Checkbox that updates dynamically - php

Question

I have a checkbox that dynamically updates a MySQL database when it is checked/unchecked using PHP and Ajax.
I am now trying to pass the users name so that the Ajax script can update the database with the users full name.
I have the name held in a variable called $full_name. I cannot seem to get this working though. Please see the code below:
Javascript:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function chkit(uid, chk) {
chk=document.getElementById("chk").checked;
$.ajax({
type: 'GET',
url: 'ajax.php',
data: { chkYesNo: chk, record_id: uid, full_name: user},
success:function(data){
// successful request; do something with the div background color
if(data==1)
{
$("#replace").addClass("checked_div_status").removeClass("replace");//removing first class and adding second class
}
else
{
$("#replace").addClass("replace").removeClass("checked_div_status");//removing second class and adding first class
}
}
});
}
</script>
HTML:
<?php
$record_id = $_GET['veh_id'];
include '../dbconnect.php';
//fetching data from database
$select=mysql_fetch_array(mysql_query("select invoice_checked from vehicle_details where veh_id = '$record_id' "));
?>
<!--The checkbox whose enable to change div his background color and onclick call function to update database-->
<table width=“100%”>
<td id="replace2" class="<?php if($select['invoice_checked']==1) { echo 'checked_div_status2'; } else{ echo 'replace2'; } ?>">
<input name="chk2" type="checkbox" id="chk2" value="1" onclick="chkit2(<?php echo $record_id;?>,'chk2');" <?php if($select['invoice_checked']==1) { echo 'checked'; } else{ echo ''; } ?> />
Invoice Checked
</td>
</table>
Ajax.php:
<?php
mysql_connect("server", "username", "password") or die("Could not connect: " . mysql_error());
mysql_select_db("database");
//here $get variable receive checkbox value true(1) either false(0)
$get=$_GET['chkYesNo'];
//here $get_id variable receive value of current id that you passed
$get_id=$_GET['record_id'];
$get_user=$_GET['full_name'];
if($get=="true")
{
$mysql_query=mysql_query("update vehicle_details set hpi_registered='1', check_user='".$get_user."' where veh_id='".$get_id."'");
$select=mysql_fetch_array(mysql_query("select hpi_registered from vehicle_details where veh_id='".$get_id."'"));
echo $select['hpi_registered'];
}
else
{
$mysql_query=mysql_query("update vehicle_details set hpi_registered='0', check_user='0' where veh_id='".$get_id."'");
$select=mysql_fetch_array(mysql_query("select hpi_registered from vehicle_details where veh_id='".$get_id."'"));
echo $select['hpi_registered'];
}
?>
Any help would be greatly received.
Thanks,
John

Answer 1

Some debug lession for you. Please, check my comments:
// Do not need to replicate your code, if the same things happens in it.
//instead, use a condition to set your variables, and use these variables later.
if ($get == "true") {
$hpi_registered = 1;
//Escape your variable to avoid sql injection
$checkUser = mysqli_real_escape_string($conn, $_GET["full_name"]);
} else {
$hpi_registered = 0;
$checkUser = 0;
}
//Store your query in a variable, so you can debug / dump it
//Let's dump it, see, what is your query, and try to run in directly in sql.
//Maybe it has syntax error.
$sql = "UPDATE vehicle_details SET"
. " hpi_registered='" . intval($hpi_registered) . "',"
. " check_user='" . $checkUser . "'"
. " WHERE veh_id='" . intval($get_id) . "'";
mysqli_query($conn, $sql);
//What happens, if you run it directly in sql? If this fails, now here is your
//error.
$sql = "SELECT hpi_registered"
. " FROM vehicle_details"
. " WHERE veh_id='" . intval($get_id) . "'";
//Do the same like previous query.
$res = mysqli_query($conn, $sql);
$select = mysqli_fetch_array($res);
echo $select['hpi_registered'];
DO NOT use mysql functions, because they are deprecated. Use mysqli or PDO instead.
Avoid sql injection by escaping your variables.