重构可选参数方法以删除“if”语句

Question

我有一个方法 select_active_buildings，其中三个参数默认为 nil，以使用提供的信息过滤我的数据库。

def select_active_buildings(category:nil, upgrader:nil, upgrade_resource:nil)
  my_active_buildings = self.buildings.active(self.townhall_level)
  if category && upgrader && upgrade_resource
    return my_active_buildings.where(category:category, upgrader:upgrader, upgrade_resource:upgrade_resource)
  elsif category && upgrader
    return my_active_buildings.where(category:category, upgrader:upgrader)
  elsif category && upgrade_resource
    return my_active_buildings.where(category:category, upgrade_resource:upgrade_resource)
  elsif upgrader && upgrade_resource
    return my_active_buildings.where(upgrader:upgrader, upgrade_resource:upgrade_resource)
  elsif category
    return my_active_buildings.where(category:category)
  elsif upgrader
    return my_active_buildings.where(upgrader:upgrader)
  elsif upgrade_resource
    return my_active_buildings.where(upgrade_resource:upgrade_resource)
  else
    return my_active_buildings
  end
end

我需要帮助重构这个方法。

Answer 1

我会这样做：

def select_active_buildings(category: nil, upgrader: nil, upgrade_resource: nil)
  my_active_buildings = self.buildings.active(self.townhall_level)
  cond = {
    category: category,
    upgrader: upgrader,
    upgrade_resource: upgrade_resource
  }.select { |k, v| v }
  my_active_buildings.where(cond)
end

或者，

def select_active_buildings(category: nil, upgrader: nil, upgrade_resource: nil)
  my_active_buildings = self.buildings.active(self.townhall_level)
  cond = {}
  cond[:category] = category if category
  cond[:upgrader] = upgrader if upgrader
  cond[:upgrade_resource] = upgrade_resource if upgrade_resource
  my_active_buildings.where(cond)
end

或者，如果我真的不关心 select_active_function 级别的语法检查，

def select_active_buildings(cond)
  my_active_buildings = self.buildings.active(self.townhall_level)
  my_active_buildings.where(cond)
end

如果你真的不想要 .where({}) 大小写，你可以在它前面加上 return my_active_buildings if cond.empty?。

Answer 2

假设您正在使用 ActiveRecord、Mongoid 或其他具有可链接 where 的东西，您可以利用以下事实：

o.where(:a => b, :c => d)

等同于：

o.where(:a => b).where(:c => d)

这让你可以这样做：

def select_active_buildings(category: nil, upgrader: nil, upgrade_resource: nil)
  buildings = self.buildings.active(self.townhall_level)
  buildings = buildings.where(category: category) if(category)
  buildings = buildings.where(upgrader: upgrader) if(upgrader)
  buildings = buildings.where(upgrade_resource: upgrade_resource) if(upgrade_resource)
  buildings
end

甚至：

def select_active_buildings(conditions)
  buildings = self.buildings.active(self.townhall_level)
  %i[category upgrader upgrade_resource]
    .select { |f| conditions.has_key?(f) }
    .inject(buildings) { |q, f| q.where(f => conditions[f]) }
end

Answer 3

def select_active_buildings(category: nil, upgrader: nil, upgrade_resource: nil)
  buildings.active(self.townhall_level).where(
    **({category: category} if category).to_h,
    **({upgrader: upgrader} if upgrader).to_h,
    **({upgrade_resource: upgrade_resource} if upgrade_resource).to_h,
  )
end

重构可选参数方法以删除“if”语句

Refactor optional parameters method to remove `if` statements

ruby-on-rails

ruby

optional-parameters