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

Posted by Derrick Johnson
Derrick Johnson
4,347 Points

My solution using ES6

This is just what I came up with before watching the video. Really enjoying AJAX!

$.getJSON('data/employees.json',(response) => {
  let html = '<ul class="bulleted">';
  $.each(response,(index,value) => {
    let name = value.name;
    let isOutOfOffice = value.inoffice ? 'in' : 'out';
    html += `<li class="${isOutOfOffice}">${name}</li>`;
  });
  html += '</ul>';
  $('#employeeList').html(html);
});

1 Answer

Angelica Hart Lindh
Angelica Hart Lindh
19,465 Points
Angelica Hart Lindh
Angelica Hart Lindh
19,465 Points

Hi Derrick,

Great work using ES6! It's brought a lot of improvements to JavaScript in my opinion.

I wanted to show you a "jQueryLESS" version in case there comes a time where you work on a project that isn't using jQuery.

fetch('data/employees.json')
   .then(response => response.json()) // Convert the response blob to JSON
   .then((data) => { // Data is now the JSON object
      let html = '<ul class="bulleted">';
      data.forEach((item) => {
         const isOutOfOffice = item.inoffice ? 'in' : 'out';
         html += `<li class="${isOutOfOffice}">${item.name}</li>`;
      })
      html += '</ul>';
      document.getElementById('employeeList').innerHTML(html)
   })
   .catch(error => console.error(error)); // Catch any errors on the fetch Promise response;

The Fetch API was used instead of jQuery getJSON. Also you can make use of the built in Array method forEach for iterating over an Array.