Question

我能够使用jquery和es5创建一个ajax请求，但我想转换我的代码，以便它的vanilla和使用es6。这个请求将如何变化。（注意：我正在查询维基百科的api）。

      var link = "https://en.wikipedia.org/w/api.php?action=query&prop=info&pageids="+ page +"&format=json&callback=?";

    $.ajax({
      type: "GET",
      url: link,
      contentType: "application/json; charset=utf-8",
      async: false,
      dataType: "json",
      success:function(re){
    },
      error:function(u){
        console.log("u")
        alert("sorry, there are no results for your search")
    }

Answer 1

您可能会使用fetch API：

fetch(link, { headers: { "Content-Type": "application/json; charset=utf-8" }})
    .then(res => res.json()) // parse response as JSON (can be res.text() for plain response)
    .then(response => {
        // here you do what you want with response
    })
    .catch(err => {
        console.log("u")
        alert("sorry, there are no results for your search")
    });

如果你想制作 async ，那是不可能的。但是你可以使用Async-Await功能看起来不像异步操作。

Answer 2

AJAX请求对于异步发送数据，获取响应，检查并通过更新其内容应用于当前网页非常有用。

function ajaxRequest()
{
    var link = "https://en.wikipedia.org/w/api.php?action=query&prop=info&pageids="+ page +"&format=json&callback=?";
    var xmlHttp = new XMLHttpRequest(); // creates 'ajax' object
        xmlHttp.onreadystatechange = function() //monitors and waits for response from the server
        {
           if(xmlHttp.readyState === 4 && xmlHttp.status === 200) //checks if response was with status -> "OK"
           {
               var re = JSON.parse(xmlHttp.responseText); //gets data and parses it, in this case we know that data type is JSON. 
               if(re["Status"] === "Success")
               {//doSomething}
               else 
               {
                   //doSomething
               }
           }

        }
        xmlHttp.open("GET", link); //set method and address
        xmlHttp.send(); //send data

}

Answer 3

如今，您无需使用jQuery或任何API。就这么简单：

var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
  if (this.readyState === 4 && this.status === 200) {
    console.log(this.responseText);
  }
};
xmlhttp.open('GET', 'https://www.example.com');
xmlhttp.send();

es6 vanilla javascript中的Ajax请求

3 个答案: