Rails的工作执行的确切顺序delayed_job

Question

我正在使用collectiveidea版本的delayed_job：https://github.com/collectiveidea/delayed_job

任何人都可以指出工人去接下一份工作时使用的实际标准吗？我假设它类似于“SELECT id FROM delayed_jobs WHERE run_at> NOW（）ORDER BY priority ASC，run_at ASC LIMIT 1”（首先选择优先级，run_at时间秒），但是我还没有找到确切的内容考虑。我在GitHub上的代码中做了一些讨论，但是没有找到下一个工作的实际查询。对几件事感到好奇，包括'created_at'，'尝试'或'failed_at'是否会影响优先级。 （我意识到我不太可能找到实际的SQL，只是一种简单的方式来表示我认为查询的作用）。

其次，这个宝石的任何实际文档的良好来源？对于在Rails中常用的东西，我见过的文档非常稀疏。

Answer 1

通过源代码进行了一些挖掘，在backend / active_record.rb中显示了这一点：

    scope :ready_to_run, lambda {|worker_name, max_run_time|
      where(['(run_at <= ? AND (locked_at IS NULL OR locked_at < ?) OR locked_by = ?) AND failed_at IS NULL', db_time_now, db_time_now - max_run_time, worker_name])
    }
    scope :by_priority, order('priority ASC, run_at ASC')

    # Find a few candidate jobs to run (in case some immediately get locked by others).
    def self.find_available(worker_name, limit = 5, max_run_time = Worker.max_run_time)
      scope = self.ready_to_run(worker_name, max_run_time)
      scope = scope.scoped(:conditions => ['priority >= ?', Worker.min_priority]) if Worker.min_priority
      scope = scope.scoped(:conditions => ['priority <= ?', Worker.max_priority]) if Worker.max_priority

      ::ActiveRecord::Base.silence do
        scope.by_priority.all(:limit => limit)
      end
    end

此外，backend / base.rb中的这个位是有意义的：

    def reserve(worker, max_run_time = Worker.max_run_time)
      # We get up to 5 jobs from the db. In case we cannot get exclusive access to a job we try the next.
      # this leads to a more even distribution of jobs across the worker processes
      find_available(worker.name, 5, max_run_time).detect do |job|
        job.lock_exclusively!(max_run_time, worker.name)
      end
    end

工人要求reserve选择下一份工作。

Answer 2

delayed_job/lib/delayed/backend/active_record.rb

self.find_available调用ready_to_run，然后根据当前的工人最小/最大优先级对其进行限定，然后通过priority ASC, run_at ASC对其进行排序

scope :ready_to_run电话

where(['(run_at <= ? AND (locked_at IS NULL OR locked_at < ?) OR locked_by = ?) AND failed_at IS NULL', db_time_now, db_time_now - max_run_time, worker_name])

所以它产生类似的东西

SELECT * FROM jobs where run at <= ? AND locked_at IS NULL OR locked_at < ?) OR locked_by = ?) AND failed_at IS NULL AND priority >= ? AND priority <= ? ORDER BY priority ASC, run_at ASC