如何在Rails应用程序中避免竞争条件？

Question

我有一个非常简单的Rails应用程序,允许用户在一组课程中注册他们的出勤.ActiveRecord模型如下:

class Course < ActiveRecord::Base
  has_many :scheduled_runs
  ...
end

class ScheduledRun < ActiveRecord::Base
  belongs_to :course
  has_many :attendances
  has_many :attendees, :through => :attendances
  ...
end

class Attendance < ActiveRecord::Base
  belongs_to :user
  belongs_to :scheduled_run, :counter_cache => true
  ...
end

class User < ActiveRecord::Base
  has_many :attendances
  has_many :registered_courses, :through => :attendances, :source => :scheduled_run
end

ScheduledRun实例具有有限数量的可用位置,一旦达到限制,就不能再接受更多的考勤.

def full?
  attendances_count == capacity
end

attendances_count是一个计数器缓存列,包含为特定ScheduledRun记录创建的出勤关联数.

我的问题是,当一个或多个人同时尝试在课程中注册最后一个可用位置时,我不完全知道确保不会发生竞争条件的正确方法.

我的考勤控制器如下所示:

class AttendancesController < ApplicationController
  before_filter :load_scheduled_run
  before_filter :load_user, :only => :create

  def new
    @user = User.new
  end

  def create
    unless @user.valid?
      render :action => 'new'
    end

    @attendance = @user.attendances.build(:scheduled_run_id => params[:scheduled_run_id])

    if @attendance.save
      flash[:notice] = "Successfully created attendance."
      redirect_to root_url
    else
      render :action => 'new'
    end

  end

  protected
  def load_scheduled_run
    @run = ScheduledRun.find(params[:scheduled_run_id])
  end

  def load_user
    @user = User.create_new_or_load_existing(params[:user])
  end

end

如您所见,它没有考虑ScheduledRun实例已达到容量的位置.

任何有关这方面的帮助将不胜感激.

更新

在这种情况下,我不确定这是否是执行乐观锁定的正确方法,但这就是我所做的:

我在ScheduledRuns表中添加了两列 -

t.integer :attendances_count, :default => 0
t.integer :lock_version, :default => 0

我还为ScheduledRun模型添加了一个方法:

  def attend(user)
    attendance = self.attendances.build(:user_id => user.id)
    attendance.save
  rescue ActiveRecord::StaleObjectError
    self.reload!
    retry unless full? 
  end

保存Attendance模型后,ActiveRecord将继续并更新ScheduledRun模型上的计数器缓存列.这是显示发生这种情况的日志输出 -

ScheduledRun Load (0.2ms)   SELECT * FROM `scheduled_runs` WHERE (`scheduled_runs`.`id` = 113338481) ORDER BY date DESC

Attendance Create (0.2ms)   INSERT INTO `attendances` (`created_at`, `scheduled_run_id`, `updated_at`, `user_id`) VALUES('2010-06-15 10:16:43', 113338481, '2010-06-15 10:16:43', 350162832)

ScheduledRun Update (0.2ms)   UPDATE `scheduled_runs` SET `lock_version` = COALESCE(`lock_version`, 0) + 1, `attendances_count` = COALESCE(`attendances_count`, 0) + 1 WHERE (`id` = 113338481)

如果在保存新的Attendance模型之前对ScheduledRun模型进行后续更新,则应触发StaleObjectError异常.此时,如果尚未达到容量,则再次重试整个过程.

更新#2

继@ kenn的回复之后,在SheduledRun对象上更新了参与方法:

# creates a new attendee on a course
def attend(user)
  ScheduledRun.transaction do
    begin
      attendance = self.attendances.build(:user_id => user.id)
      self.touch # force parent object to update its lock version
      attendance.save # as child object creation in hm association skips locking mechanism
    rescue ActiveRecord::StaleObjectError
      self.reload!
      retry unless full?
    end
  end 
end

Answer 1

乐观锁定是可行的方法,但正如您可能已经注意到的那样,您的代码永远不会引发ActiveRecord :: StaleObjectError,因为has_many关联中的子对象创建会跳过锁定机制.看看下面的SQL:

UPDATE `scheduled_runs` SET `lock_version` = COALESCE(`lock_version`, 0) + 1, `attendances_count` = COALESCE(`attendances_count`, 0) + 1 WHERE (`id` = 113338481)

更新父对象中的属性时,通常会看到以下SQL:

UPDATE `scheduled_runs` SET `updated_at` = '2010-07-23 10:44:19', `lock_version` = 2 WHERE id = 113338481 AND `lock_version` = 1

上面的语句显示了如何实现乐观锁定:请注意lock_version = 1WHERE子句.当竞争条件发生时,并发进程尝试运行这个确切的查询,但只有第一个成功,因为第一个进程原子地将lock_version更新为2,后续进程将无法找到记录并引发ActiveRecord :: StaleObjectError,因为相同的记录lock_version = 1不再有.

因此,在您的情况下,可能的解决方法是在创建/销毁子对象之前触摸父对象,如下所示:

def attend(user)
  self.touch # Assuming you have updated_at column
  attendance = self.attendances.create(:user_id => user.id)
rescue ActiveRecord::StaleObjectError
  #...do something...
end

这并不意味着严格避免竞争条件,但实际上它应该适用于大多数情况.