如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
<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>
其他回答
使用“vanilla”(普通)JavaScript:
function loadXMLDoc() {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == XMLHttpRequest.DONE) { // XMLHttpRequest.DONE == 4
if (xmlhttp.status == 200) {
document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
}
else if (xmlhttp.status == 400) {
alert('There was an error 400');
}
else {
alert('something else other than 200 was returned');
}
}
};
xmlhttp.open("GET", "ajax_info.txt", true);
xmlhttp.send();
}
jQuery:
$.ajax({
url: "test.html",
context: document.body,
success: function() {
$(this).addClass("done");
}
});
<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>
XMLHttpRequest ()
您可以使用XMLHttpRequest()构造函数创建一个新的XMLHttpRequest(XHR)对象,该对象将允许您使用标准的HTTP请求方法(如GET和POST)与服务器交互:
const data = JSON.stringify({
example_1: 123,
example_2: 'Hello, world!',
});
const request = new XMLHttpRequest();
request.addEventListener('load', function () {
if (this.readyState === 4 && this.status === 200) {
console.log(this.responseText);
}
});
request.open('POST', 'example.php', true);
request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
request.send(data);
fetch ()
你也可以使用fetch()方法获取一个Promise,它解析为响应对象,表示对请求的响应:
const data = JSON.stringify({
example_1: 123,
example_2: 'Hello, world!',
});
fetch('example.php', {
method: 'POST',
headers: {
'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8',
},
body: data,
}).then(response => {
if (response.ok) {
response.text().then(response => {
console.log(response);
});
}
});
领航员sendBeacon()。
另一方面,如果你只是试图POST数据,不需要服务器的响应,最短的解决方案是使用navigator.sendBeacon():
const data = JSON.stringify({
example_1: 123,
example_2: 'Hello, world!',
});
navigator.sendBeacon('example.php', data);
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)
};
这个版本在普通ES6/ES2015中怎么样?
function get(url) {
return new Promise((resolve, reject) => {
const req = new XMLHttpRequest();
req.open('GET', url);
req.onload = () => req.status === 200 ? resolve(req.response) : reject(Error(req.statusText));
req.onerror = (e) => reject(Error(`Network Error: ${e}`));
req.send();
});
}
函数返回一个promise。下面是一个关于如何使用该函数并处理它返回的承诺的示例:
get('foo.txt')
.then((data) => {
// Do stuff with data, if foo.txt was successfully loaded.
})
.catch((err) => {
// Do stuff on error...
});
如果你需要加载一个json文件,你可以使用json .parse()将加载的数据转换为JS对象。
您还可以集成req。responseType='json'到函数中,但不幸的是,没有IE支持它,所以我将坚持使用json .parse()。
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