Question

我正在尝试实现一个WinForm RabbitMQ客户端，我从服务器接收消息如下 -

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方法。相反，如果有OnMessageReceived事件会更好。

您对RabbiMQ中基于事件的消息接收有什么好的例子吗？

Answer 1

你这样做的方式是正确的。 consumer.Queue.Dequeue();本质上是一个OnMessageReceived，它是一个阻塞调用，它等待从兔子发送消息。

在RabbitMQ C＃API中收到消息事件

1 个答案: