jQuery:

$.get(
    "somepage.php",
    {paramOne : 1, paramX : 'abc'},
    function(data) {
       alert('page content: ' + data);
    }
);

要做到这一点，建议使用Fetch API，使用JavaScript Promises。XMLHttpRequest (XHR)、IFrame对象或动态<script>标记是较旧(且较笨重)的方法。

<script type=“text/javascript”> 
    // Create request object 
    var request = new Request('https://example.com/api/...', 
         { method: 'POST', 
           body: {'name': 'Klaus'}, 
           headers: new Headers({ 'Content-Type': 'application/json' }) 
         });
    // Now use it! 

   fetch(request) 
   .then(resp => { 
         // handle response 
   }) 
   .catch(err => { 
         // handle errors 
    });
</script>

这里有一个很棒的获取演示和MDN文档

<button type="button" onclick="loadXMLDoc()"> GET CONTENT</button>

 <script>
        function loadXMLDoc() {
            var xmlhttp = new XMLHttpRequest();
            var url = "<Enter URL>";``
            xmlhttp.onload = function () {
                if (xmlhttp.readyState == 4 && xmlhttp.status == "200") {
                    document.getElementById("demo").innerHTML = this.responseText;
                }
            }
            xmlhttp.open("GET", url, true);
            xmlhttp.send();
        }
    </script>

在纯javascript和返回一个承诺:

  httpRequest = (url, method = 'GET') => {
    return new Promise((resolve, reject) => {
      const xhr = new XMLHttpRequest();
      xhr.open(method, url);
      xhr.onload = () => {
        if (xhr.status === 200) { resolve(xhr.responseText); }
        else { reject(new Error(xhr.responseText)); }
      };
      xhr.send();
    });
  }

短的、干净的:

const http = new XMLHttpRequest() http。打开(“得到”,“https://api.lyrics.ovh/v1/toto/africa”) http.send () http。onload = () => console.log(http.response)

现代、干净、简洁

fetch('https://baconipsum.com/api/?type=1')

让url = 'https://baconipsum.com/api/?type=all-meat&paras=1&start-with-lorem=2'; //只发送GET请求而不等待响应 fetch (url); //使用then来等待结果 获取(url)。然后(r = > r.json()。then(j=> console.log('\nREQUEST 2'，j))); //或async/await (异步()= > console.log('\nREQUEST 3'， await(await fetch(url)).json()) ) (); 打开Chrome控制台网络选项卡查看请求