我的API允许用户购买某些独特的项目,每个项目只能出售给一个用户.所以当多个用户尝试购买相同的项目时,一个用户应该得到响应：好的,另一个用户应该得到这个响应too_late.
现在,我的代码似乎有错误.比赛条件.如果两个用户同时尝试购买相同的项目,他们都可以得到答案.这个问题在生产中是显而易见的.现在我写了一个简单的测试,尝试通过rspec来重现它：
context "when I try to provoke a race condition" do
# ...
before do
@concurrent_requests = 2.times.map do
Thread.new do
Thread.current[:answer] = post "/api/v1/item/buy.json",:id => item.id
end
end
@answers = @concurrent_requests.map do |th|
th.join
th[:answer].body
end
end
it "should only sell the item to one user" do
@answers.sort.should == ["ok","too_late"].sort
end
end
似乎不同时执行查询.为了测试这个,我把下面的代码放到我的控制器动作中：
puts "Is it concurrent?" sleep 0.2 puts "Oh Noez."
预期的输出将是,如果请求是并发的：
Is it concurrent? Is it concurrent? Oh Noez. Oh Noez.
但是,我得到以下输出：
Is it concurrent? Oh Noez. Is it concurrent? Oh Noez.
这告诉我,水瓶要求不是同时运行,而是一次一个.如何使我的卡巴拉要求并发？
解决方法
多线程和capybara不起作用,因为Capabara使用单独的服务器线程来顺序处理连接.但是,如果你叉,它是有效的.
我正在使用退出代码作为进程间通信机制.如果你做更复杂的东西,你可能想使用套接字.
这是我的快速肮脏的黑客：
before do
@concurrent_requests = 2.times.map do
fork do
# ActiveRecord explodes when you do not re-establish the sockets
ActiveRecord::Base.connection.reconnect!
answer = post "/api/v1/item/buy.json",:id => item.id
# Calling exit! instead of exit so we do not invoke any rspec's `at_exit`
# handlers,which cleans up,measures code coverage and make things explode.
case JSON.parse(answer.body)["status"]
when "accepted"
exit! 128
when "too_late"
exit! 129
end
end
end
# Wait for the two requests to finish and get the exit codes.
@exitcodes = @concurrent_requests.map do |pid|
Process.waitpid(pid)
$?.exitstatus
end
# Also reconnect in the main process,just in case things go wrong...
ActiveRecord::Base.connection.reconnect!
# And reload the item that has been modified by the seperate processs,# for use in later `it` blocks.
item.reload
end
it "should only accept one of two concurrent requests" do
@exitcodes.sort.should == [128,129]
end
我使用非常异乎寻常的退出代码,如128和129,因为如果未达到大小写块,则进程退出代码0,如果异常发生则为1.两者都不应该发生.所以通过使用更高的代码,我注意到事情出错.