Create variables for getJSON response

Question

I'm very new to Javascript and JQuery, I've been trying different ways to pull and manipulate yahoo finance data, and have decided on jquery. My first basic attempt was this:

$.getJSON("http://query.yahooapis.com/v1/public/yql?q=select%20Name%2C%20LastTradePriceOnly%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22RHT%22%29&format=json&env=http%3A%2F%2Fdatatables.org%2Falltables.env" + "?callback=?", function(json) {

var lastprice = json[0].results.quote.LastTradePriceOnly

console.log(lastprice)

It did not work, and the error console was not of any help. I searched here and found this question: load json into variable and tried this after thinking maybe the response hadn't yet been received from yahoo:

var json = (function () {
    var json = null;
    $.ajax({
        'async': false,
        'global': false,
        'url': "http://query.yahooapis.com/v1/public/yql?q=select%20Name%2C%20LastTradePriceOnly%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22RHT%22%29&format=json&env=http%3A%2F%2Fdatatables.org%2Falltables.env",
        'dataType': "json",
        'success': function (data) {
            json = data;
        }
    });
    return json;
})(); 

    var lastprice = json.results.quote.LastTradePriceOnly

    console.log(lastprice)

});

This too wasn't right. I feel like I'm close. Any help would be greatly appreciated

juan.facorro · Accepted Answer · 2012-10-26T19:14:37.490

Remove the callback for the url and you get the following JSON result from the query, which is not an array but an object.

{
   "query":{
      "count":1,
      "created":"2012-10-26T19:00:18Z",
      "lang":"en-US",
      "results":{
         "quote":{
            "LastTradePriceOnly":"50.26",
            "Name":"Red Hat, Inc. Com"
         }
      }
   }
}

You should acces its data in the following way:

var lastprice = json.query.results.quote.LastTradePriceOnly;

Here's a working demo for the modified code: DEMO.

EDIT (WARNING)

I was testing that call and I'm seeing that sometimes the result from that service with the same URL returns the following error:

{
   "error":{
      "lang":"en-US",
      "description":"No definition found for Table yahoo.finance.quotes"
   }
}

I was half-way through the same answer when you beat me too it. For reference to the OP, try pasting the url into [JSONLint](http://jsonlint.com/#) to check the format. — Matt Burland, Oct 26 '12 at 19:05

score 0 · Answer 2 · answered Oct 26 '12 at 19:05

data will come back in this format:

{"query":
       {"count":1,"created":"2012-10-26T19:00:42Z","lang":"en-US","results":
          {"quote":
              {"LastTradePriceOnly":"50.28","Name":"Red Hat, Inc. Com"}}}}

so the object you get back will have to be traversed : data.query.results.quote.LastTradePriceOnly

>     $.ajax({
>         'async': false,
>         'global': false,
>         'url': "http://query.yahooapis.com/v1/public/yql?q=select%20Name%2C%20LastTradePriceOnly%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22RHT%22%29&format=json&env=http%3A%2F%2Fdatatables.org%2Falltables.env&callback=?",
>         'dataType': "json",
>         'success': function (data) {
>             alert(data.query.results.quote.LastTradePriceOnly);
>         }
>     });

Here is a jsfiddle which demonstrates it: http://jsfiddle.net/erick382/Qghtu/

Thank you (feel dumb for getting that path wrong), I tried adding a variable to what you wrote by changing the end to this (but it didn't work): 'success': function (data) { json = data; } var last = data.query.results.quote.LastTradePriceOnly console.log(last) }); — majordomo, Oct 26 '12 at 19:39
Hmmmmm... Did you look at the jsfiddle? I know I simplified it.. But it might help you troubleshoot.. — ek_ny, Oct 30 '12 at 16:34

Create variables for getJSON response

2 Answers2