Question

如果处理结果不适合我，如何将消息返回队列。只找到有关消息确认的信息，但我认为它不适合我。我需要，如果作为处理结果我得到参数RETRY消息被添加回队列。然后这个工人或另一个工人再次拿起它并试图处理它。

例如：

<?php
use PhpAmqpLib\Connection\AMQPStreamConnection;

echo ' [*] Waiting for messages. To exit press CTRL+C', "\n";

$connection = new AMQPStreamConnection($AMQP);
$channel = $connection->channel();

$channel->queue_declare('test', false, false, false, false);

$callback = function($msg) {
    $condition = json_decode($msg->body);

    if (!$condition) {
        # return to the queue
    }
};

$channel->basic_consume('test', '', false, true, false, false, $callback);

while(count($channel->callbacks)) {
    $channel->wait();
}

$channel->close();
$connection->close();
?>

Answer 1

解决方案比我想象的更容易，事实证明任务不是专门针对RabbitMQ，而是关于变量的范围。如果有人对解决方案感兴趣，请点击此处：

<?php
use PhpAmqpLib\Connection\AMQPStreamConnection;
use PhpAmqpLib\Message\AMQPMessage;

echo ' [*] Waiting for messages. To exit press CTRL+C', "\n";

$connection = new AMQPStreamConnection($AMQP);
$channel = $connection->channel();

$channel->queue_declare('test', false, false, false, false);

$callback = function($msg) {
  global $channel;

  $condition = json_decode($msg->body);

  if (!$condition) {
    $msg = new AMQPMessage(json_encode(array(
      'condition' => false
    )));

    $channel->basic_publish($msg, '', 'test');
  }
};

$channel->basic_consume('test', '', false, true, false, false, $callback);

while(count($channel->callbacks)) {
  $channel->wait();
}

$channel->close();
$connection->close();
?>

Answer 2

将自动 no_ack 标志设为false

queue：从哪里获取消息的队列
      consumer_tag：消费者标识符
      no_local：不接收此消费者发布的消息       no_ack：告诉服务器消费者是否会确认消息       exclusive：请求独占的消费者访问权限，这意味着只有这个消费者可以访问队列
      NOWAIT：
      回调：PHP回调

$ channel-＆gt; basic_consume（'test'，''，false，false，false，false，$ callback）;

你必须使用确认，如果你的过程不起作用，你可以忽略确认

<?php
use PhpAmqpLib\Connection\AMQPStreamConnection;

echo ' [*] Waiting for messages. To exit press CTRL+C', "\n";

$connection = new AMQPStreamConnection($AMQP);
$channel = $connection->channel();

$channel->queue_declare('test', false, false, false, false);

$callback = function($msg) {
    $condition = json_decode($msg->body);

    if (!$condition) {
        // return to the queue 
        $message->delivery_info['channel']->basic_nack($message->delivery_info['delivery_tag']);
    }else{
        // send ack , remove from queue
        $message->delivery_info['channel']->basic_ack($message->delivery_info['delivery_tag']);
    }
};

$channel->basic_consume('test', '', false, false, false, false, $callback);

while(count($channel->callbacks)) {
    $channel->wait();
}

$channel->close();
$connection->close();
?>

RabbitMQ和PHP如何将任务返回队列？

2 个答案: