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JavaScript AJAX Basics Programming AJAX Programming AJAX Solution

Nicolás Melgarejo
Nicolás Melgarejo
14,154 Points
Posted by Nicolás Melgarejo
Nicolás Melgarejo
14,154 Points

MY SOLUTION

var xhr = new XMLHttpRequest();

xhr.onreadystatechange = function () {
  if (xhr.readyState === 4) {
      var employees = JSON.parse(xhr.responseText);
      let ul = document.createElement('ul');
      let employeeList = document.getElementById('employeeList');
      ul.className = 'bulleted';
       for (let i = 0; i < employees.length; i++){
         let li = document.createElement('li')
         let name = employees[i].name; 
         let inoffice = employees[i].inoffice;

          if (inoffice === true) {
            li.className = 'in';
          } else {
            li.className = 'out';
          };

         li.textContent = name;
         ul.appendChild(li);
       };
      employeeList.append(ul);
  };
};

xhr.open('GET', 'data/employees.json');

xhr.send();

//


var xhrRooms = new XMLHttpRequest();

xhrRooms.onreadystatechange = function () {
  if (xhrRooms.readyState === 4) {
      var rooms = JSON.parse(xhrRooms.responseText);
      let ul = document.createElement('ul');
      let roomList = document.getElementById('roomList');
      ul.className = 'rooms';
       for (let i = 0; i < rooms.length; i++){
         let li = document.createElement('li')
         let room = rooms[i].room; 
         let available = rooms[i].available;

          if (available === true) {
            li.className = 'empty';
          } else {
            li.className = 'full';
          };

         li.textContent = room;
         ul.appendChild(li);
       };
      roomList.append(ul);
  };
};

xhrRooms.open('GET', 'data/rooms.json');

xhrRooms.send();