重构可选参数删除`if`语句的方法

我有一个方法select_active_buildings有三个参数，默认为nil来使用提供的信息过滤我的数据库。重构可选参数删除`if`语句的方法

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

我需要帮助重构此方法。

来源

2015-12-15 thedanotto

这个问题将在http://codereview.stackexchange.com/ –

更好这个问题可能是适用于[代码审查（http://codereview.stackexchange.com/help）只要（a）您的代码按预期工作，（b）您的代码是真实的代码，而不是示例代码，并且（c）您的代码包含在问题的正文中。如果您希望通过同行评审来改进代码的各个方面，请将其发布在代码评审中。 – Phrancis

谢谢，你是对的。 – thedanotto

假设你用ActiveRecord，Mongoid，或用可链接where做别的事情，你可以采取的事实，即这样的：

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

是一样的：

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

这可以让你做到这一点：

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

来源

2015-12-15 04:48:48

我和你的第一个选择一起去了。非常感谢。 – thedanotto

我应该这样做：

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，如果我真的不关心语法在select_active_function水平检查，

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({})的情况下，你可以把它放在return my_active_buildings if cond.empty?之前。

来源

2015-12-15 04:22:27 Amadan

最后一个变体是'where（cond.slice（：category，...））'。 –

-1

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

来源

2015-12-15 05:02:33 sawa

重构可选参数删除`if`语句的方法

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