Question

执行this code kata并且我必须从以下方法中尽可能多地删除if语句：

#  returns permissions if user is set in security context
def get_user_permissions 
  user_permissions = Set.new
  if (@user != nil)
    user_permissions << :DEFAULT_PERMISSION
    if (has_cm_team_role) 
      user_permissions << :CM_TEAM_ROLE_PERMISSION
    end
    if (has_cm_invoice_view_role || has_invoice_finance_role)
      user_permissions << :CM_INVOICE_USER_PERMISSION
      user_permissions << :INVOICE_VIEW_PERMISSION
      user_permissions << :ACCESS_ALL_INVOICE_PERMISSION
    end
    if (has_invoice_finance_role) 
      user_permissions << :FINANCE_INVOICE_PERMISSION
    end
    if (has_application_access)
      user_permissions << :CM_INVOICE_USER_PERMISSION
    end
    if (has_application_access(:CM_INVOICE_ROLE)) 
      user_permissions << :CM_ANY_INVOICE_PERMISSION
    end
    if (has_application_access(:PA_INVOICE_ROLE)) 
      user_permissions << :PA_ANY_INVOICE_PERMISSION
    end
    if (has_application_access(:SDT_INVOICE_ROLE))
      user_permissions << :SDT_ANY_INVOICE_PERMISSION
    end
  end
  user_permissions
end

我的第一次尝试几乎有效，但有些测试失败了：

def get_user_permissions 
  user_permissions = Set.new
  if (@user != nil)
    user_permissions << :DEFAULT_PERMISSION
    # add the permissions in another method
    add_permissions(user_permissions)
  end
  user_permissions
end

def add_permissions(user_permissions)
  # a hash where each key is a condition, and each value is a permission
  hash = {
    has_cm_team_role => :CM_TEAM_ROLE_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :CM_INVOICE_USER_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :INVOICE_VIEW_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :ACCESS_ALL_INVOICE_PERMISSION,
    has_invoice_finance_role => :FINANCE_INVOICE_PERMISSION,
    has_application_access => :CM_INVOICE_USER_PERMISSION,
    has_application_access(:CM_INVOICE_ROLE) => :CM_ANY_INVOICE_PERMISSION,
    has_application_access(:PA_INVOICE_ROLE) => :PA_ANY_INVOICE_PERMISSION,
    has_application_access(:SDT_INVOICE_ROLE) => :SDT_ANY_INVOICE_PERMISSION
  }

  # loop through the hash and add permissions if the key is true
  hash.each do |condition, permission|
    if (condition)
      user_permissions << permission
    end
  end

这种方法的问题在于，每个不会评估三个 OR 语句（在我的哈希的第二，第三和第四个键中）。所以我通过使用Procs解决这个问题，如下：

def add_permissions(user_permissions)
  hash = {
    Proc.new{has_cm_team_role} => :CM_TEAM_ROLE_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :INVOICE_VIEW_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :ACCESS_ALL_INVOICE_PERMISSION,
    Proc.new{has_invoice_finance_role} => :FINANCE_INVOICE_PERMISSION,
    Proc.new{has_application_access} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_application_access(:CM_INVOICE_ROLE)} => :CM_ANY_INVOICE_PERMISSION,
    Proc.new{has_application_access(:PA_INVOICE_ROLE)} => :PA_ANY_INVOICE_PERMISSION,
    Proc.new{has_application_access(:SDT_INVOICE_ROLE)} => :SDT_ANY_INVOICE_PERMISSION
  }
  hash.each do |condition, permission|
    if (condition.call)
      user_permissions << permission
    end
  end
end

