如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
使用XMLHttpRequest。
简单的GET请求
httpRequest = new XMLHttpRequest()
httpRequest.open('GET', 'http://www.example.org/some.file')
httpRequest.send()
简单的POST请求
httpRequest = new XMLHttpRequest()
httpRequest.open('POST', 'http://www.example.org/some/endpoint')
httpRequest.send('some data')
我们可以通过可选的第三个参数指定请求应该是异步(true)(默认值)或同步(false)。
// Make a synchronous GET request
httpRequest.open('GET', 'http://www.example.org/some.file', false)
我们可以在调用httpRequest.send()之前设置头信息
httpRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
我们可以通过设置httpRequest来处理响应。在调用httpRequest.send()之前，onreadystatechange函数
httpRequest.onreadystatechange = function(){
// Process the server response here.
if (httpRequest.readyState === XMLHttpRequest.DONE) {
if (httpRequest.status === 200) {
alert(httpRequest.responseText);
} else {
alert('There was a problem with the request.');
}
}
}
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);
}
});
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";
?>
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
alert(this.responseText);
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
使用XMLHttpRequest。
简单的GET请求
httpRequest = new XMLHttpRequest()
httpRequest.open('GET', 'http://www.example.org/some.file')
httpRequest.send()
简单的POST请求
httpRequest = new XMLHttpRequest()
httpRequest.open('POST', 'http://www.example.org/some/endpoint')
httpRequest.send('some data')
我们可以通过可选的第三个参数指定请求应该是异步(true)(默认值)或同步(false)。
// Make a synchronous GET request
httpRequest.open('GET', 'http://www.example.org/some.file', false)
我们可以在调用httpRequest.send()之前设置头信息
httpRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
我们可以通过设置httpRequest来处理响应。在调用httpRequest.send()之前，onreadystatechange函数
httpRequest.onreadystatechange = function(){
// Process the server response here.
if (httpRequest.readyState === XMLHttpRequest.DONE) {
if (httpRequest.status === 200) {
alert(httpRequest.responseText);
} else {
alert('There was a problem with the request.');
}
}
}
现在在现代浏览器中有一个更好的本机Fetch API可用。fetch()方法允许您发出web请求。 例如，从/get-data请求一些JSON:
let options = {
method: 'GET',
headers: {}
};
fetch('/get-data', options)
.then(response => response.json())
.then(body => {
// Do something with body
});
更多细节请参见MDN Web Docs: Using Fetch API。
