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);
        }
    }
}

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);

xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

XMLHttpRequest ()

您可以使用XMLHttpRequest()构造函数创建一个新的XMLHttpRequest(XHR)对象，该对象将允许您使用标准的HTTP请求方法(如GET和POST)与服务器交互:

const data = JSON.stringify({
  example_1: 123,
  example_2: 'Hello, world!',
});

const request = new XMLHttpRequest();

request.addEventListener('load', function () {
  if (this.readyState === 4 && this.status === 200) {
    console.log(this.responseText);
  }
});

request.open('POST', 'example.php', true);
request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
request.send(data);

fetch ()

你也可以使用fetch()方法获取一个Promise，它解析为响应对象，表示对请求的响应:

const data = JSON.stringify({
  example_1: 123,
  example_2: 'Hello, world!',
});

fetch('example.php', {
  method: 'POST',
  headers: {
    'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8',
  },
  body: data,
}).then(response => {
  if (response.ok) {
    response.text().then(response => {
      console.log(response);
    });
  }
});

领航员sendBeacon()。

另一方面，如果你只是试图POST数据，不需要服务器的响应，最短的解决方案是使用navigator.sendBeacon():

const data = JSON.stringify({
  example_1: 123,
  example_2: 'Hello, world!',
});

navigator.sendBeacon('example.php', data);

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));

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。