Question

我在运行Ubuntu的LAMP堆栈服务器上使用Laravel 5。我正在使用队列系统：

http://laravel.com/docs/master/queues

我可以从文档中看到，有能力运行队列：listen将永远监听并处理队列中的过去和未来项目。

还有queue：work，它只处理队列中的第一个项目。

有没有办法只处理队列中的每个项目，然后停止收听？

我只想定期处理队列，那么我如何设置一个将处理队列的cron作业，然后只要队列中的所有内容都完成就退出？

Answer 1

我也只是看着这个。我修改了内置的queue:work命令来处理整个队列并退出。

php artisan make：console ProcessQueueAndExit

您可以在https://gist.github.com/jdforsythe/b8c9bd46250ee23daa9de15d19495f07

获取代码

或者在这里，为了永久性：

namespace App\Console\Commands;

use Illuminate\Console\Command;
use Carbon\Carbon;
use Illuminate\Queue\Worker;
use Illuminate\Contracts\Queue\Job;
use Symfony\Component\Console\Input\InputOption;
use Symfony\Component\Console\Input\InputArgument;

class ProcessQueueAndExit extends Command {

  protected $signature = 'queue:workall {connection?} {--queue=} {--daemon} {--delay=} {--force} {--memory=} {--sleep=} {--tries=}';

  protected $description = 'Process all jobs on a queue and exit';
  protected $worker;

  public function __construct(Worker $worker) {
    parent::__construct();

    $this->worker = $worker;
  }

  public function handle() {
    if ($this->downForMaintenance() && ! $this->option('daemon')) {
      return $this->worker->sleep($this->option('sleep'));
    }

    $queue = $this->option('queue');
    $delay = $this->option('delay');
    $memory = $this->option('memory');
    $connection = $this->argument('connection');

    // keep processing until there are no more jobs returned
    do {
      $response = $this->runWorker(
        $connection, $queue, $delay, $memory, $this->option('daemon')
      );

      if (! is_null($response['job'])) {
        $this->writeOutput($response['job'], $response['failed']);
      }
    } while (! is_null($response['job']));
  }

  protected function runWorker($connection, $queue, $delay, $memory, $daemon = false) {
    if ($daemon) {
      $this->worker->setCache($this->laravel['cache']->driver());

      $this->worker->setDaemonExceptionHandler(
        $this->laravel['Illuminate\Contracts\Debug\ExceptionHandler']
      );

      return $this->worker->daemon(
        $connection, $queue, $delay, $memory,
        $this->option('sleep'), $this->option('tries')
      );
    }

    return $this->worker->pop(
      $connection, $queue, $delay,
      $this->option('sleep'), $this->option('tries')
    );
  }

  protected function writeOutput(Job $job, $failed) {
    if ($failed) {
      $this->output->writeln('<error>['.Carbon::now()->format('Y-m-d H:i:s').'] Failed:</error> '.$job->getName());
    }
    else {
      $this->output->writeln('<info>['.Carbon::now()->format('Y-m-d H:i:s').'] Processed:</info> '.$job->getName());
    }
  }

  protected function downForMaintenance() {
    if ($this->option('force')) {
        return false;
    }
    return $this->laravel->isDownForMaintenance();
  }
}

Answer 2

我在命令文件中使用它：

$queue = Queue::connection('yourqueueconnection');

while ($entry = $queue->pop()) {
    // your task
}

如何处理队列中的所有项目并退出

2 个答案: