我想制作一个分片,帮助逐步处理组织成管道的工作人员的数据。我取得了一些成果,但在我看来,它可以使它更容易,更方便。在当前的解决方案中,我不喜欢需要指定每个Worker的类型,即使Worker被包装在另一个类(online example)中:
class Worker
def do_it(data : String)
puts "#{data} from #{self.class}"
end
end
class Worker1 < Worker; end
class Worker2 < Worker; end
class Worker3 < Worker; end
class Wrapper(T)
property item : T
def initialize(@item : T); end
def worker
@item
end
end
class Pipeline(T)
@stack = [] of T
def self.build
with self.new yield
end
def add(wrapper : T)
@stack << wrapper
end
def run(data : String)
@stack.each do |wrapper|
wrapper.worker.do_it(data)
end
end
end
alias WrapperObject = Wrapper(Worker1) | Wrapper(Worker2) | Wrapper(Worker3)
Pipeline(WrapperObject).build do
add Wrapper(Worker1).new(Worker1.new)
add Wrapper(Worker2).new(Worker2.new)
add Wrapper(Worker3).new(Worker3.new)
run("Some string data")
end
我尝试用从块中提取类型的宏来解决这个问题并将其传递给Pipeline类的常量,然后我可以将类型用于内部方法:
macro collect_types(workers)
{% worker_types = workers.map{|k| k.id }.join(" | ") %}
alias worker_types = {{ worker_types }}
STACK = [] of {{ worker_types }}
end
但是失去了继承的可能性Pipeline ......根据我的梦想,碎片应该像这样使用:
require 'pipeline'
class Worker < Pipeline::Worker
def process(data : String)
# some business logic
end
end
pipeline = Pipeline::Line.build do
add Worker1
add Worker2
add Worker3
end
result = pipeline.run(data)
有可能的?
@veelenga用Command pattern解决了这个问题。 Example on carc.in