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Ramesh Naidu
simple ajax Request and Response code...
0 Answer(s)      3 years and 2 months ago
Posted in : Ajax


var request=null; if (window.XMLHttpRequest) { request = new XMLHttpRequest(); } else if (window.ActiveXObject) { request = new ActiveXObject("Microsoft.XMLHTTP"); }

if(request) { var url="deleteStudentInfo.jsp"; url+="?sId="+chosen+"";

request.open("POST",url);
request.onreadystatechange = function()
{

if(request.readyState==4) { // alert("visible"); document.getElementById("theResults").innerHTML=request.responseText; document.getElementById("theResults").style.visibility="visible"; }

} request.send(null);

} else { alert("Your browser don't support AJax !"); } } else { //alert("hidden"); document.getElementById("theResults").innerHTML=""; document.getElementById("theResults").style.visibility="hidden";

}

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