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我需要一个函数,单击一个链接,发送一个Ajax请求到2个页面。ajax请求逐个
首先发送数据到page1.php
,当完成page1的工作后,将数据返回到主页面,然后发送数据到page2.php
做另一个ajax进程并返回数据到主页面。
我写page1.php
document.getElementById("div1").innerHTML = HttPRequest.responseText;
page2.php
要求,但这只会返回page2's
数据和firebug
错过回从page1.php
数据,consule:
page1.php (always show loading.gif)
page2.php 200 OK 2199ms
如何做AJAX一个一个好?谢谢。
function myajax(text) {
var HttPRequest = false;
if (window.XMLHttpRequest) {
HttPRequest = new XMLHttpRequest();
if (HttPRequest.overrideMimeType) {
HttPRequest.overrideMimeType('text/html');
}
} else if (window.ActiveXObject) {
try {
HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {}
}
}
if (!HttPRequest) {
alert('Cannot create XMLHTTP instance');
return false;
}
var url = 'page1.php';
var pmeters = "text=" + text;
HttPRequest.open('POST',url,true);
HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
HttPRequest.send(pmeters);
HttPRequest.onreadystatechange = function()
{
if(HttPRequest.readyState == 4)
{
document.getElementById("div1").innerHTML = HttPRequest.responseText;
var url = 'page2.php';
var pmeters = "text=" + text;
HttPRequest.open('POST',url,true);
HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
HttPRequest.send(pmeters);
HttPRequest.onreadystatechange = function()
{
if(HttPRequest.readyState == 4)
{
document.getElementById("div2").innerHTML = HttPRequest.responseText;
}
}
}
}
}
我会建议使用像jQuery的AJAX框架。 – TMS