How to get data from window.fetch() response?

Question

I'm trying to use window.fetch() to get json from the server, but can't get the data from the response. I have this code:

let url =
'https://api.flightstats.com/flex/schedules/rest/v1/json/from/SYD/to/MEL/departing/2016/3/28?appId=f4b1b6c9&appKey=59bd8f0274f2ae88aebd2c1db7794f7f';
let request = new Request (url, {
        method: 'GET',
        mode: 'no-cors'
    });

 fetch(request)
        .then(function(response){
            console.log(response)
        });

It seems that this request is successfull, I see status 200 and response body with json in network tab - status and response. But in console.log I dont see json object - console log image

I cant understand why I dont see json in console.log

Answer 1

The host site you are requesting from does not appear to support CORS. As such, you can't use fetch() to make a cross origin request and get the data back. If, you change your fetch() request to mode: 'cors', the debug console will show that the host site does not offer CORS headers to allow the browser to show you the result of the request.

When you are using mode: 'no-cors', the browser is hiding the result from you (because you don't have permission to see it) and you can see the response is tagged as opaque.

In a little poking around on the api.flightstats.com site, I did see that it supports JSONP which will allow you to work around the lack of CORS support issue and successfully complete a cross origin request.

For simplicity of showing that it can work, I used jQuery to just prove that a JSONP request can be made. Here's that code in a working snippet. Note I changed the URL from /json/ to /jsonp/ and specific dataType: "jsonp" in the jQuery request. This causes jQuery to add the callback=xxxxx query parameter and to fetch the response via that corresponding script (the JSONP method).

var url =
'https://api.flightstats.com/flex/schedules/rest/v1/jsonp/from/SYD/to/MEL/departing/2016/3/28?appId=f4b1b6c9&appKey=59bd8f0274f2ae88aebd2c1db7794f7f';

$.ajax(url, {dataType: "jsonp"}).then(function(response) {
	log(response);
}, function(err) {
    log("$.ajax() error:")
	log(err);
})

<script src="http://files.the-friend-family.com/log.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>