使用God(godrb.com)我正在尝试编写一个配方来启动新流程,而不管部署应用程序时现有流程的状态如何。现有流程需要长时间运行超时才能完成当前任务,但新流程应立即使用新部署的代码启动。
我现在拥有的东西在停止时设置超时300秒,但在启动新进程之前等待整整300秒。
God.watch do |w|
w.name = "sidekiq"
w.interval = 30.seconds
w.start = "bash -lc 'cd /path/to/current && ./bin/sidekiq -P /path/to/shared/pids/sidekiq.pid'"
w.stop = "bash -lc 'kill -USR1 `cat /path/to/shared/pids/sidekiq.pid`'"
w.stop_timeout = 300.seconds
w.pid_file = "/path/to/shared/pids/sidekiq.pid"
w.behavior(:clean_pid_file)
end
在这种情况下,kill -USR1
告诉sidekiq完成当前任务的处理,但不再需要工作。
我想在现有工作程序上保持300秒超时,但是一旦运行kill命令就启动新进程。
答案 0 :(得分:0)
我认为你必须定义一些过渡。
这是我的god.rb
:
# I'm using Rails
rails_env = ENV['RAILS_ENV'] || 'development'
rails_root = '/path/to/current'
pid_file = "#{rails_root}/tmp/pids/sidekiq.pid"
God.watch do |w|
w.dir = rails_root
w.name = "sidekiq"
w.interval = 30.seconds
w.env = {'RAILS_ENV' => rails_env, 'BUNDLE_GEMFILE' => "#{rails_root}/Gemfile"}
w.uid = 'deployer'
w.gid = 'staff'
w.start = "cd #{rails_root}; bundle exec sidekiq -e #{rails_env} -C #{rails_root}/config/sidekiq.yml -i #{i} -P #{pid_file}&"
w.stop = "cd #{rails_root}; bundle exec sidekiqctl stop #{pid_file} 10"
w.restart = "#{w.stop} && #{w.start}"
w.start_grace = 15.seconds
w.restart_grace = 15.seconds
w.pid_file = pid_file
w.log = "#{rails_root}/log/sidekiq.log"
# clean pid files before start if necessary
w.behavior(:clean_pid_file)
# determine the state on startup
w.transition(:init, {true => :up, false => :start}) do |on|
on.condition(:process_running) do |c|
c.running = true
end
end
# determine when process has finished starting
w.transition([:start, :restart], :up) do |on|
on.condition(:process_running) do |c|
c.running = true
end
# failsafe
on.condition(:tries) do |c|
c.times = 5
c.transition = :start
end
end
# start if process is not running
w.transition(:up, :start) do |on|
on.condition(:process_exits)
end
# lifecycle
w.lifecycle do |on|
on.condition(:flapping) do |c|
c.to_state = [:start, :restart]
c.times = 5
c.within = 5.minute
c.transition = :unmonitored
c.retry_in = 10.minutes
c.retry_times = 5
c.retry_within = 2.hours
end
end
end