如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
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");
}
});
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";
?>
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.');
}
}
}
From youMightNotNeedJquery.com + JSON.stringify
var request = new XMLHttpRequest();
request.open('POST', '/my/url', true);
request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
request.send(JSON.stringify(data));
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
