Ruby 简单:需要帮助重构这个看起来笨拙的方法吗
我该怎么清理?它读起来很尴尬,而且太长了Ruby 简单:需要帮助重构这个看起来笨拙的方法吗,ruby,ruby-on-rails-3,refactoring,Ruby,Ruby On Rails 3,Refactoring,我该怎么清理?它读起来很尴尬,而且太长了 def report_total(feed_event, advisor) count = 0 advisor.activity_feed_events.each do |lead| if lead == SignupFeedEvent count += 1 else if lead.is_a?(feed_event) if lead.event_date > (Time.now -
def report_total(feed_event, advisor)
count = 0
advisor.activity_feed_events.each do |lead|
if lead == SignupFeedEvent
count += 1
else
if lead.is_a?(feed_event)
if lead.event_date > (Time.now - 7.days)
count += 1
end
end
end
end
return count
end
也许你可以使用:
def report_total(feed_event, advisor)
advisor.activity_feed_events.count do |lead|
lead == SignupFeedEvent ||
(lead.is_a?(feed_event) && lead.event_date > (Time.now - 7.days))
end
end
做同样的事情,代码更少。读一本好书,比如重构Ruby版 良好的OO实践建议不要检查类的相等性或is_a?,而是检查对象的功能,例如使用respond_to
if lead.respond_to?(:event_date) ...
这应该被问到。