如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
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
其他回答
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";
?>
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()。
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。
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
alert(this.responseText);
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
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)
};
