如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
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。
其他回答
快速代码获取没有jQuery
async function product_serach(word) {
var response = await fetch('<?php echo base_url(); ?>home/product_search?search='+word);
var json = await response.json();
for (let [key, value] of Object.entries(json))
{
console.log(json)
}
}
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.');
}
}
}
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。
如果您不想包含JQuery,我建议您尝试一些轻量级AJAX库。
我最喜欢的是reqwest。它只有3.4kb,构建得非常好:https://github.com/ded/Reqwest
下面是一个带有reqwest的GET请求示例:
reqwest({
url: url,
method: 'GET',
type: 'json',
success: onSuccess
});
现在,如果您想要更轻量级的东西,我将尝试仅需0.4kb的microAjax: https://code.google.com/p/microajax/
这是所有的代码:
function microAjax(B,A){this.bindFunction=function(E,D){return function(){return E.apply(D,[D])}};this.stateChange=function(D){if(this.request.readyState==4){this.callbackFunction(this.request.responseText)}};this.getRequest=function(){if(window.ActiveXObject){return new ActiveXObject("Microsoft.XMLHTTP")}else{if(window.XMLHttpRequest){return new XMLHttpRequest()}}return false};this.postBody=(arguments[2]||"");this.callbackFunction=A;this.url=B;this.request=this.getRequest();if(this.request){var C=this.request;C.onreadystatechange=this.bindFunction(this.stateChange,this);if(this.postBody!==""){C.open("POST",B,true);C.setRequestHeader("X-Requested-With","XMLHttpRequest");C.setRequestHeader("Content-type","application/x-www-form-urlencoded");C.setRequestHeader("Connection","close")}else{C.open("GET",B,true)}C.send(this.postBody)}};
下面是一个示例调用:
microAjax(url, onSuccess);
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();