ActiveRecord 按匹配条件数排序

Question

我在 Rails 中有一个查询，它根据一系列 OR 语句获取一组记录。

@properties = policy_scope(Property).withdrawn_query(@hotlist_preference.withdrawn?)
              .or(policy_scope(Property).fallen_through_query(@hotlist_preference.fallen_through?))
              .or(policy_scope(Property).time_on_market_query(@hotlist_preference.time_on_market?))
              .includes(:listings).paginate(page: params[:page], per_page: 20)

基本上有 3 个查询，更简单的版本可能如下所示：

@properties = Property.where(state = withdrawn).or(where(state = fallen_through)).or(where(age = 4.weeks.ago))

我想要做的是按每个结果满足的 OR 条件的数量对该查询进行排序。假设有一个属性处于撤销状态，并且它的年龄是 2 周前，它的订单值为 2。如果可能的话，我还想将订单值添加到返回的属性哈希中，这样我就可以使用它在视图中。

我一直在努力寻找一种干净的方法来执行此操作，而无需再次通过一组相同的查询运行结果。

Answer 1

这看起来很脏，但这是实现这一目标的唯一方法

@properties = Property.where(state = withdrawn).or(where(state = fallen_through)).or(where(age = 4.weeks.ago)).order('case when properties.state = 'withdrawn' && properties.state = 'fallen_through' && properties.age = 'your date' then 10 when hen properties.state = 'withdrawn' && properties.state = 'fallen_through' then 9 when properties.state = 'fallen_through' && properties.age = 'your date' then 9 when properties.state = 'withdrawn'  && properties.age = 'your date' then 9 when properties.state = 'withdrawn' && properties.state = 'fallen_through' && properties.age = 'your date' then 10 when properties.state = 'withdrawn' then 8 when properties.state = 'fallen_through' then 8 when properties.age = 'your date' then 8 else 0 end)