Cannot assign JSON value to JS variable

Question

I am building a YT mp3 downloader. I'm using API from youtubeinmp3. This is how I am getting the download link in JSON format.

I am using this to assign the JSON value for "link" to JS Variable ytdlink. But it's not working, the value for ytdlink is getting set to "function link() { [native code] }".

function getyt()
            {
            var a = document.getElementById("mquery").value;
                $.get(
                    "http://www.youtubeinmp3.com/fetch/?format=JSON",
                    {
                        video:a
                    },function(data) {
                    console.log(data);
                    var ytdlink = data.link;
                    });

            }

How do I assign the link to a JS variable?

$.get has a fourth parameter "dataType" which you should set to "json". Right now you're getting a string. — James
– James, Commented Apr 12, 2017 at 18:02
@James It should be intelligently guessing and converting it to JSON already as per the docs — mhodges
– mhodges, Commented Apr 12, 2017 at 18:06
@mhodges It should be, but it's not. Why in your answer are you suggesting to separately parse the string to an object if it should already be done? — James
– James, Commented Apr 12, 2017 at 18:13
@James My previous comment was more-so inquisitive of why it was not automatically converting it. I wonder if it has to do with the escaped characters in the link value? I'm honestly not sure.. — mhodges
– mhodges, Commented Apr 12, 2017 at 18:18
youtubeinmp3.com is returning a Content-Type of text/html instead of application/json, is probably why. Note that setting json using the fourth parameter will force the conversion. — James
– James, Commented Apr 12, 2017 at 18:30

mhodges · Accepted Answer · 2017-04-12 18:11:58Z

2

Since link is a function on the String prototype and you are working with a JSON string, that is why it is spitting out

function link() { [native code] }

You must use JSON.parse() on your data in order to access the properties like so:

var obj = JSON.parse(data);
var ytdlink = obj.link;

edited Apr 12, 2017 at 18:11

answered Apr 12, 2017 at 18:00

mhodges

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