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JavaScript AJAX Basics (retiring) jQuery and AJAX The Office Status Project Revisited

Posted by Eugene Letkin
Eugene Letkin
11,737 Points

Why another each() method (which we were told about in "jQuery Basics" doesn't work here?

The console tells me: Uncaught TypeError: $employees.each is not a function. What's wrong with my code?

$(document).ready(function () {
    $.getJSON('data/employees.json', function(response){
        var $ul = $("<ul class='bulleted'></ul>");
        var $employees = response;
        $employees.each(function(){
            var $li = $('<li></li>');
            $li.text($(this).name);
            if ( $(this).inoffice){
                $li.addClass('in');
            }else{
                $li.addClass('out');
            };
            $ul.append($li);
        });
        $('#employeeList').append($ul);
    });
}) // end read

2 Answers

Emmanuel Molina
Emmanuel Molina
9,268 Points
Emmanuel Molina
Emmanuel Molina
9,268 Points

Because $employees.each(function(){}); isn't the correct synthax :) Try $.each($employees, function(){}); instead.

Justin Roberts
Justin Roberts
14,583 Points
Justin Roberts
Justin Roberts
14,583 Points

Try logging the response. One thing I think it might be is that the response is not a jQuery object (which it has to be to use a jQuery function). JavaScript itself does have an each type function, it's just "forEach" rather than just "each".