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
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
这只是一个简单的4步过程,
我希望这对你们有帮助
步骤1。存储对XMLHttpRequest对象的引用
var xmlHttp = createXmlHttpRequestObject();
步骤2。检索XMLHttpRequest对象
function createXmlHttpRequestObject() {
// will store the reference to the XMLHttpRequest object
var xmlHttp;
// if running Internet Explorer
if (window.ActiveXObject) {
try {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
xmlHttp = false;
}
}
// if running Mozilla or other browsers
else {
try {
xmlHttp = new XMLHttpRequest();
} catch (e) {
xmlHttp = false;
}
}
// return the created object or display an error message
if (!xmlHttp)
alert("Error creating the XMLHttpRequest object.");
else
return xmlHttp;
}
步骤3。使用XMLHttpRequest对象进行异步HTTP请求
function process() {
// proceed only if the xmlHttp object isn't busy
if (xmlHttp.readyState == 4 || xmlHttp.readyState == 0) {
// retrieve the name typed by the user on the form
item = encodeURIComponent(document.getElementById("input_item").value);
// execute the your_file.php page from the server
xmlHttp.open("GET", "your_file.php?item=" + item, true);
// define the method to handle server responses
xmlHttp.onreadystatechange = handleServerResponse;
// make the server request
xmlHttp.send(null);
}
}
步骤4。当从服务器接收消息时自动执行
function handleServerResponse() {
// move forward only if the transaction has completed
if (xmlHttp.readyState == 4) {
// status of 200 indicates the transaction completed successfully
if (xmlHttp.status == 200) {
// extract the XML retrieved from the server
xmlResponse = xmlHttp.responseText;
document.getElementById("put_response").innerHTML = xmlResponse;
// restart sequence
}
// a HTTP status different than 200 signals an error
else {
alert("There was a problem accessing the server: " + xmlHttp.statusText);
}
}
}
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));
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");
}
});
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();