PHP返回空数组,MySQL返回集合

我在phpMyAdmin中直接运行SQL代码与使用PHP的PDO之间存在差异。这是我的PHP代码:

<?php
    $configs_database = include(dirname( __DIR__, 2) . '/php/config_database.php');

    // Database variables
    $dbHost = $configs_database['host'];
    $dbUsername = $configs_database['username'];
    $dbPassword = $configs_database['password'];
    $dbName = $configs_database['name'];

    /* Create connection */ 
    $dsn = "mysql:dbname=$dbName;host=$dbHost;charset=utf8mb4";
    $db = new PDO($dsn, $dbUsername, $dbPassword);
    $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $db->setAttribute(PDO::ATTR_EMULATE_PREPARES, true);

    // Get the business information where the user_id equals the session value
    $stmt = $db->prepare("
        SELECT
            customer.*
        FROM
            customer
        WHERE
            customer.business_id = (SELECT business_id FROM `user` WHERE user_id = :userId)
        LIMIT :limit
        OFFSET :offset;");

    // Parameterize the query
    $userId = ($_POST["userId"] ? $_POST["userId"] : $_SESSION["id"]);
    $limit = ($_POST["rowCount"] ? $_POST["rowCount"] : 99999);
    $offset = ($_POST["current"] && $_POST["rowCount"] ? intval($_POST["current"]) * $_POST["rowCount"] : 0);
    $stmt->bindValue(':userId', $userId, PDO::PARAM_INT);
    $stmt->bindValue(':limit', $limit, PDO::PARAM_INT);
    $stmt->bindValue(':offset', $offset, PDO::PARAM_INT);

    // Execute the query
    $stmt->execute();

    // Explicitly close the connection
    $db = null;

    // Get the results
    $customers = $stmt->fetchAll(PDO::FETCH_ASSOC);

    echo json_encode($customers);
    return json_encode($customers);
?>

如果我在失眠时提交请求,并使用以下JSON:

{
    "userId": 2,
    "current": 1,
    "rowCount": 100
}

它打印一个空数组。但是,如果我直接在phpMyAdmin中运行SQL,它将返回预期的集合。我甚至不知道如何调试它,因为如果我回显以下值 $userId , $limit ,以及 $offset (我在参数中传递的值)它们都是预期值(2、1和100)。是什么导致了这种差异?