JavaScript AJAX Basics (retiring) Programming AJAX Processing JSON Data

Loy Yee Ko
Loy Yee Ko
5,814 Points

I cannot see the employees list status, what's the problem?

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function ()
{
  if(xhr.readyState === 4)
  {
    var employees = JSON.parse(xhr.reposonseText);
    var statusHTML = '<ul class="bulleted">';
    for(var 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();

I was comparing mine with the teacher's code as well as others, the status didn't show up at all. from https://teamtreehouse.com/community/does-not-work-5

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+=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();

please let me know if you find any problems! thanks!

1 Answer

Loy Yee Ko
Loy Yee Ko
5,814 Points

nevermind guys I have misspelled responseText....