStart(2，'0'); const dd = String(date.getDate()).padStart(2，'0'); 返回“$ {yyyy} $ {mm} $ {dd} ' ｝ console.log (getDate_yyyymmdd ())

new Date('Jun 5 2016').
  toLocaleString('en-us', {year: 'numeric', month: '2-digit', day: '2-digit'}).
  replace(/(\d+)\/(\d+)\/(\d+)/, '$3-$1-$2');

// => '2016-06-05'