var date = new Date();

alert( date.getFullYear() + ("0" + (date.getMonth() + 1)).slice(-2) + ("0" + date.getDate()).slice(-2) + ("0" + date.getHours() ).slice(-2) + ("0" + date.getMinutes()).slice(-2) + ("0" + date.getSeconds()).slice(-2) );

edit

function pad2(n) { return n < 10 ? '0' + n : n }

var date = new Date();
    
alert( date.getFullYear().toString() + pad2(date.getMonth() + 1) + pad2( date.getDate()) + pad2( date.getHours() ) + pad2( date.getMinutes() ) + pad2( date.getSeconds() ) );

Just leverage the built-in toISOString method that brings your date to the ISO 8601 format:

let yourDate = new Date()
yourDate.toISOString().split('T')[0]

Where yourDate is your date object.

Edit: @exbuddha wrote this to handle time zone in the comments:

const offset = yourDate.getTimezoneOffset()
yourDate = new Date(yourDate.getTime() - (offset*60*1000))
return yourDate.toISOString().split('T')[0]

[Addendum 12/2022]: Here's a library to format dates using Intl.DateTimeFormat.

[Addendum 01/2024]: And here is a (ES-)Date manipulation library

Try something like this

var d = new Date,
    dformat = [d.getMonth()+1,
               d.getDate(),
               d.getFullYear()].join('/')+' '+
              [d.getHours(),
               d.getMinutes(),
               d.getSeconds()].join(':');

If you want leading zero's for values < 10, use this number extension

Number.prototype.padLeft = function(base,chr){
    var  len = (String(base || 10).length - String(this).length)+1;
    return len > 0? new Array(len).join(chr || '0')+this : this;
}
// usage
//=> 3..padLeft() => '03'
//=> 3..padLeft(100,'-') => '--3' 

Applied to the previous code:

var d = new Date,
    dformat = [(d.getMonth()+1).padLeft(),
               d.getDate().padLeft(),
               d.getFullYear()].join('/') +' ' +
              [d.getHours().padLeft(),
               d.getMinutes().padLeft(),
               d.getSeconds().padLeft()].join(':');
//=> dformat => '05/17/2012 10:52:21'

See this code in [jsfiddle][1]

[edit 2019] Using ES20xx, you can use a template literal and the new padStart string extension.

const dt = new Date();
const padL = (nr, len = 2, chr = `0`) => `${nr}`.padStart(2, chr);

console.log(`${
    padL(dt.getMonth()+1)}/${
    padL(dt.getDate())}/${
    dt.getFullYear()} ${
    padL(dt.getHours())}:${
    padL(dt.getMinutes())}:${
    padL(dt.getSeconds())}`
);

From ES6 onwards you can use template strings to make it a little shorter:

var now = new Date();
var todayString = `${now.getFullYear()}-${now.getMonth()}-${now.getDate()}`;

This solution does not zero pad. Look to the other good answers to see how to do that.

The code looks good and works well. Since it provides a useful bit of work, you should convert it into a function. Then you can copy that function to another program if you need so.

function yyyymmdd() {
    var x = new Date();
    var y = x.getFullYear().toString();
    var m = (x.getMonth() + 1).toString();
    var d = x.getDate().toString();
    (d.length == 1) && (d = '0' + d);
    (m.length == 1) && (m = '0' + m);
    var yyyymmdd = y + m + d;
    return yyyymmdd;
}

To make this code a little easier to read, you should rename x to now.

You could also omit the calls to toString(), which makes the code a little shorter. Plus, you should introduce the variables mm and dd, so that you don't reassign to the d and m variables. This is a generally useful pattern, because when stepping through the code you can always look at the variable definition to see how it was computed. This is not possible for variables that change their value during execution.

The modified code looks like this:

function yyyymmdd() {
    var now = new Date();
    var y = now.getFullYear();
    var m = now.getMonth() + 1;
    var d = now.getDate();
    var mm = m < 10 ? '0' + m : m;
    var dd = d < 10 ? '0' + d : d;
    return '' + y + mm + dd;
}

Or, you could inline the last few lines:

function yyyymmdd() {
    var now = new Date();
    var y = now.getFullYear();
    var m = now.getMonth() + 1;
    var d = now.getDate();
    return '' + y + (m < 10 ? '0' : '') + m + (d < 10 ? '0' : '') + d;
}

That last variant is harder to read though, therefore I prefer the previous one.

Another possibility is to define a helper function that produces a two-digit string:

function yyyymmdd() {
    function twoDigit(n) { return (n < 10 ? '0' : '') + n; }

    var now = new Date();
    return '' + now.getFullYear() + twoDigit(now.getMonth() + 1) + twoDigit(now.getDate());
}

As new Date().toISOString() will return current UTC time, to get local time in ISO String format we have to get time from new Date() function like the following method

document.write(new Date(new Date().toString().split('GMT')[0]+' UTC').toISOString().split('.')[0]);