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>I’ve created the stuff_data database and populated it with 4 rows. My dropdown list does nothing. Please help.
Thanks.
Neo
source https://stackoverflow.com/questions/71256438/easy-ajax-mysql-and-php
Hello, my name is Muhammad Athar. I'm a 17 year old student. I'm a Web Developer, Designer and Freelancer.
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