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);
     }
 });

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);

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。

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);

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();