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。

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

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)
            }                                 
        }

我正在寻找一种方法，包括承诺与ajax和排除jQuery。HTML5 Rocks上有一篇文章谈到了ES6的承诺。(您可以使用像Q这样的承诺库填充)您可以使用我从文章中复制的代码片段。

function get(url) {
  // Return a new promise.
  return new Promise(function(resolve, reject) {
    // Do the usual XHR stuff
    var req = new XMLHttpRequest();
    req.open('GET', url);

    req.onload = function() {
      // This is called even on 404 etc
      // so check the status
      if (req.status == 200) {
        // Resolve the promise with the response text
        resolve(req.response);
      }
      else {
        // Otherwise reject with the status text
        // which will hopefully be a meaningful error
        reject(Error(req.statusText));
      }
    };

    // Handle network errors
    req.onerror = function() {
      reject(Error("Network Error"));
    };

    // Make the request
    req.send();
  });
}

注意:我还写了一篇关于这方面的文章。

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);

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();