ruby超时和系统命令

Question

我有一个ruby超时,调用这样的系统(bash)命令..

Timeout::timeout(10) {
  `my_bash_command -c12 -o text.txt`
}

但我认为即使ruby线程被中断,实际命令仍然在后台运行..这是正常的吗？我怎么能杀了它？

Answer 1

我想你必须kill手动:

require 'timeout'

puts 'starting process'
pid = Process.spawn('sleep 20')
begin
  Timeout.timeout(5) do
    puts 'waiting for the process to end'
    Process.wait(pid)
    puts 'process finished in time'
  end
rescue Timeout::Error
  puts 'process not finished in time, killing it'
  Process.kill('TERM', pid)
end

Answer 2

为了正确地停止生成的进程树(不仅仅是父进程),应该考虑这样的事情:

def exec_with_timeout(cmd, timeout)
  pid = Process.spawn(cmd, {[:err,:out] => :close, :pgroup => true})
  begin
    Timeout.timeout(timeout) do
      Process.waitpid(pid, 0)
      $?.exitstatus == 0
    end
  rescue Timeout::Error
    Process.kill(15, -Process.getpgid(pid))
    false
  end
end

这也允许您跟踪进程状态

Answer 3

处理进程、信号和计时器并不是很容易。timeout这就是您可能考虑委派此任务的原因：在新版本的 Linux 上使用以下命令：

timeout --kill-after=5s 10s my_bash_command -c12 -o text.txt

Answer 4

也许这会帮助其他人寻求实现类似的超时功能,但需要从shell命令收集输出.

我已经使用@ shurikk的方法来处理Ruby 2.0和来自Fork子进程的一些代码,使用超时和捕获输出来收集输出.

def exec_with_timeout(cmd, timeout)
  begin
    # stdout, stderr pipes
    rout, wout = IO.pipe
    rerr, werr = IO.pipe
    stdout, stderr = nil

    pid = Process.spawn(cmd, pgroup: true, :out => wout, :err => werr)

    Timeout.timeout(timeout) do
      Process.waitpid(pid)

      # close write ends so we can read from them
      wout.close
      werr.close

      stdout = rout.readlines.join
      stderr = rerr.readlines.join
    end

  rescue Timeout::Error
    Process.kill(-9, pid)
    Process.detach(pid)
  ensure
    wout.close unless wout.closed?
    werr.close unless werr.closed?
    # dispose the read ends of the pipes
    rout.close
    rerr.close
  end
  stdout
 end