5

I am using Beanstalkd with laravel for handling queue job. How can i prevent from adding same job if job is already in queue. I am using laravel 5.3 with Beanstalkd 3.1

Deepak Singh Basnal
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3 Answers3

0

There is no such concept of preventing a job to be part of a message queue. Simply you cannot do this.

Make sure that your code is written in a manner than it will handle duplicates. If you still need something to work out, maybe check Redis's SortedSet, and use that to store permanently your job.

Pentium10
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0

This workaround will help:

class ClientSendInvoice implements ShouldQueue {

use Dispatchable, InteractsWithQueue, Queueable, SerializesModels;

public $uuid, $cacheKey;
/**
 * Create a new job instance.
 *
 * @return void
 * @throws InvalidArgumentException
*/
public function __construct()
{
   $this->delay = 5;
   $this->uuid = Str::uuid();
   $this->cacheKey = "ClientSendInvoice_{$this->invoice->id}";
   cache()->put($this->cacheKey, $this->uuid, $this->delay);
}

/**
 * Execute the job.
 *
 * @return void
 */
public function handle(Invoice $invoice)
{
    if (cache()->get($this->cacheKey, $this->uuid) === $this->uuid) {
        Mail::to($invoice->client)
            ->send(new ClientInvoice(['invoice' => $invoice]));
    }
}
nolan23
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-1

There is a workaround solution you can try to add below code before dispatch the queue

$queue = \DB::table(config('queue.connections.database.table'))->first();
    if($queue){
        $payload = json_decode($queue->payload,true);
        if($payload['displayName'] == 'App\Jobs\ProcessReport'){
            \flash('The report in process','info');
            return back()->withInput();
        }
    }
//dispatch the queue
ProcessReport::dispatch();
vipmaa
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  • OP specifically mentioned using beanstalkd, not a database, as their queue driver. Databases should not be used for queues in production. – hackel May 06 '20 at 15:00