How to Convert MySQL Code to a PDO Prepared Statement: A Step-by-Step Guide

Convert MySQL Code to PDO Statement: A Step-by-Step Guide

Understanding the Issue

You wish to replace the first if statement with a PDO statement to retrieve a user's email from the database using their ID.

Connecting with PDO

To utilize PDO, you must first establish a connection to the MySQL database:

$db_host = "127.0.0.1";
$db_user = "root";
$db_pass = "";
$db_database = "my_database";

$pdo = new PDO("mysql:host=$db_host;dbname=$db_database", $db_user, $db_pass, [
    PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_EMULATE_PREPARES => false
]);

Updating the Code

Prepared Statements

Prepared statements offer increased security and readability, especially when working with user input.

$sql = "SELECT email FROM users WHERE u_id = ?";
$stmt = $pdo->prepare($sql);
$stmt->bindParam(1, $id, PDO::PARAM_INT);
$stmt->execute();
$email = $stmt->fetchColumn();

Updated Code

The updated code with PDO and prepared statements:

$id = $_SESSION['u_id'] ?? NULL;

if ($id) {
    $sql = "SELECT email FROM users WHERE u_id = ?";
    $stmt = $pdo->prepare($sql);
    $stmt->bindParam(1, $id, PDO::PARAM_INT);
    $stmt->execute();
    $email = $stmt->fetchColumn();
}

$email = $email ?? "";  // To avoid PHP notices
$suggestions = selectAll($table);

$optionOne = $_POST['optionOne'] ?? "";
$optionTwo = $_POST['optionTwo'] ?? "";
$newSuggestion = $_POST['new-suggestion'] ?? "";

if ($newSuggestion && $id && $email && $optionOne && $optionTwo) {
    $sql = "INSERT INTO suggestions (user_id, email, option_1, option_2) VALUES (?, ?, ?, ?)";
    $stmt = $pdo->prepare($sql);
    $stmt->bindParam(1, $id, PDO::PARAM_INT);
    $stmt->bindParam(2, $email, PDO::PARAM_STR);
    $stmt->bindParam(3, $optionOne, PDO::PARAM_STR);
    $stmt->bindParam(4, $optionTwo, PDO::PARAM_STR);
    $stmt->execute();
} else {
    echo "All options must be entered";
}

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