确定，这是有效的，测试全部通过，但我将所有的密钥转换为Procs，只是为了让其中3个正确评估。我更喜欢只在需要时使用Procs，即第二，第三和第四个键，如下所示：

def add_permissions(user_permissions)
  hash = {
    # just use the method name unless a Proc is required
    has_cm_team_role => :CM_TEAM_ROLE_PERMISSION,
    # use a Proc here, so that the OR is evaluated later
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :INVOICE_VIEW_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :ACCESS_ALL_INVOICE_PERMISSION,
    has_invoice_finance_role => :FINANCE_INVOICE_PERMISSION,
    has_application_access => :CM_INVOICE_USER_PERMISSION,
    has_application_access(:CM_INVOICE_ROLE) => :CM_ANY_INVOICE_PERMISSION,
    has_application_access(:PA_INVOICE_ROLE) => :PA_ANY_INVOICE_PERMISSION,
    has_application_access(:SDT_INVOICE_ROLE) => :SDT_ANY_INVOICE_PERMISSION
  }

然后用某种方法确定密钥是否为Proc，如果是，则调用它，但如果没有，只需将条件视为任何其他密钥：

  hash.each do |condition, permission|
    if condition.is_proc?
      if (condition.call)
        user_permissions << permission
      end
    elsif condition
      user_permissions << permission
    end
  end
end

任何提示？还有比我试图做的更好的想法吗？通过这样做，我是否使它更具周期性复杂性？我应该坚持使用我的工作解决方案，其中所有的密钥都是Procs？

Answer 1

condition.is_a? Proc满足您的条件吗？

Answer 2

没有关于过程问题的事情。我从未做过代码kata，但我认为你可以改进的东西很少。

首先，作为哈希键的条件对我来说是一个坏主意。你将有许多重复的密钥，只有虚假和真实我会以另一种方式执行此操作并将您的权限设置为键，值将为true或false。

你也有3次完全相同的指令，每次只改变2个字母。

if (has_application_access(:CM_INVOICE_ROLE)) 
  user_permissions << :CM_ANY_INVOICE_PERMISSION
end
if (has_application_access(:PA_INVOICE_ROLE)) 
  user_permissions << :PA_ANY_INVOICE_PERMISSION
end
if (has_application_access(:SDT_INVOICE_ROLE))
  user_permissions << :SDT_ANY_INVOICE_PERMISSION
end

可以减少到

%(cm pa sdt).each do |key|
    user_permissions << :"#{key}_ANY_INVOICE_PERMISSION" if has_application_access(:"#{key}_INVOICE_ROLE")
end

我试着这样做，得到了两个结果。一个使用哈希，一个只是尽可能地切割。

# Hash version
def get_user_permissions
  return Set.new if !!@user

  user_permissions_hash = {:DEFAULT_PERMISSION => true,
                           :CM_TEAM_ROLE_PERMISSION => has_cm_team_role,
                           :CM_INVOICE_USER_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role || has_application_access,
                           :INVOICE_VIEW_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role,
                           :ACCESS_ALL_INVOICE_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role,
                           :FINANCE_INVOICE_PERMISSION => has_invoice_finance_role
  }
  %(cm pa sdt).each do |key|
    user_permissions_hash[:"#{key}_ANY_INVOICE_PERMISSION"] = has_application_access(:"#{key}_INVOICE_ROLE")
  end

  return user_permissions_hash.map {|k, v| k if v}.compact.to_set
end

# Normal version
def get_user_permissions
  return (user_permissions = Set.new) if !!@user

  user_permissions << :DEFAULT_PERMISSION
  user_permissions << :CM_TEAM_ROLE_PERMISSION if has_cm_team_role
  user_permissions << :CM_INVOICE_USER_PERMISSION if has_cm_invoice_view_role || has_invoice_finance_role || has_application_access
  if (has_cm_invoice_view_role || has_invoice_finance_role)
    user_permissions << :INVOICE_VIEW_PERMISSION
    user_permissions << :ACCESS_ALL_INVOICE_PERMISSION
  end
  user_permissions << :FINANCE_INVOICE_PERMISSION if has_invoice_finance_role
  %(cm pa sdt).each do |key|
    user_permissions << :"#{key}_ANY_INVOICE_PERMISSION" if has_application_access(:"#{key}_INVOICE_ROLE")
  end

  user_permissions
end

Ruby是否能够确定密钥（或方法参数）是否为proc？

2 个答案: