PHP Prepared Statements handle duplicate entry - php

Question

I have a problem, I am not able to handle a duplicate entry with prepared statements.
I want to end the program when a duplicate entry appears. This is what I have been trying to do:
function insert_vulnerability ($CVE, $Description, $Date, $Score, $Type){
$conn = connection();
$stmt = $conn->prepare("INSERT INTO Vulnerabilities (CVE, Description, Date, Score, Type)
VALUES (?, ?, ?, ?, ?)");
$stmt->bind_param("sssis", $CVE, $Description, $Date, $Score, $Type);
if ( false === $stmt ) {
die('prepare() failed: ' . htmlspecialchars($mysqli->error));
}
$stmt->execute();
$conn->close();
}
When not using prepared statements I handled the error this way and everything worked perfectly:
function insert_vulnerability ($CVE, $Description, $Date, $Score, $Type){
$conn = connection();
$Description = htmlspecialchars($Description);
$sql = "INSERT INTO Vulnerabilities (CVE, Description, Date, Score, Type)
VALUES ('".$CVE."', '".$Description."', '".$Date."', '".$Score."', '".$Type."')";
if ($conn->query($sql) === TRUE) {
//echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
$conn->close();
die();
}
$conn->close();
}
So how do I get the same result with prepared statements ¿?
Thank you in advance.

Answer 1

You could just connect the checking inside the $stmt->execute() to see if the prepared statement did work properly.
function insert_vulnerability ($CVE, $Description, $Date, $Score, $Type){
$conn = connection();
$stmt = $conn->prepare('
INSERT INTO Vulnerabilities (CVE, Description, Date, Score, Type)
VALUES (?, ?, ?, ?, ?)
');
$stmt->bind_param('sssis', $CVE, $Description, $Date, $Score, $Type);
if($stmt->execute()) { // true, success, else error
echo 'New record created successfully';
} else {
echo $conn->error;
}
$conn->close();
}
Just to note, you have an undefined variable on your prepared statement side:
$mysqli->error