如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
HTML:
<!DOCTYPE html>
<html>
<head>
<script>
function loadXMLDoc()
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","1.php?id=99freebies.blogspot.com",true);
xmlhttp.send();
}
</script>
</head>
<body>
<div id="myDiv"><h2>Let AJAX change this text</h2></div>
<button type="button" onclick="loadXMLDoc()">Change Content</button>
</body>
</html>
PHP:
<?php
$id = $_GET[id];
print "$id";
?>
其他回答
你可以根据浏览器获得正确的对象
function getXmlDoc() {
var xmlDoc;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlDoc = new XMLHttpRequest();
}
else {
// code for IE6, IE5
xmlDoc = new ActiveXObject("Microsoft.XMLHTTP");
}
return xmlDoc;
}
有了正确的对象，GET可以被抽象为:
function myGet(url, callback) {
var xmlDoc = getXmlDoc();
xmlDoc.open('GET', url, true);
xmlDoc.onreadystatechange = function() {
if (xmlDoc.readyState === 4 && xmlDoc.status === 200) {
callback(xmlDoc);
}
}
xmlDoc.send();
}
并将邮件发送到:
function myPost(url, data, callback) {
var xmlDoc = getXmlDoc();
xmlDoc.open('POST', url, true);
xmlDoc.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlDoc.onreadystatechange = function() {
if (xmlDoc.readyState === 4 && xmlDoc.status === 200) {
callback(xmlDoc);
}
}
xmlDoc.send(data);
}
您可以使用以下函数:
function callAjax(url, callback){
var xmlhttp;
// compatible with IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function(){
if (xmlhttp.readyState == 4 && xmlhttp.status == 200){
callback(xmlhttp.responseText);
}
}
xmlhttp.open("GET", url, true);
xmlhttp.send();
}
你可以在这些链接上尝试类似的解决方案:
https://www.w3schools.com/xml/tryit.asp?filename=tryajax_first https://www.w3schools.com/xml/tryit.asp?filename=tryajax_callback
使用下面的代码片段，你可以很容易地完成类似的事情，就像这样:
ajax.get('/test.php', {foo: 'bar'}, function() {});
以下是片段:
var ajax = {};
ajax.x = function () {
if (typeof XMLHttpRequest !== 'undefined') {
return new XMLHttpRequest();
}
var versions = [
"MSXML2.XmlHttp.6.0",
"MSXML2.XmlHttp.5.0",
"MSXML2.XmlHttp.4.0",
"MSXML2.XmlHttp.3.0",
"MSXML2.XmlHttp.2.0",
"Microsoft.XmlHttp"
];
var xhr;
for (var i = 0; i < versions.length; i++) {
try {
xhr = new ActiveXObject(versions[i]);
break;
} catch (e) {
}
}
return xhr;
};
ajax.send = function (url, callback, method, data, async) {
if (async === undefined) {
async = true;
}
var x = ajax.x();
x.open(method, url, async);
x.onreadystatechange = function () {
if (x.readyState == 4) {
callback(x.responseText)
}
};
if (method == 'POST') {
x.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
}
x.send(data)
};
ajax.get = function (url, data, callback, async) {
var query = [];
for (var key in data) {
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
}
ajax.send(url + (query.length ? '?' + query.join('&') : ''), callback, 'GET', null, async)
};
ajax.post = function (url, data, callback, async) {
var query = [];
for (var key in data) {
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
}
ajax.send(url, callback, 'POST', query.join('&'), async)
};
HTML:
<!DOCTYPE html>
<html>
<head>
<script>
function loadXMLDoc()
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","1.php?id=99freebies.blogspot.com",true);
xmlhttp.send();
}
</script>
</head>
<body>
<div id="myDiv"><h2>Let AJAX change this text</h2></div>
<button type="button" onclick="loadXMLDoc()">Change Content</button>
</body>
</html>
PHP:
<?php
$id = $_GET[id];
print "$id";
?>
这是一个没有JQuery的JSFiffle
http://jsfiddle.net/rimian/jurwre07/
function loadXMLDoc() {
var xmlhttp = new XMLHttpRequest();
var url = 'http://echo.jsontest.com/key/value/one/two';
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == XMLHttpRequest.DONE) {
if (xmlhttp.status == 200) {
document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
} else if (xmlhttp.status == 400) {
console.log('There was an error 400');
} else {
console.log('something else other than 200 was returned');
}
}
};
xmlhttp.open("GET", url, true);
xmlhttp.send();
};
loadXMLDoc();
