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。

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();

您可以使用以下函数:

function callAjax(url, callback){
    var xmlhttp;
    // compatible with IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function(){
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200){
            callback(xmlhttp.responseText);
        }
    }
    xmlhttp.open("GET", url, true);
    xmlhttp.send();
}

你可以在这些链接上尝试类似的解决方案:

https://www.w3schools.com/xml/tryit.asp?filename=tryajax_first https://www.w3schools.com/xml/tryit.asp?filename=tryajax_callback

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();

使用下面的代码片段，你可以很容易地完成类似的事情，就像这样:

ajax.get('/test.php', {foo: 'bar'}, function() {});

以下是片段:

var ajax = {};
ajax.x = function () {
    if (typeof XMLHttpRequest !== 'undefined') {
        return new XMLHttpRequest();
    }
    var versions = [
        "MSXML2.XmlHttp.6.0",
        "MSXML2.XmlHttp.5.0",
        "MSXML2.XmlHttp.4.0",
        "MSXML2.XmlHttp.3.0",
        "MSXML2.XmlHttp.2.0",
        "Microsoft.XmlHttp"
    ];

    var xhr;
    for (var i = 0; i < versions.length; i++) {
        try {
            xhr = new ActiveXObject(versions[i]);
            break;
        } catch (e) {
        }
    }
    return xhr;
};

ajax.send = function (url, callback, method, data, async) {
    if (async === undefined) {
        async = true;
    }
    var x = ajax.x();
    x.open(method, url, async);
    x.onreadystatechange = function () {
        if (x.readyState == 4) {
            callback(x.responseText)
        }
    };
    if (method == 'POST') {
        x.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    }
    x.send(data)
};

ajax.get = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url + (query.length ? '?' + query.join('&') : ''), callback, 'GET', null, async)
};

ajax.post = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url, callback, 'POST', query.join('&'), async)
};

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));