如何使用JavaScript进行AJAX调用，而不使用jQuery?
当前回答
这里有一个非常好的纯javascript解决方案
/*create an XMLHttpRequest object*/
let GethttpRequest=function(){
let httpRequest=false;
if(window.XMLHttpRequest){
httpRequest =new XMLHttpRequest();
if(httpRequest.overrideMimeType){
httpRequest.overrideMimeType('text/xml');
}
}else if(window.ActiveXObject){
try{httpRequest =new ActiveXObject("Msxml2.XMLHTTP");
}catch(e){
try{
httpRequest =new ActiveXObject("Microsoft.XMLHTTP");
}catch(e){}
}
}
if(!httpRequest){return 0;}
return httpRequest;
}
/*Defining a function to make the request every time when it is needed*/
function MakeRequest(){
let uriPost ="myURL";
let xhrPost =GethttpRequest();
let fdPost =new FormData();
let date =new Date();
/*data to be sent on server*/
let data = {
"name" :"name",
"lName" :"lName",
"phone" :"phone",
"key" :"key",
"password" :"date"
};
let JSONdata =JSON.stringify(data);
fdPost.append("data",JSONdata);
xhrPost.open("POST" ,uriPost, true);
xhrPost.timeout = 9000;/*the time you need to quit the request if it is not completed*/
xhrPost.onloadstart = function (){
/*do something*/
};
xhrPost.onload = function (){
/*do something*/
};
xhrPost.onloadend = function (){
/*do something*/
}
xhrPost.onprogress =function(){
/*do something*/
}
xhrPost.onreadystatechange =function(){
if(xhrPost.readyState < 4){
}else if(xhrPost.readyState === 4){
if(xhrPost.status === 200){
/*request succesfull*/
}else if(xhrPost.status !==200){
/*request failled*/
}
}
}
xhrPost.ontimeout = function (e){
/*you can stop the request*/
}
xhrPost.onerror = function (){
/*you can try again the request*/
};
xhrPost.onabort = function (){
/*you can try again the request*/
};
xhrPost.overrideMimeType("text/plain; charset=x-user-defined-binary");
xhrPost.setRequestHeader("Content-disposition", "form-data");
xhrPost.setRequestHeader("X-Requested-With","xmlhttprequest");
xhrPost.send(fdPost);
}
/*PHP side
<?php
//check if the variable $_POST["data"] exists isset() && !empty()
$data =$_POST["data"];
$decodedData =json_decode($_POST["data"]);
//show a single item from the form
echo $decodedData->name;
?>
*/
/*Usage*/
MakeRequest();
其他回答
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
alert(this.responseText);
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
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。
<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>
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();
}
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");
}
});
