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JavaScript AJAX Basics Programming AJAX Processing JSON Data

Zenzi Ali
PLUS
Zenzi Ali
Courses Plus Student 2,412 Points
Posted by Zenzi Ali
Zenzi Ali
Courses Plus Student 2,412 Points

My widget also doesn't appear on the page. I'm not sure where my error is.

I checked other answers and confirmed that widget.js is linked to my HTML page.

var xhr = new XMLHttpRequest(); xhr.onreadystatechange = function() { if(xhr.readyState === 4){ var employees = JSON.parse(xhr.responseText) var statusHTML = '<ul class="bulleted">' for (i =0: i < employees.length; i+=1){ if(employees[i].inoffice === true) { statusHTML += '<li class="in">'; } else { statusHTML += '<li class="out">'; } statusHTML += employees[i].name; statusHTML += '</li>'; } statusHTML += '</ul>'; document.getElementById('employeeList').innerHTML = statusHTML; } }; xhr.open('GET','data/employees.json'); xhr.send();

1 Answer

James Croxford
James Croxford
12,723 Points
James Croxford
James Croxford
12,723 Points

Hi there, just checked through your code:

//seem to be missing a semi-colon:
 var employees = JSON.parse(xhr.responseText) 
 var statusHTML = '<ul class="bulleted">' 

//used a colon instead of a semi-colon
 for (i =0: i < employees.length; i+=1)

Those should be your issues with it not working. Don't forget you should switch your variables over to the 'let' and 'const' keywords too!