Question

How can I build a minimal working sample on a site like codepen showing a location and it's temperature using the Yahoo weather API. I need specifically San Diego, CA. And using only HTML and Javascript, not PHP.

Background

I did check the site for a similar question but it only addressed temperature Getting only temperature from Yahoo Weather but it's only answer linked to an overplicated tutorial with excessive code.

Other answers on the site only have YML but don't show how to integrate an entire working example.

I was following along to the documentation from Yahoo but there is no working example like how NASA has a live example

Code

I have this CodePen demo

HTML

<div id="output"></div>

Javascript

$(document).ready(function () {
    $.getJSON(".condition%20from%20weather.forecast%20where%20woeid%20%3D%202487889&format=json&env=store%3A%2F%2Fdatatables%2Falltableswithkeys/", function (data) {
        console.log(data);
        console.log(query)
        $('#output').append( 'The temperature in' + result.location.["location"] + 'is' + result.condition.["temp"] );
    })
})

Question

How can I build a minimal working sample on a site like codepen showing a location and it's temperature using the Yahoo weather API. I need specifically San Diego, CA. And using only HTML and Javascript, not PHP.

Background

I did check the site for a similar question but it only addressed temperature Getting only temperature from Yahoo Weather but it's only answer linked to an overplicated tutorial with excessive code.

Other answers on the site only have YML but don't show how to integrate an entire working example.

I was following along to the documentation from Yahoo but there is no working example like how NASA has a live example

Code

I have this CodePen demo

HTML

<div id="output"></div>

Javascript

$(document).ready(function () {
    $.getJSON("https://query.yahooapis./v1/public/yql?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20%3D%202487889&format=json&env=store%3A%2F%2Fdatatables%2Falltableswithkeys/", function (data) {
        console.log(data);
        console.log(query)
        $('#output').append( 'The temperature in' + result.location.["location"] + 'is' + result.condition.["temp"] );
    })
})

• javascript
• yahoo-weather-api

• You have lots of problems with your example like accessing variables that dont exist – Kevin Jantzer Commented Feb 6, 2017 at 19:02
• Check your console and evaluate data. Also, there are a few errors in your demo... First, your syntax is incorrect. result.location.["location"] would throw a parsing error. Also, query is never defined. – Dom Commented Feb 6, 2017 at 19:03

3 Answers 3

Here's a working example based on your original code.

Something to note: you were doing this result.location.["location"] Which is invalid. You could use result.location["location"] or result.location.location (neither of which are returned in your result btw)

var queryURL = "https://query.yahooapis./v1/public/yql?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20%3D%202487889&format=json&env=store%3A%2F%2Fdatatables%2Falltableswithkeys/";

$.getJSON(queryURL, function (data) {
  
  var results = data.query.results
  var firstResult = results.channel.item.condition
  console.log(firstResult);
  
  var location = 'Unknown' // not returned in response
  var temp = firstResult.temp
  var text = firstResult.text
  
  $('#output').append('The temperature is ' + temp + '. Forecast calls for '+text);
  
})

<script src="https://ajax.googleapis./ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="output"></div>

Update

Your API query doesn't return location because you have it limited to select item.condition

Change q=select%20item.condition to q=select%20* andd you get a lot more data returned, including location.

Couple of things here:

• You are trying to access the location and weather data incorrectly. You should be using data.location and data.weather since you are passing the JSON into the function as data in the function (data) section.
• Your API call is not being made properly. Review the documentation here and try to make the call again. https://developer.yahoo./weather/

This example does not have any excessive code and would be a great place to start: https://developer.yahoo./weather/#get-started

Based on the accepted answer I made one modification to account for the location. It's woeid has to be looked up using something like http://woeid.rosselliot.co.nz/ and then defined as a variable, in my case it was San Diego.

The resulting Javascript was

var queryURL = "https://query.yahooapis./v1/public/yql?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20%3D%202487889&format=json&env=store%3A%2F%2Fdatatables%2Falltableswithkeys/";

$.getJSON(queryURL, function (data) {

  var results = data.query.results
  var firstResult = results.channel.item.condition
  console.log(firstResult);

  var location = 'San Diego'
  var temp = firstResult.temp
  var text = firstResult.text

  $('#output').append('The temperature in ' + location + ' is ' + temp + '. Forecast looks '+ text);

})

full working demo is at http://codepen.io/JGallardo/pen/XpBMRX