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);
        }
    }
}

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。

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。

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);
     }
 });