如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
尝试使用Fetch Api (Fetch Api)
fetch('http://example.com/movies.json').then(response => response.json()).then(data => console.log(data));
非常清澈,100%香草味。
其他回答
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);
}
});
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)
};
这可能会有帮助:
function doAjax(url, callback) {
var xmlhttp = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
callback(xmlhttp.responseText);
}
}
xmlhttp.open("GET", url, true);
xmlhttp.send();
}
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
alert(this.responseText);
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
<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>
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