使用javascript从URL读取JSON数据

Question

I want to read data from this link http://starlord.hackerearth.com/gamesext . I went through this https://developer.mozilla.org/en-US/docs/Learn/JavaScript/Objects/JSON and was able to obtain data from https://mdn.github.io/learning-area/javascript/oojs/json/superheroes.json . Trying similar approach for getting data from http://starlord.hackerearth.com/gamesext is not working for me. This is how I tried:

var requestURL = 'http://starlord.hackerearth.com/gamesext';
    var request = new XMLHttpRequest();
    request.open('GET', requestURL);
    request.responseType = 'json';
    request.send();
    request.onload = function() {
    var games = request.response;
    document.getElementById("para").innerHTML = "for check";//para is a paragraph id
    fun1(games);
    }
    function fun1(jsonObj){
        //getting first title
        document.getElementById("para").innerHTML = jsonObj[0]["title"];
    }

I would want to know is that data in JSON and how to get it?

Answer 1

Try using the JSON.parse() method:

function fun1(jsonObj){
    //getting first title
    jsonObj = JSON.parse(jsonObj);
    document.getElementById("para").innerHTML = jsonObj[0]["title"];
}

This will turn valid JSON into a javascript object that can be accessed as you are trying to do below.

Answer 2

This works perfectly fine for me:

var requestURL = 'http://starlord.hackerearth.com/gamesext';
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        console.log(xhttp.response[0].title)  # LittleBigPlanet PS Vita
    }
};
xhttp.open("GET", requestURL);
xhttp.responseType = 'json';
xhttp.send();

Give it a try!

Answer 3

Using fetch this is pretty simply.

Below is an example.

 const url = 'https://mdn.github.io/learning-area/javascript/oojs/json/superheroes.json'; async function getData() { const json = await (await fetch(url)).json(); console.log(json); } getData(); 

Answer 4

just put request.send(); after all the code you provided.