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JavaScript AJAX Basics Programming AJAX Processing JSON Data

Anton Kilk
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Anton Kilk
Full Stack JavaScript Techdegree Student 11,256 Points
Posted by Anton Kilk
.a{fill-rule:evenodd;}techdegree
Anton Kilk
Full Stack JavaScript Techdegree Student 11,256 Points

Fails to display

I couldn't see any results during the exercises here in processing JSON data. I couldn't see results on console log. Now it did not go through innerHTML either. Workspace snapshot: https://w.trhou.se/z4c05nbmyj

I use Google Chrome.

2 Answers

Emily Rodriguez
Emily Rodriguez
12,572 Points

As Kris said, you're not referencing widget.js in index.html. Also, you have an error in your for loop on line 6. This: for (var i = 0; var < employees.length; i++){ Should be this: for (var i = 0; i < employees.length; i++){

You're referencing var instead of i.

Hope this helps!

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points
KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points

You have two issues:

(1) There is no reference to the widget.js script in your index.html. This is why you don't see anything.

(2) This one will display in the console once you fix #1.