RabbitMQ C#API中的消息接收事件

我正在尝试实现一个WinForm RabbitMQ客户机，在该客户机中，我从服务器接收消息，如下所示-

private void Form1_Load(object sender, System.EventArgs e)
{
    backgroundWorker1.RunWorkerAsync();
}

private void backgroundWorker1_DoWork(object sender, DoWorkEventArgs e)
{
    var factory = new ConnectionFactory() { HostName = "192.168.100.6", Password = "pass", UserName = "username" };
    using (var connection = factory.CreateConnection())
    {
        using (var channel = connection.CreateModel())
        {
            channel.QueueDeclare("CallCenter", false, false, false, null);  
            var consumer = new QueueingBasicConsumer(channel);
            channel.BasicConsume("CallCenter", true, consumer);
            while (true)
            {
                var ea = (BasicDeliverEventArgs)consumer.Queue.Dequeue();
                var body = ea.Body;
                var message = Encoding.UTF8.GetString(body);
                MessageBox.Show(message);
            }
        }
    }
}

private void Form1_FormClosing(object sender, FormClosingEventArgs e)
{
    if (backgroundWorker1.IsBusy)
    {
        backgroundWorker1.CancelAsync();
    }

    backgroundWorker1.Dispose();
}

我很确定这不是一个好办法。相反，如果有一个

OnMessageReceived

事件会更好

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您有关于RabbiMQ中基于事件的消息接收的好例子吗

你这样做是正确的<代码>consumer.Queue.Dequeue（）本质上是一个OnMessageReceived，它是一个阻塞调用，它坐在那里等待rabbit向它发送消息