ruby-on-rails - 如何重构这段 Rails 代码？

Question

只是出于好奇：

这个（相当丑陋的）Rails 代码如何被美化/重构：

def section_link(name, path)    
  link = link_to(name, path) 
  if name != controller.controller_name.titlecase
    link
  else
    link_to(name, path, :class => 'current')
  end
end  

Answer 1

def section_link(name, path)
  options = {}
  options[:class] = 'current' if name == controller_name.titlecase
  link_to name, path, options
end

Answer 2

我会写：

def section_link(name, path)
  is_current = (name == controller.controller_name.titlecase)
  link_to(name, path, :class => ('current' if is_current))
end

理由： 1) 变量is_current使代码更具声明性。2）link_to假设这nil 意味着空类（我们在这里想要的）。

Answer 3

你可以这样做：

def section_link(name, path)
  link_to(name, path, class: name == controller.controller_name.titlecase ? "current" : nil)
end

但这有点难以阅读。我会将班级确定拆分为另一种方法：

def section_link(name, path)
  link_to(name, path, class: class_for(name) )
end

def class_for(name)
  name == controller.controller_name.titlecase ? "current" : nil
end

Answer 4

def section_link(name, path)     
  if name != controller_name.titlecase
    link_to(name, path)
  else
    link_to(name, path, :class => 'current')
  end
end

或类似的东西

def section_link(name, path)
  link_to(name, path, :class => "#{"current" if name == controller_name.titlecase }")
end

不要认为它真的需要重构，如果它有效......

Answer 5

def section_link(name, path)    
  link_to(name, path,
  *({class: "current"} if name == controller.controller_name.titlecase))
end