I will begin with an apology: I'm bad with AJAX. I'm trying to learn, but this particular project is giving me an absolute headache. I've read through a lot of examples, but can't find anything that helps with this issue.
PROBLEM: Using an "old-fashioned" PHP/MySQL/HTML environment, I'm trying to send data from a simple HTML form using AJAX; send the ajax request to php for updating a record in a MySQL db. Unfortunately, my code just does not work. It seems that I am effectively sending data, but somewhere in all of this I am coming up short. NOTE: I've confirmed that there is no problem with my Database connection, I use "$db" when inputting my name, password, db_name as indicated in my "show_client.php" file.
MY GOAL: I'm hoping to: (a) Update the existing variable using this AJAX request (no refresh of page); AND (b) Display output in the div titled "case_activity_id2" after the form database row is updated and my ajax request has been submitted.
I have two relevant files: (1) index.php; and (2) show_client.php.
index.php
<html>
<td>
<form id= "ajaxForm" action = "" method = "POST" >
<input type = "hidden" name='case_activity_id' id = 'case_activity_id' value = '<?php echo $case_activity_id?>'>
<select name ='show_client_id' id = 'show_client_id' class="form-control">
<?php
showClient();
?>
</select>
</td>
<td><input type="submit" value="send" id = "btnClick" class="btn btn-primary" /> </form></td>
<td><div id = "case_activity_id2"></div></td>
<script>
(function(){
$("#btnClick").on("click", function(){ submitForm();});
})();
function submitForm(){
$(document).ready(function() {
$("#ajaxForm").submit(function(event){
var case_activity_id = $("#case_activity_id").val();
var show_client_id = $("#show_client_id").val();
$.ajax( {
type: "POST",
url:'show_client.php',
dataType: 'json',
data: {
case_activity_id:case_activity_id,
show_client_id:show_client_id
},
success:function(data) {
$('#case_activity_id2').html(data);
console.log(data);
}
});
});
});
}
</script>
</html>
(2) show_client.php
<?php
session_start();
require 'db/connect.php';
$show_client_id = $_POST['show_client_id'];
$case_activity_id = $_POST['case_activity_id'];
$sql = "UPDATE case_activity
SET show_client_id = '$show_client_id'
WHERE case_activity_id = '$case_activity_id'";
mysqli_query($db, $sql);
?>
