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running rabbitmq without Console.ReadLine();

Determined Dolphin answered on November 13, 2020 Popularity 1/10 Helpfulness 1/10

answer running rabbitmq without Console.ReadLine();

running rabbitmq without Console.ReadLine();

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public class RabbitMQManager : IDisposable
{
    private bool _disposed = false;
    private IModel _channel;
    private IConnection _connection;

    public event EventHandler<string> MessageReceived;

    protected virtual void Dispose(bool disposing)
    {
        if (!_disposed)
        {
            if (disposing)
            {
                _channel?.Dispose();
                _connection?.Dispose();
            }

            _disposed = true;
        }
    }

    public void Dispose()
    {
        Dispose(true);
    }

    public void Connect()
    {
        var factory = new ConnectionFactory { HostName = "xxx", UserName = "xxx", Password = "xxx" };
        _connection = factory.CreateConnection();
        _channel = _connection.CreateModel();

        _channel.ExchangeDeclare(exchange: "call_notify", type: "fanout");

        string queueName = _channel.QueueDeclare().QueueName;
        _channel.QueueBind(queue: queueName,
                          exchange: "call_notify",
                          routingKey: "");

        var consumer = new EventingBasicConsumer(_channel);
        consumer.Received += (model, ea) =>
        {
            byte[] body = ea.Body;
            string message = Encoding.UTF8.GetString(body);
            MessageReceived?.Invoke(this, message);
        };

        _channel.BasicConsume(queue: queueName,
                             autoAck: true,
                             consumer: consumer);
    }
}

Popularity 1/10 Helpfulness 1/10 Language whatever

Source: stackoverflow.com

Tags: rabbitmq whatever

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Contributed on Nov 13 2020

Determined Dolphin

0 Answers Avg Quality 2/10

running rabbitmq without Console.ReadLine();

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