Parsing JSON in jquery Yahoo Weather API

Question

I cannot figure this out at all.

my current json results the following using console.log(json)

I was wondering if someone could help me retrieve the Condition part?

EDIT

                       var geoFORECAST = 'http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20weather.forecast%20where%20woeid=%2726344339%27%20and%20u=%27c%27&format=json';
                    $.getJSON(geoFORECAST, function(json) {
                       console.log(json);
                       console.log(json.query.results.item.condition.text);
                     });

score 1 · Answer 1 · answered Aug 21 '12 at 22:11

1

You need to use parseJSON: http://api.jquery.com/jQuery.parseJSON/

Then just access the Condition var like you would any object property.

answered Aug 21 '12 at 22:11

Madbreaks

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score 1 · Accepted Answer · answered Aug 21 '12 at 22:15

1

Use native JSON object.

var obj = JSON.parse(json);
var condition = obj.query.results.channel.item.condition;

answered Aug 21 '12 at 22:15

Shiplu Mokaddim

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@Madbreaks yes, Its supported in [most modern browser](http://stackoverflow.com/a/891306/376535). – Shiplu Mokaddim Aug 21 '12 at 22:19

Lemex · Answer 3 · 2012-08-21T22:26:33.410

0

Item is already parsed so save what that pages sends back to varible test and code below iwll work

and access things like..

test.query.results.item.condition.text

Which will return a value

To do conditions you could do

var b =  test.query.results.item.condition.text ;

if(b == ???) or even > or < or what ever is required

edited Aug 21 '12 at 22:26

answered Aug 21 '12 at 22:15

Lemex

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I just receive this. SyntaxError: JSON.parse: unexpected character – ngplayground Aug 21 '12 at 22:24
Sorry its still not working.ReferenceError: test is not defined – ngplayground Aug 21 '12 at 22:29

Parsing JSON in jquery Yahoo Weather API

3 Answers3