如何使用 C# 从 RabbitMQ 中的队列接收单个消息

Question

I would like to know how I can receive only one message at a time this is basic code for that

var factory = new ConnectionFactory() { HostName = "localhost" };
var connection = factory.CreateConnection()
var channel = connection.CreateModel()
channel.QueueDeclare("hello", false, false, false, null);
var consumer = new QueueingBasicConsumer(channel);
channel.BasicConsume("hello", true, consumer);

BasicDeliverEventArgs ea = (BasicDeliverEventArgs)consumer.Queue.Dequeue();
var body = ea.Body;
var message = Encoding.UTF8.GetString(body);
Response.Write(message + " Received.");

Answer 1

If you want wait for 1 message, then keep using channel.BasicConsume , but leave consumer method after single message.

If you just get 1 message (if at least 1 exists in queue) then channel.BasicGet

var data = channel.BasicGet(queueName, true);

PS:

There are nice example on CloudAMQP with .NET: Getting started page.

Answer 2

Complementing the last message if you need the text from the message you need to convert:

var data = channel.BasicGet(QueueName, true);
var message = System.Text.Encoding.UTF8.GetString(data.Body);

Answer 3

For other languages if you get an error using the channel.BasicGet method you can use channel.get method of the library This was the case with me when i was trying to achieve this using Node.Js

my Node.Js code:

const open = require('amqplib').connect('amqp://admin:password@172.17.0.1:5672?heartbeat:30')
open
  .then(conn => conn.createChannel())
  .then(async channel => {
    console.log('channel created')
    const msg = await channel.get('myQueue')
    if (msg) {
      console.log(msg.content.toString())
      channel.ack(msg)
    }
  })
  .catch(console.warn)