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JavaScript AJAX Basics (retiring) Programming AJAX Processing JSON Data

Humberto Ventura
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Humberto Ventura
Front End Web Development Techdegree Graduate 14,734 Points
Posted by Humberto Ventura
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Humberto Ventura
Front End Web Development Techdegree Graduate 14,734 Points

Does not work

I did as Dave told me, my widget.js script does not have any error - I check it by using the Chrome console - and still the list just won't appear. What is going on?

Steven Parker
Steven Parker
231,172 Points
Steven Parker
Steven Parker
231,172 Points

If you show your code perhaps someone can help find the issue. A good way to do that is to make a snapshot of your workspace (if you're using one) and post the link to it here.

Dean Paterson
Dean Paterson
12,996 Points
Dean Paterson
Dean Paterson
12,996 Points

Did this course recently and pretty sure i had the same issue! As Steven says ... show us your code And also, are you using workspaces or your own editor? ( i was using my own editor and that's why i had an issue)

2 Answers

Carlos Lantigua
Carlos Lantigua
5,938 Points
Carlos Lantigua
Carlos Lantigua
5,938 Points

At first I thought he was using backticks rather than regular quotes due to the color of his editor syntax. Maybe that is the issue you are having? Here is my code, it ran fine for me.. maybe you can spot the difference in yours.

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();