Manuel Astudillo
Manuel Astudillo
Yes, but they are not really the same job, one job is the one being processed and the other one is the one you add afterwards.
@jtomaszewski it is quite trivial actually, the first starts processing, you add the second job and then if there is a free worker you will have two workers working in...
again, you may get confused by the nomenclature. These are not the same job, they are different jobs.
I haven't studied the your code, just assuming by this conversation that you are adding the job to the queue if the current job is still being processed. If this...
> * **At-least-once guarantee**: Every push eventually gets deployed I think I fully understand the case now, however this sentence confuses me, I would say that the last Push will...
Or in other words and to simplify the requirement, what we would like to achieve is that as long as one job is active, no other job will be active,...
What do you mean with "add support" ?
Have you tried omitting the third generic parameter? (not a fix but a workaround): ``` export const importDeletionQueue = new Queue( IMPORT_DELETE_QUEUE, { connection: getRedisClient(RedisClient.BullMQ) as IORedis.Redis, defaultJobOptions, }, );...
We cannot say it is the same issue. You should provide reproducible code, or we will not be able to solve it.
@miridih-jujang as you are just mimicking the default behaviour, i.e. manualRegistration: false, it feels quite safe to solve the issue like this. However it is concerning that this issue existed...