Php code to update or insert values - php

Question

I am trying to write an API with php which will check the values a user has entered and if the trackid and userid are already in the database it will update the rating. However if the track and user id doesn't exist then the API should insert them along with a rating. How can this be done. Please note I keep getting the error:
Call to a member function rowCount() on a non-object
on line 14. Any help will be appreciated.
<?php
$id = $_GET['TrackId'];
$user = $_GET['UserId'];
$rating = $_GET['Rating'];
include("connect.php");
$query = "UPDATE `rating` SET `rating` = '$rating' WHERE track_id = '$id' AND user_id = '$user' AND rating_set=NOW()";
$sth = $dbc->query($query);
$rows = $sth->rowCount();
if ($rows != 2) {
$query = "INSERT INTO `rating` values ('$id','$rating','$user',NOW())";
$results = $dbc->query($query);
$rows = $results->rowCount();
header('Content-type: application/json; charset=utf-8');
echo $_GET['onJSONPLoad'];
echo "(" . json_encode($rows) . ")";
}
else
{
header('Content-type: application/json; charset=utf-8');
echo $_GET['onJSONPLoad'];
echo "(" . json_encode($rows) . ")";
}
$dbc = null;
?>

Answer 1

The error is because your $results is null / false, which is probably due to a failed SQL execution.
Assuming that your $dbc->query($sql) is a wrapper of mysqli_query(), the function will return false on failed query. To debug this behavior, you can var_dump($results).

Answer 2

In the update query part of the condition I am looking for is NOW() which will never be found. So I moved it to the update section of the query instead of the conditional area.

Answer 3

Use the ON DUPLICATE KEY clause
INSERT INTO table (column1, ... )
VALUES (’value1’, ...)
ON DUPLICATE KEY
UPDATE column2 = =value2, ...
http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html