Welcome to the Treehouse Community

The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

JavaScript AJAX Basics (retiring) AJAX and APIs Call the jQuery $.getJSON method

Raeed Sabree
Raeed Sabree
10,664 Points

Everything's in place to send the AJAX request: you've specified the URL, a data object, and a callback function. Use jQ

Everything's in place to send the AJAX request: you've specified the URL, a data object, and a callback function. Use jQuery's $.getJSON() method to make the AJAX call.

what am i doing wrong?

weather.js
$(document).ready(function() {

  var weatherAPI = 'http://api.openweathermap.org/data/2.5/weather';
  var data = {
    q : "Portland,OR",
    units : "metric"
  };
  function showWeather(weatherReport) {
    $('#temperature').text(weatherReport.main.temp);
  }
  $.getJSON(weatherAPI, data,  function )
});
index.html
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>What's the Weather Like?</title>
  <script src="jquery.js"></script>
  <script src="weather.js"></script>
</head>
<body>
  <div id="main">
    <h1>Current temperature: <span id="temperature"></span>&deg;</h1>
  </div>
</body>
</html>

1 Answer

Simon Coates
Simon Coates
28,693 Points

$.getJSON(weatherAPI, data, function ) - function is a keyword, i think. YOu probably want to be passing in a function variable. the following seems to pass test

$(document).ready(function() {

  var weatherAPI = 'http://api.openweathermap.org/data/2.5/weather';
  var data = {
    q : "Portland,OR",
    units : "metric"
  };
  function showWeather(weatherReport) {
    $('#temperature').text(weatherReport.main.temp);
  }
  $.getJSON(weatherAPI, data,  showWeather )
});
Raeed Sabree
Raeed Sabree
10,664 Points

so what do i put for the last thing? i put function and that didnt work then i put function variable and then that didnt work

Simon Coates
Simon Coates
28,693 Points

the code i posted above (you may need to refresh) works.