How can I solve the problem of ajax technology always sending status?

Ajax technology always sends out the status ajax. js code: & nbsp; var & nbsp; xmlHttp; function & nbsp; S_xmlhttprequest () {& nbs ajax technology always goes out of the sending status
Ajax. js code:
Var xmlHttp;
Function S_xmlhttprequest (){
If (window. ActiveXObject ){
XmlHttp = new ActiveXObject ("Microsoft. XMLHTTP ");
}
Else if (widow. XMLHttpRequest ){
XmlHttp = new XMLHttpRequest ();
}
}
Function funphp100 (url ){
S_xmlhttprequest ();
XmlHttp. open ("GET", "for. php? Id = "+ url, true );
XmlHttp. onreadystatechange = byphp;
XmlHttp. send (null );
}
Function byphp (){
If (xmlHttp. readyState = 1 ){
Document. getElementById ('php100'). innerHTML = "loading ....";
}
If (xmlHttp. readyState = 4 ){

Var byphp100 = xmlHttp. responseText;
Document. getElementById ('php100'). innerHTML = byphp100;

}

}
For. php code
If ($ id = $ _ GET [id]) {
For ($ I = 1; $ I <= 5; $ I ++ ){

Echo $ I;
Sleep (2 );
}
Exit ();
}
?>
Index. php code:
Script
O
T
X



------ Solution --------------------

if($id=$_GET[id]){
  for($i=1;$i<=5;$i++){

echo $i;
//sleep(2);
   }
   exit();
}

 

  
   《script》
   

   

   

   
    
var xmlHttp;
    
function S_xmlhttprequest(){
    
if(window.ActiveXObject){
    
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
    
}
    
else if(window.XMLHttpRequest){
    
xmlHttp=new XMLHttpRequest();
    
}
    
}
    
function funphp100(url){
    
S_xmlhttprequest();
    
xmlHttp.open("GET","test.php?id="+url,true);
    
xmlHttp.onreadystatechange=byphp;
    
xmlHttp.send(null);
    
}
    
function byphp(){
    
if(xmlHttp.readyState==1){
    
document.getElementById('php100').innerHTML="loading....";
    
}
    
if(xmlHttp.readyState==4){
    

    
var byphp100 = xmlHttp.responseText;
    
document.getElementById('php100').innerHTML=byphp100;
    

    
}
    

    
}
    

    
《script》
    

    
o
    
t
    
x
    

    

    

    

    

   
  

------ Solution --------------------
Else if (widow. XMLHttpRequest ){
XmlHttp = new XMLHttpRequest ();
}
Window