如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
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);
}
}
}
其他回答
在浏览器中使用纯JavaScript:
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState == XMLHttpRequest.DONE ) {
if(xhr.status == 200){
console.log(xhr.responseText);
} else if(xhr.status == 400) {
console.log('There was an error 400');
} else {
console.log('something else other than 200 was returned');
}
}
}
xhr.open("GET", "mock_data.json", true);
xhr.send();
或者如果你想使用Browserify使用node.js来捆绑你的模块。你可以使用超级代理:
var request = require('superagent');
var url = '/mock_data.json';
request
.get(url)
.end(function(err, res){
if (res.ok) {
console.log('yay got ' + JSON.stringify(res.body));
} else {
console.log('Oh no! error ' + res.text);
}
});
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();
}
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);
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。
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");
}
});