Laravel 查询日志中存在大量作业查询是否可以？

Question

最近，我们的 Laravel 查询日志变得明显更大。每天约为 50-100mb，一天后增加到约 1-1.5gb。有很多对jobs表的查询，像这样[05.08.2020 00:00:02] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574802, 1596571202][05.08.2020 00:00:05] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574805, 1596571205][05.08.2020 00:00:09] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574809, 1596571209][05.08.2020 00:00:12] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574812, 1596571212][05.08.2020 00:00:15] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574815, 1596571215][05.08.2020 00:00:18] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574818, 1596571218][05.08.2020 00:00:21] select * from `jobs` where `queue` = ? and ((`reserved_at` is null and `available_at` <= ?) or (`reserved_at` <= ?)) order by `id` asc limit 1 for update [default, 1596574821, 1596571221]他们整天都在这样询问。我怎样才能减少这个金额？或者有那么多疑问可以吗jobs？

Answer 1

是的，如果您有积压的工作需要几秒钟才能完成，这似乎很正常。

来自文档：

当队列中有可用作业时，工作人员将继续处理作业，作业之间不会有任何延迟。但是，该sleep选项确定如果没有新的工作可用，工作人员将“睡眠”多长时间（以秒为单位）。睡眠期间，worker 不会处理任何新作业 - 这些作业将在worker 再次醒来后处理。

https://laravel.com/docs/7.x/queues#job-expirations-and-timeouts

因此，在您的示例中，队列工作线程有积压的作业需要处理，并且在每个完成的作业后立即获取下一个作业。