Question : PHP Updating Databases error handling.

i wrote this code to code to update a database, . the error im getting is

Warning: mysqli_query() expects parameter 1 to be mysqli,

it is refering to this line. but i dont see what the problem is.

$queryResult = mysqli_query($con,$queryString);

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<?php

$Milg = $_POST[mileage];
$NMilg = $_POST[Nmileage];
$Mdl=$_POST[model];
$Mke=$_POST[make];

$con = mysql_connect("localhost","phpuser","phppass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_cars", $con);

$queryString="UPDATE Cars SET Mileage = '$NMilg'
WHERE Model = '$Mdl' AND Make = '$Mke' AND Mileage ='$Milg'";

$queryResult = mysqli_query($con,$queryString);
  		    if ($queryResult) {
            echo "Car <b>$Mke $Mdl</b> been updated";
		      } else {
			     ?>
             Car not found
            <form action="update.php" method="post"><br>
	    Make: <input name="make" type="text"/><br>
	    Model: <input name="model" type="text"/><br>
	Old	Mileage: <input name="mileage" type="text"/><br>
	New Mileage:<input name="Nmileage" type="text"/><br>
	    <input type="Submit" value="Update" />
	           
	          </form>
			  <?php
}
mysql_close($con);
?>

Question : PHP Updating Databases error handling.

Answer : PHP Updating Databases error handling.

Oops, I should have caught that, you had it backwards. Should be "mysql_query($queryString,$con)" .