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Simple and flexible ajax You can use [GET]or [POST] method and you can Specify the function to process the response text

Submitted By: ahmad_511

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Rating:

Views: 954

Language: JavaScript

Last Modified: June 17, 2007

Instructions: call the Ajax_Send() function and pass the parameters Ex:
Ajax_Send("POST","www.Mysite.com/myurl","user=Ahmad&pass=1234&memtype=admin",myfunction)

function myfunction(response){
// code to process the response
}

Snippet

// Here is The Ajax_Send Function
function Ajax_Send(GP,URL,PARAMETERS,RESPONSEFUNCTION){
var xmlhttp //Declaring Variable
//Trying to catch xmlhttp
try{xmlhttp=new ActiveXObject("Msxml2.XMLHTTP")}
catch(e){
try{xmlhttp=new ActiveXObject("Microsoft.XMLHTTP")}
catch(e){
try{xmlhttp=new XMLHttpRequest()}
catch(e){
alert("Your Browser Does Not Support AJAX")}}}
// Waiting the response text
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readyState == 4)eval(RESPONSEFUNCTION(xmlhttp.responseText))// calling the function we want and passing the response text to it when ready
}
// GET Method
if (GP=="GET"){
URL+="?"+PARAMETERS
xmlhttp.open("GET",URL,true)
xmlhttp.send(null)
}
// POST Method
if (GP="POST"){
PARAMETERS=encodeURI(PARAMETERS)
xmlhttp.open("POST",URL,true)
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded")
xmlhttp.setRequestHeader("Content-length",PARAMETERS.length)
xmlhttp.setRequestHeader("Connection", "close")
xmlhttp.send(PARAMETERS)
}
}