JS版的date函数（和PHP的date函数一样）

// 和PHP一样的时间戳格式化函数
// @param  {string} format    格式
// @param  {int}    timestamp 要格式化的时间 默认为当前时间
// @return {string}           格式化的时间字符串
function date ( format, timestamp ) {
    var a, jsdate=((timestamp) ? new Date(timestamp*1000) : new Date());
    var pad = function(n, c){
        if( (n = n + "").length < c ) {
            return new Array(++c - n.length).join("0") + n;
        } else {
            return n;
        }
    };

var txt_weekdays = ["Sunday","Monday","Tuesday","Wednesday", "Thursday","Friday","Saturday"];        

var txt_ordin = {1:"st",2:"nd",3:"rd",21:"st",22:"nd",23:"rd",31:"st"};

var txt_months = ["", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];

var f = {
        // Day
            d: function(){
                return pad(f.j(), 2);
            },
            D: function(){
                t = f.l(); return t.substr(0,3);
            },
            j: function(){
                return jsdate.getDate();
            },
            l: function(){
                return txt_weekdays[f.w()];
            },
            N: function(){
                return f.w() + 1;
            },
            S: function(){
                return txt_ordin[f.j()] ? txt_ordin[f.j()] : 'th';
            },
            w: function(){
                return jsdate.getDay();
            },
            z: function(){
                return (jsdate - new Date(jsdate.getFullYear() + "/1/1")) / 864e5 >> 0;
            },

// Week
            W: function(){
                var a = f.z(), b = 364 + f.L() - a;
                var nd2, nd = (new Date(jsdate.getFullYear() + "/1/1").getDay() || 7) - 1;

if(b <= 2 && ((jsdate.getDay() || 7) - 1) <= 2 - b){
                    return 1;
                } else{

if(a <= 2 && nd >= 4 && a >= (6 - nd)){
                        nd2 = new Date(jsdate.getFullYear() - 1 + "/12/31");
                        return date("W", Math.round(nd2.getTime()/1000));
                    } else{
                        return (1 + (nd <= 3 ? ((a + nd) / 7) : (a - (7 - nd)) / 7) >> 0);
                    }
                }
            },