如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
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)
};
其他回答
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)
};
我正在寻找一种方法，包括承诺与ajax和排除jQuery。HTML5 Rocks上有一篇文章谈到了ES6的承诺。(您可以使用像Q这样的承诺库填充)您可以使用我从文章中复制的代码片段。
function get(url) {
// Return a new promise.
return new Promise(function(resolve, reject) {
// Do the usual XHR stuff
var req = new XMLHttpRequest();
req.open('GET', url);
req.onload = function() {
// This is called even on 404 etc
// so check the status
if (req.status == 200) {
// Resolve the promise with the response text
resolve(req.response);
}
else {
// Otherwise reject with the status text
// which will hopefully be a meaningful error
reject(Error(req.statusText));
}
};
// Handle network errors
req.onerror = function() {
reject(Error("Network Error"));
};
// Make the request
req.send();
});
}
注意:我还写了一篇关于这方面的文章。
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";
?>
下面的几个例子的一个小组合，创造了这个简单的作品:
function ajax(url, method, data, async)
{
method = typeof method !== 'undefined' ? method : 'GET';
async = typeof async !== 'undefined' ? async : false;
if (window.XMLHttpRequest)
{
var xhReq = new XMLHttpRequest();
}
else
{
var xhReq = new ActiveXObject("Microsoft.XMLHTTP");
}
if (method == 'POST')
{
xhReq.open(method, url, async);
xhReq.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
xhReq.send(data);
}
else
{
if(typeof data !== 'undefined' && data !== null)
{
url = url+'?'+data;
}
xhReq.open(method, url, async);
xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
xhReq.send(null);
}
//var serverResponse = xhReq.responseText;
//alert(serverResponse);
}
// Example usage below (using a string query):
ajax('http://www.google.com');
ajax('http://www.google.com', 'POST', 'q=test');
或者如果你的参数是object(s) -轻微的额外代码调整:
var parameters = {
q: 'test'
}
var query = [];
for (var key in parameters)
{
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(parameters[key]));
}
ajax('http://www.google.com', 'POST', query.join('&'));
两者都应该完全兼容浏览器+版本。
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);
}
