php左连接if else以显示特定的html按钮标记

<?php
// 假设有两个表：users和buttons
// users表包含用户信息，buttons表包含按钮信息

// 连接数据库
$servername = "数据库服务器地址";
$username = "数据库用户名";
$password = "数据库密码";
$dbname = "数据库名";

$conn = new mysqli($servername, $username, $password, $dbname);

// 检查连接是否成功
if ($conn->connect_error) {
    die("连接数据库失败: " . $conn->connect_error);
}

// 执行左连接查询
$sql = "SELECT users.username, buttons.button_name
        FROM users
        LEFT JOIN buttons ON users.user_id = buttons.user_id";

$result = $conn->query($sql);

// 检查查询结果
if ($result->num_rows > 0) {
    // 输出数据
    while ($row = $result->fetch_assoc()) {
        $username = $row["username"];
        $buttonName = $row["button_name"];

        // 根据条件显示不同的HTML按钮标记
        if ($buttonName == "Button A") {
            echo '<button type="button" class="button-a">Button A</button>';
        } elseif ($buttonName == "Button B") {
            echo '<button type="button" class="button-b">Button B</button>';
        } else {
            echo '<button type="button" class="default-button">Default Button</button>';
        }
    }
} else {
    echo "没有匹配的记录";
}

// 关闭数据库连接
$conn->close();
?>