使用php求两个时间戳的值

是个自动倒计时，具体代码如下
var serverTime 和 var Htime 两个值的php代码怎么写，目前是固定的，时间到了无法重新开始，三天重复一次。高手看清问题和代码了回答
serverTime 和 Htime值是php代码获取的，在查看源代码里看不到，高手可以把获取这两个值的php代码写出来

 <script type="text/javascript">
//var dateTimezz = new Date();
//alert(dateTimezz);
var serverTime = 1470970070 * 1000;
var Htime = 220330000;
jQuery(document).ready(function(){
//var dateTime = new Date('Sun Dec 04 2015 00:00:00');
//alert(dateTime.getTime());
var dateTime = new Date();
var difference = dateTime.getTime() - serverTime;
var endTime = new Date().getTime()+Htime-difference;

setInterval(function(){
jQuery(".tlimit").each(function(){
    var obja = jQuery(this);

    var dateTimez = new Date();
    var strDateList = daysBetween('2015-12-06',(dateTimez.getYear()+1900)+'-'+(dateTimez.getMonth()+1)+'-'+dateTimez.getDate()).toLocaleString();
    //var strDateList = daysBetween('2015-12-06','2015-12-19').toLocaleString();
    var chaday = Math.ceil(strDateList/3)*3;
    //alert(chaday);
    var str2 = 'TIME LIMIT: '+dateAdd("d", chaday-2, '2015/12/06').toLocaleString()+' - '+dateAdd("d", chaday, '2015/12/06').toLocaleString();
obja.html(str2);
});
}, 10);
setInterval(function(){
jQuery(".t3").each(function(){
var obj = jQuery(this);

var dateTimea = new Date();
var nMS=endTime - dateTimea.getTime();
var myD=Math.floor(nMS/(1000 * 60 * 60 * 24));
var myH=Math.floor(nMS/(1000*60*60)) % 24;
var myM=Math.floor(nMS/(1000*60)) % 60;
var myS=Math.floor(nMS/1000) % 60;

if(myD&gt;= 0){
    myD = ( (  myD  &lt;  10  ) ?  "0"  :  "")+myD;
    myH = ( (  myH  &lt;  10  ) ?  "0"  :  "")+myH;
    myM = ( (  myM  &lt;  10  ) ?  "0"  :  "")+myM;
    myS = ( (  myS  &lt;  10  ) ?  "0"  :  "")+myS;
    var str = '&lt;i class="d"&gt;Day&lt;br&gt;&lt;b&gt;' + myD+'&lt;/b&gt;&lt;/i&gt;&lt;i class="h"&gt;Hou&lt;br&gt;&lt;b&gt;'+myH+'&lt;/b&gt;&lt;/i&gt;&lt;i class="m"&gt;Min&lt;br&gt;&lt;b&gt;'+myM+'&lt;/b&gt;&lt;/i&gt;&lt;i class="s"&gt;Sec&lt;br&gt;&lt;b&gt;'+myS+'&lt;/b&gt;&lt;/i&gt;';
}else{
    var str = '&lt;i class="d"&gt;Day&lt;br&gt;&lt;b&gt;00&lt;/b&gt;&lt;/i&gt;&lt;i class="h"&gt;Hou&lt;br&gt;&lt;b&gt;00&lt;/b&gt;&lt;/i&gt;&lt;i class="m"&gt;Min&lt;br&gt;&lt;b&gt;00&lt;/b&gt;&lt;/i&gt;&lt;i class="s"&gt;Sec&lt;br&gt;&lt;b&gt;00&lt;/b&gt;&lt;/i&gt;';   
}
obj.html(str);
});
}, 10);


});

function daysBetween(DateOne,DateTwo) 
{  
var OneMonth = DateOne.substring(5,DateOne.lastIndexOf ('-')); 
var OneDay = DateOne.substring(DateOne.length,DateOne.lastIndexOf ('-')+1); 
var OneYear = DateOne.substring(0,DateOne.indexOf ('-')); 
var TwoMonth = DateTwo.substring(5,DateTwo.lastIndexOf ('-')); 
var TwoDay = DateTwo.substring(DateTwo.length,DateTwo.lastIndexOf ('-')+1); 
var TwoYear = DateTwo.substring(0,DateTwo.indexOf ('-')); 
var cha=((Date.parse(OneMonth+'/'+OneDay+'/'+OneYear)- Date.parse(TwoMonth+'/'+TwoDay+'/'+TwoYear))/86400000);    
return Math.abs(cha);
}
function  dateAdd(strInterval,  NumDay,  dtDate)  { 
var  dtTmp  =  new  Date(dtDate); 
if  (isNaN(dtTmp))  dtTmp  =  new  Date(); 
var ddTmp = new  Date(Date.parse(dtTmp)  +  (86400000  *  NumDay));
return  (ddTmp.getYear()+1900)+'.'+(ddTmp.getMonth()+1)+'.'+ddTmp.getDate(); 
/*switch  (strInterval)  { 
case  "s":return  new  Date(Date.parse(dtTmp)  +  (1000  *  NumDay)); 
case  "n":return  new  Date(Date.parse(dtTmp)  +  (60000  *  NumDay)); 
case  "h":return  new  Date(Date.parse(dtTmp)  +  (3600000  *  NumDay)); 
case  "d":return  ((new  Date(Date.parse(dtTmp)  +  (86400000  *  NumDay))).getYear()+1900)+'.'+((new  Date(Date.parse(dtTmp)  +  (86400000  *  NumDay))).getMonth()+1)+'.'+(new  Date(Date.parse(dtTmp)  +  (86400000  *  NumDay))).getDate(); 
case  "w":return  new  Date(Date.parse(dtTmp)  +  ((86400000  *  7)  *  NumDay)); 
case  "m":return  new  Date(dtTmp.getFullYear(),  (dtTmp.getMonth())  +  NumDay,  dtTmp.getDate(),  dtTmp.getHours(),  dtTmp.getMinutes(),  dtTmp.getSeconds()); 
case  "y":return  new  Date((dtTmp.getFullYear()  +  NumDay),  dtTmp.getMonth(),  dtTmp.getDate(),  dtTmp.getHours(),  dtTmp.getMinutes(),  dtTmp.getSeconds()); 
} */
} 
</script>

为什么不利用服务器的计时器实现