How to parse a date in format "YYYYmmdd" in JavaScript?

This is a noob question:

How to parse a date in format "YYYYmmdd" without external libraries ? If the input string is not in this format I would like to get invalid Date (or undefined if it will be easier).

Solution 1:

function parse(str) {
    if(!/^(\d){8}$/.test(str)) return "invalid date";
    var y = str.substr(0,4),
        m = str.substr(4,2),
        d = str.substr(6,2);
    return new Date(y,m,d);
}

Usage:

parse('20120401');

UPDATE:

As Rocket said, months are 0-based in js...use this if month's aren't 0-based in your string

function parse(str) {
    if(!/^(\d){8}$/.test(str)) return "invalid date";
    var y = str.substr(0,4),
        m = str.substr(4,2) - 1,
        d = str.substr(6,2);
    return new Date(y,m,d);
}

UPDATE:

More rigorous checking for validity of date. Adopted HBP's way to validate date.

function parse(str) {
    var y = str.substr(0,4),
        m = str.substr(4,2) - 1,
        d = str.substr(6,2);
    var D = new Date(y,m,d);
    return (D.getFullYear() == y && D.getMonth() == m && D.getDate() == d) ? D : 'invalid date';
}

Solution 2:

A more robust version validating the numbers :

 function parse (str) {
        // validate year as 4 digits, month as 01-12, and day as 01-31 
        if ((str = str.match (/^(\d{4})(0[1-9]|1[0-2])(0[1-9]|[12]\d|3[01])$/))) {
           // make a date
           str[0] = new Date (+str[1], +str[2] - 1, +str[3]);
           // check if month stayed the same (ie that day number is valid)
           if (str[0].getMonth () === +str[2] - 1)
              return str[0];
        }
        return undefined;
 }

See fiddle at : http://jsfiddle.net/jstoolsmith/zJ7dM/

I recently wrote a much more capable version you can find here : http://jsfiddle.net/jstoolsmith/Db3JM/

Solution 3:

simplistic answer maybe, without checks, but fast...

var date = parseInt(date);
new Date(date / 10000, date % 10000 / 100, date % 100);

or, if months are not zero based in the source,

new Date(date / 10000, (date % 10000 / 100) - 1, date % 100);