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

Leo Marco Corpuz
Leo Marco Corpuz
18,975 Points

Why is my $.getJSON not working?

For the code challenge, the url and data are stored in variables.I stored the callback function in a variable also. Now I just put these variables as arguments to $.getJSON() and I get an error saying I need to use the $.getJSON() storing the arguments in proper order. I'm lost.

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

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

  $.getJSON(weatherAPI, data, response);

});
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>

2 Answers

Try this:

$(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);

});

The response will be invalid because openweathermap requires an API key now. But it will get you through the quiz.

hamdi taboubi
hamdi taboubi
3,627 Points

remove $.getJSON(weatherAPI, data, response);