'Easy AJAX, MySQL, and PHP

I’m trying to get this example to work. Here is my HTML as familyguy.html.

<html>
<head>
<script>
function showUser(str) {
  if (str == "") {
    document.getElementById("txtHint").innerHTML = "";
    return;
  } else { 
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
        document.getElementById("txtHint").innerHTML = this.responseText;
      }
    };
    xmlhttp.open("GET","getuser.php?q="+str,true);
    xmlhttp.send();
  }
}
</script>
</head>
<body>

<form>
<select name="users" onchange="showUser(this.value)">
  <option value="">Select a person:</option>
  <option value="1">Peter Griffin</option>
  <option value="2">Lois Griffin</option>
  <option value="3">Joseph Swanson</option>
  <option value="4">Glenn Quagmire</option>
  </select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>

</body>
</html>
And here is my php code getuser.php.

I’ve created the stuff_data database and populated it with 4 rows. My dropdown list does nothing. Please help.

Thanks.

Neo

javascriptphpmysqlajax


Solution 1:[1]

I changed the host field to localhost and this solved the problem.

Sources

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Source: Stack Overflow

Solution Source
Solution 1 Neo

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