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

使用XMLHttpRequest。

简单的GET请求

httpRequest = new XMLHttpRequest()
httpRequest.open('GET', 'http://www.example.org/some.file')
httpRequest.send()

简单的POST请求

httpRequest = new XMLHttpRequest()
httpRequest.open('POST', 'http://www.example.org/some/endpoint')
httpRequest.send('some data')

我们可以通过可选的第三个参数指定请求应该是异步(true)(默认值)或同步(false)。

// Make a synchronous GET request
httpRequest.open('GET', 'http://www.example.org/some.file', false)

我们可以在调用httpRequest.send()之前设置头信息

httpRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

我们可以通过设置httpRequest来处理响应。在调用httpRequest.send()之前，onreadystatechange函数

httpRequest.onreadystatechange = function(){
  // Process the server response here.
  if (httpRequest.readyState === XMLHttpRequest.DONE) {
    if (httpRequest.status === 200) {
      alert(httpRequest.responseText);
    } else {
      alert('There was a problem with the request.');
    }
  }
}

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();
}

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

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。