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();
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();
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.');
}
}
}
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。
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");
}
});
