RabbitMQ - PHP。消费者必须在生产者之前 运行 吗?
RabbitMQ - PHP. Must the consumer be ran before producer?
大家好,我已经开始(我真的是新手)一些使用 php 的 rabbitMQ 编码。
目前,我刚刚在 Rabbitmq 网站上测试了 examples。
他们工作。
问题是,如果我在启动消费者之前向生产者发送一些消息,后者将不会接收之前的消息。它只接收从它开始时发送的消息。
那么我真的必须总是先启动制作人吗?
有没有一种方法可以实现一个队列,该队列会等到消费者可用后再刷新消息?
或者代码可能需要额外的设置?
感谢您的帮助
(来自 rabbitMQ 教程的 PHP 代码)
Producer.php
require_once __DIR__ . '/vendor/autoload.php';
use PhpAmqpLib\Connection\AMQPStreamConnection;
use PhpAmqpLib\Message\AMQPMessage;
$connection = new AMQPStreamConnection('localhost', 5672, 'guest', 'guest');
$channel = $connection->channel();
$channel->exchange_declare('direct_logs', 'direct', false, false, false);
$severity = isset($argv[1]) && !empty($argv[1]) ? $argv[1] : 'info';
$data = implode(' ', array_slice($argv, 2));
if(empty($data)) $data = "Hello World!";
$msg = new AMQPMessage($data);
$channel->basic_publish($msg, 'direct_logs', $severity);
echo " [x] Sent ",$severity,':',$data," \n";
$channel->close();
$connection->close();
?>
消费者:
receive_logs_direct.php 的代码:
<?php
require_once __DIR__ . '/vendor/autoload.php';
use PhpAmqpLib\Connection\AMQPStreamConnection;
$connection = new AMQPStreamConnection('localhost', 5672, 'guest', 'guest');
$channel = $connection->channel();
$channel->exchange_declare('direct_logs', 'direct', false, false, false);
list($queue_name, ,) = $channel->queue_declare("", false, false, true, false);
$severities = array_slice($argv, 1);
if(empty($severities )) {
file_put_contents('php://stderr', "Usage: $argv[0] [info] [warning] [error]\n");
exit(1);
}
foreach($severities as $severity) {
$channel->queue_bind($queue_name, 'direct_logs', $severity);
}
echo ' [*] Waiting for logs. To exit press CTRL+C', "\n";
$callback = function($msg){
echo ' [x] ',$msg->delivery_info['routing_key'], ':', $msg->body, "\n";
};
$channel->basic_consume($queue_name, '', false, true, false, false, $callback);
while(count($channel->callbacks)) {
$channel->wait();
}
$channel->close();
$connection->close();
?>
简而言之,不,不能。然而,队列需要存在并且需要 durable/auto-delete 设置为 false。
我知道这是为了测试,但在发布没有人消费的消息时要小心 - 队列最终会处于流状态。
大家好,我已经开始(我真的是新手)一些使用 php 的 rabbitMQ 编码。
目前,我刚刚在 Rabbitmq 网站上测试了 examples。 他们工作。 问题是,如果我在启动消费者之前向生产者发送一些消息,后者将不会接收之前的消息。它只接收从它开始时发送的消息。 那么我真的必须总是先启动制作人吗? 有没有一种方法可以实现一个队列,该队列会等到消费者可用后再刷新消息?
或者代码可能需要额外的设置?
感谢您的帮助
(来自 rabbitMQ 教程的 PHP 代码)
Producer.php
require_once __DIR__ . '/vendor/autoload.php';
use PhpAmqpLib\Connection\AMQPStreamConnection;
use PhpAmqpLib\Message\AMQPMessage;
$connection = new AMQPStreamConnection('localhost', 5672, 'guest', 'guest');
$channel = $connection->channel();
$channel->exchange_declare('direct_logs', 'direct', false, false, false);
$severity = isset($argv[1]) && !empty($argv[1]) ? $argv[1] : 'info';
$data = implode(' ', array_slice($argv, 2));
if(empty($data)) $data = "Hello World!";
$msg = new AMQPMessage($data);
$channel->basic_publish($msg, 'direct_logs', $severity);
echo " [x] Sent ",$severity,':',$data," \n";
$channel->close();
$connection->close();
?>
消费者: receive_logs_direct.php 的代码:
<?php
require_once __DIR__ . '/vendor/autoload.php';
use PhpAmqpLib\Connection\AMQPStreamConnection;
$connection = new AMQPStreamConnection('localhost', 5672, 'guest', 'guest');
$channel = $connection->channel();
$channel->exchange_declare('direct_logs', 'direct', false, false, false);
list($queue_name, ,) = $channel->queue_declare("", false, false, true, false);
$severities = array_slice($argv, 1);
if(empty($severities )) {
file_put_contents('php://stderr', "Usage: $argv[0] [info] [warning] [error]\n");
exit(1);
}
foreach($severities as $severity) {
$channel->queue_bind($queue_name, 'direct_logs', $severity);
}
echo ' [*] Waiting for logs. To exit press CTRL+C', "\n";
$callback = function($msg){
echo ' [x] ',$msg->delivery_info['routing_key'], ':', $msg->body, "\n";
};
$channel->basic_consume($queue_name, '', false, true, false, false, $callback);
while(count($channel->callbacks)) {
$channel->wait();
}
$channel->close();
$connection->close();
?>
简而言之,不,不能。然而,队列需要存在并且需要 durable/auto-delete 设置为 false。
我知道这是为了测试,但在发布没有人消费的消息时要小心 - 队列最终会处于流状态。