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JavaScript AJAX Basics Programming AJAX Processing JSON Data

Need a 2nd pair of eyes. No errors on console. But no widget list is created.

`var xhr = new XMLHttpRequest(); xhr.onreadystatechange = function () { if(xhr.readyState ===4){ var employees = JSON.parse(xhr.responseText); var statusHTML = '<ul class = "bulleted">'; for(var i= 0; i < employees.length; i++){ 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

Remigiusz Szostak
Remigiusz Szostak
7,827 Points

everything should be ok but you need to remove these ```characters at the end and begining because they are reason why your code is treated like a string ('template literal $[expression]) therefore it does not show errors in the console