如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
alert(this.responseText);
}
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();
其他回答
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();
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();
}
下面的几个例子的一个小组合,创造了这个简单的作品:
function ajax(url, method, data, async)
{
method = typeof method !== 'undefined' ? method : 'GET';
async = typeof async !== 'undefined' ? async : false;
if (window.XMLHttpRequest)
{
var xhReq = new XMLHttpRequest();
}
else
{
var xhReq = new ActiveXObject("Microsoft.XMLHTTP");
}
if (method == 'POST')
{
xhReq.open(method, url, async);
xhReq.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
xhReq.send(data);
}
else
{
if(typeof data !== 'undefined' && data !== null)
{
url = url+'?'+data;
}
xhReq.open(method, url, async);
xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
xhReq.send(null);
}
//var serverResponse = xhReq.responseText;
//alert(serverResponse);
}
// Example usage below (using a string query):
ajax('http://www.google.com');
ajax('http://www.google.com', 'POST', 'q=test');
或者如果你的参数是object(s) -轻微的额外代码调整:
var parameters = {
q: 'test'
}
var query = [];
for (var key in parameters)
{
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(parameters[key]));
}
ajax('http://www.google.com', 'POST', query.join('&'));
两者都应该完全兼容浏览器+版本。
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。
这是一个没有JQuery的JSFiffle
http://jsfiddle.net/rimian/jurwre07/
function loadXMLDoc() {
var xmlhttp = new XMLHttpRequest();
var url = 'http://echo.jsontest.com/key/value/one/two';
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == XMLHttpRequest.DONE) {
if (xmlhttp.status == 200) {
document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
} else if (xmlhttp.status == 400) {
console.log('There was an error 400');
} else {
console.log('something else other than 200 was returned');
}
}
};
xmlhttp.open("GET", url, true);
xmlhttp.send();
};
loadXMLDoc();