紅寶石複雜的查詢優雅的解決方案

Question

在我的軌道4,5的項目，我有以下表

在這種SO question，我搜索了一個SQL查詢與實際項目狀態ID獲取項目= XX。實際上，我的意思是最大（created_at）。

我得到了我的查詢答案是

select p.* from projects p 
    left join projects_status_projects psp on (psp.project_id = p.id) 
    where created_at = (select max(created_at) from projects_status_projects 
      where project_id = p.id)  
    and project_status_id = XX 

我的模型定義

class Project < ActiveRecord::Base 
    has_many :projects_status_projects 
    has_many :projects_statuses, :through => :projects_status_projects 
end 

class ProjectStatus < ActiveRecord::Base 
    has_many :projects_status_projects 
    has_many :projects, :through => :projects_status_projects 
end 

class ProjectsStatusType < ActiveRecord::Base 
    belongs_to :project 
    belongs_to :project_status 
end 

在我項目模型，我有以下方法

def self.with_status(status) 
    joins(:projects_status_projects) 
     .where('projects_status_projects.created_at = (select max(created_at) from 
        projects_status_projects where project_id = p.id)') 
     .where('projects_status_projects.project_status_id = ?', status) 
end 

雖然查詢是正確的，但收到的結果已被很好地過濾，I f ind這個解決方案可怕並且不夠優雅。

有什麼辦法可以得到與範圍相同的結果嗎？

感謝您的幫助。

Answer 1

怎麼樣？

scope :with_status, ->(status = "default_status") { 
    joins(:projects_status_projects). 
    where('projects_status_projects.project_status_id = ?', status). 
    order("projects_status_projects.created_at DESC").first 
    } 

Answer 2

你基於評論認爲

scope :with_status, -> (status) { 
    ProjectsStatusType.where(:project_status_id, status).order(:created_at).last.project 
} 

編輯的內容：

正如sockmonk說，範圍應該是可鏈接。這是一個更乾淨的方式來做到這一點，如果沒有找到項目，這也解決了問題。

# ProjectStatusType model 
scope :with_ordered_status, -> (status) { 
    where(:project_status_id, status).order(:created_at) 
} 

# Project model 
def self.find_by_status(status) 
    project_statuses = ProjectsStatusType.with_ordered_status(status) 
    project_statuses.any? ? project_statuses.last.project : nil 
end 

Answer 3

scope :with_status, ->(status = "default_status") { 
    joins(:projects_status_projects) 
    .where('projects_status_projects.project_status_id = ?', status) 
    .order("projects_status_projects.created_at DESC")  
} 

你打電話時，你會想釘一個「首先」到它的結束;不能在範圍本身中包含.first，因爲這會使其無法鏈接。