如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
这只是一个简单的4步过程，
我希望这对你们有帮助
步骤1。存储对XMLHttpRequest对象的引用
var xmlHttp = createXmlHttpRequestObject();
步骤2。检索XMLHttpRequest对象
function createXmlHttpRequestObject() {
// will store the reference to the XMLHttpRequest object
var xmlHttp;
// if running Internet Explorer
if (window.ActiveXObject) {
try {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
xmlHttp = false;
}
}
// if running Mozilla or other browsers
else {
try {
xmlHttp = new XMLHttpRequest();
} catch (e) {
xmlHttp = false;
}
}
// return the created object or display an error message
if (!xmlHttp)
alert("Error creating the XMLHttpRequest object.");
else
return xmlHttp;
}
步骤3。使用XMLHttpRequest对象进行异步HTTP请求
function process() {
// proceed only if the xmlHttp object isn't busy
if (xmlHttp.readyState == 4 || xmlHttp.readyState == 0) {
// retrieve the name typed by the user on the form
item = encodeURIComponent(document.getElementById("input_item").value);
// execute the your_file.php page from the server
xmlHttp.open("GET", "your_file.php?item=" + item, true);
// define the method to handle server responses
xmlHttp.onreadystatechange = handleServerResponse;
// make the server request
xmlHttp.send(null);
}
}
步骤4。当从服务器接收消息时自动执行
function handleServerResponse() {
// move forward only if the transaction has completed
if (xmlHttp.readyState == 4) {
// status of 200 indicates the transaction completed successfully
if (xmlHttp.status == 200) {
// extract the XML retrieved from the server
xmlResponse = xmlHttp.responseText;
document.getElementById("put_response").innerHTML = xmlResponse;
// restart sequence
}
// a HTTP status different than 200 signals an error
else {
alert("There was a problem accessing the server: " + xmlHttp.statusText);
}
}
}
其他回答
老了，但我会尝试，也许有人会发现这个信息有用。
这是执行GET请求并获取一些JSON格式数据所需的最小代码量。这只适用于现代浏览器，如最新版本的Chrome, FF, Safari, Opera和Microsoft Edge。
const xhr = new XMLHttpRequest();
xhr.open('GET', 'https://example.com/data.json'); // by default async
xhr.responseType = 'json'; // in which format you expect the response to be
xhr.onload = function() {
if(this.status == 200) {// onload called even on 404 etc so check the status
console.log(this.response); // No need for JSON.parse()
}
};
xhr.onerror = function() {
// error
};
xhr.send();
还可以查看新的Fetch API，它是XMLHttpRequest API的基于承诺的替代品。
<html>
<script>
var xmlDoc = null ;
function load() {
if (typeof window.ActiveXObject != 'undefined' ) {
xmlDoc = new ActiveXObject("Microsoft.XMLHTTP");
xmlDoc.onreadystatechange = process ;
}
else {
xmlDoc = new XMLHttpRequest();
xmlDoc.onload = process ;
}
xmlDoc.open( "GET", "background.html", true );
xmlDoc.send( null );
}
function process() {
if ( xmlDoc.readyState != 4 ) return ;
document.getElementById("output").value = xmlDoc.responseText ;
}
function empty() {
document.getElementById("output").value = '<empty>' ;
}
</script>
<body>
<textarea id="output" cols='70' rows='40'><empty></textarea>
<br></br>
<button onclick="load()">Load</button>
<button onclick="empty()">Clear</button>
</body>
</html>
在浏览器中使用纯JavaScript:
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState == XMLHttpRequest.DONE ) {
if(xhr.status == 200){
console.log(xhr.responseText);
} else if(xhr.status == 400) {
console.log('There was an error 400');
} else {
console.log('something else other than 200 was returned');
}
}
}
xhr.open("GET", "mock_data.json", true);
xhr.send();
或者如果你想使用Browserify使用node.js来捆绑你的模块。你可以使用超级代理:
var request = require('superagent');
var url = '/mock_data.json';
request
.get(url)
.end(function(err, res){
if (res.ok) {
console.log('yay got ' + JSON.stringify(res.body));
} else {
console.log('Oh no! error ' + res.text);
}
});
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。
