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Start your free trialManoj Gurung
5,318 Pointswhy is this returning undefined
const astrosUrl = 'http://api.open-notify.org/astros.json'; const wikiUrl = 'https://en.wikipedia.org/api/rest_v1/page/summary/'; const peopleList = document.getElementById('people'); const btn = document.querySelector('button');
// Make an AJAX request function getJSON(url) { const xhr = new XMLHttpRequest(); xhr.open('GET', url); xhr.onload = () => { if(xhr.status === 200) { let data = JSON.parse(xhr.responseText); return data; } }; xhr.send(); }
var dataP=getJSON(astrosUrl); console.log(dataP);
1 Answer
Robert Manolis
Treehouse Guest TeacherHi Manoj Gurung, I can't say for certain without running the code myself, but if you're asking why the log statement at the end is printing undefined to the console, I would say it probably has to do with the asynchronous nature of making those http requests. At the moment the log statement runs, the dataP variable is undefined. You can use promises, async/await or callbacks to delay the printing of the log statement till after the requests/response cycle resolves. Or you could even just use a setTimeout function to delay the printing for a bit, but that is a pretty hacky way to go about it, and not really something you would want to get in the habit of doing.
Hope that helps!