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();
}

尝试使用Fetch Api (Fetch Api)

fetch('http://example.com/movies.json').then(response => response.json()).then(data => console.log(data));

非常清澈，100%香草味。

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

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);
     }
 });

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。

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");
    }
});