mas de una funcion ajax en simultaneo...

[rome96]

Gracias por responder!

Me falla solo con el explorer.

Estas son las funciones tal cual las uso y me traen conflicto:

//------------------------------------------------ajax video progrma

lo= 1;


function ajaxFunction1(valor00){
//alert(valor00);
try{
//Firefox, Opera 8.0+, Safari
xml=new XMLHttpRequest();
}
catch (e){
// Internet Explorer
try{
xml=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e){
try{
xml=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("Tu navegador no soporta Ajax");
}
}
}
xml.onreadystatechange=function(){
if(xml.readyState==4){

// document.write(xml.responseText);
document.getElementById("video_programa").innerHTM L=xml.responseText
}
}







/*
if(i==4){
i=1;
}
*/
//alert(lo);
xml.onreadystatechange=function(){
if(xml.readyState==4){

// document.write(xml.responseText);
document.getElementById("video_programa").innerHTM L=xml.responseText
}
}

loo = valor00;


xml.open("GET","video_programa.php?clave="+ loo);
xml.send(null);



}


//------------------------------------------------fin ajax video programa

//------------------------------------------------ajax videito

num_video= 1;


function ajaxFunction2(valor1,valor2){
try{
//Firefox, Opera 8.0+, Safari
xml=new XMLHttpRequest();
}
catch (e){
// Internet Explorer
try{
xml=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e){
try{
xml=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("Tu navegador no soporta Ajax");
}
}
}
xml.onreadystatechange=function(){
if(xml.readyState==4){

// document.write(xml.responseText);
document.getElementById("videito").innerHTML=xml.r esponseText
}
}



xml.onreadystatechange=function(){
if(xml.readyState==4){

// document.write(xml.responseText);
document.getElementById("videito").innerHTML=xml.r esponseText
}
}






if(valor1 == "avanza"){
num_video= num_video + 1;
}

if(valor1 == "retrocede"){
num_video= num_video - 1;
}



if(num_video > valor2){
num_video=1
}

if(num_video == 0){
num_video= valor2;
}

xml.open("GET","videitos.php?numero=" + num_video);
xml.send(null);

//ajaxFunction1(num_video);

}


//------------------------------------------------fin ajax videito