Albert Collazo-Delvalle
7,223 Pointscan anyone spot the issue with my $.getJSON() method on line 8?
Bummer: You didn't correctly call the $.getJSON() method. To use it, pass it three arguments: the URL, the data you want to send to the server, and the callback function.
$(document).ready(function() {
var weatherAPI = "http://api.openweathermap.org/data/2.5/weather";
var data = {
q : "Portland,OR",
units : "metric"
};
$.getJSON(weatherAPI, data, function (weatherReport) {
$('#temperature').text(weatherReport.main.temp);
});
});
<!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>°</h1>
</div>
</body>
</html>
2 Answers
Clayton Perszyk
Treehouse Moderator 46,143 PointsYou're code is correct, but to complete the challenge pass the function that is already present to getJSON (i.e. do not use an anonymous function).
Clayton Perszyk
Treehouse Moderator 46,143 Pointsstrange this passes for me
$(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);
});
Albert Collazo-Delvalle
7,223 Pointsi called the call back function wrong! i had: $.getJSON(weatherAPI, data, showWeather()); Thanks a lot Clayton, much appreciated!
Albert Collazo-Delvalle
7,223 PointsAlbert Collazo-Delvalle
7,223 Pointshey Clayton, thanks for the response. Unfortunately I have tried that approach as well but still got an error message.