为什么我不能抛出自己的 sql 连接异常？

问：

我正在尝试创建一个简单的 WPF 应用程序。我希望应用程序指示它是否未连接到数据库，以便它无法发送请求并执行其连接检查。只需抛出并捕获异常，该异常将打印在屏幕上

我的方法：IsConnected

public static bool IsConnected(SqlConnection conn)
{
    bool isConnected = false;

    try
    {
        if (conn == null)
        {
            throw new ConnectionException("It is not possible to connect to the database. Please check your settings and try again");
        }
        else
        {
            isConnected = true;
        }
    }
    catch (ConnectionException ex)
    {
        MessageBox.Show(ex.Message);
    }

    return isConnected;
}

我在哪里使用这种方法：IsConnected()

public User UserLogin(string email, string password)
{
    query = @"SELECT * FROM [User] WHERE email = @email AND password = @password";`

    User user = null;

    try
    {
        using (SqlConnection conn = new SqlConnection(DatabaseSingleton.connString))
        {
            if (DatabaseSingleton.IsConnected(conn))
            {
                using (SqlCommand command = new SqlCommand(query, conn))
                {
                    conn.Open();
                    command.Parameters.Add("@email", SqlDbType.VarChar, 50).Value = email;
                    command.Parameters.Add("@password", SqlDbType.VarChar, 50).Value = password;

                    SqlDataReader reader = command.ExecuteReader();

                    while (reader.Read())
                    {
                        user = new User
                            {
                                Id = reader.GetInt32(0),
                                Name = reader.GetString(1),
                                Second_name = reader.GetString(2),
                                Email = reader.GetString(3),
                                Password = reader.GetString(4),
                                User_type = (Type_Of_User)Enum.ToObject(typeof(Type_Of_User), reader.GetInt32(5))
                            };
                    }

                    reader.Close();
                }
            }
        }
    }
    catch (InvalidInput e)
    {
        MessageBox.Show(e.Message);
    }

    return user;
}