一种支持旧浏览器的解决方案:

function httpRequest() {
    var ajax = null,
        response = null,
        self = this;

    this.method = null;
    this.url = null;
    this.async = true;
    this.data = null;

    this.send = function() {
        ajax.open(this.method, this.url, this.asnyc);
        ajax.send(this.data);
    };

    if(window.XMLHttpRequest) {
        ajax = new XMLHttpRequest();
    }
    else if(window.ActiveXObject) {
        try {
            ajax = new ActiveXObject("Msxml2.XMLHTTP.6.0");
        }
        catch(e) {
            try {
                ajax = new ActiveXObject("Msxml2.XMLHTTP.3.0");
            }
            catch(error) {
                self.fail("not supported");
            }
        }
    }

    if(ajax == null) {
        return false;
    }

    ajax.onreadystatechange = function() {
        if(this.readyState == 4) {
            if(this.status == 200) {
                self.success(this.responseText);
            }
            else {
                self.fail(this.status + " - " + this.statusText);
            }
        }
    };
}

这段代码可能有点过分，但绝对是安全的。

用法:

//create request with its porperties
var request = new httpRequest();
request.method = "GET";
request.url = "https://example.com/api?parameter=value";

//create callback for success containing the response
request.success = function(response) {
    console.log(response);
};

//and a fail callback containing the error
request.fail = function(error) {
    console.log(error);
};

//and finally send it away
request.send();

简单的异步请求:

function get(url, callback) {
  var getRequest = new XMLHttpRequest();

  getRequest.open("get", url, true);

  getRequest.addEventListener("readystatechange", function() {
    if (getRequest.readyState === 4 && getRequest.status === 200) {
      callback(getRequest.responseText);
    }
  });

  getRequest.send();
}

为了刷新来自joann的最佳答案，这是我的代码:

let httpRequestAsync = (method, url) => {
    return new Promise(function (resolve, reject) {
        var xhr = new XMLHttpRequest();
        xhr.open(method, url);
        xhr.onload = function () {
            if (xhr.status == 200) {
                resolve(xhr.responseText);
            }
            else {
                reject(new Error(xhr.responseText));
            }
        };
        xhr.send();
    });
}

Ajax

你最好使用Prototype或jQuery这样的库。

function get(path) {
    var form = document.createElement("form");
    form.setAttribute("method", "get");
    form.setAttribute("action", path);
    document.body.appendChild(form);
    form.submit();
}


get('/my/url/')

同样的事情也可以为post request做。 看看这个链接JavaScript post请求像一个表单提交

现在使用异步js，我们可以使用fetch()方法以更简洁的方式做出承诺。所有现代浏览器都支持异步函数。

async函数funcName(url){ Const response = await fetch(url); Var data = await response.json(); ｝