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

Marcin Robert Kaźmierczak
Marcin Robert Kaźmierczak
33,570 Points
Posted by Marcin Robert Kaźmierczak
Marcin Robert Kaźmierczak
33,570 Points

I'm stucked on the Final AJAX challenge...

$(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, weatherReport, data);
  }

});

1 Answer

Marcin Robert Kaźmierczak
Marcin Robert Kaźmierczak
33,570 Points
Marcin Robert Kaźmierczak
Marcin Robert Kaźmierczak
33,570 Points

Ok. I've got it :) There was an error with the JSON function and placement of this function. Final code is:

$(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);  
});
Alan Fidelino
Alan Fidelino
12,846 Points
Alan Fidelino
Alan Fidelino
12,846 Points

Thanks. I actually made a mistake of passing the 3rd parameter - showWeather as showWeather(), hence my code didn't work.