ruby-on-rails – 如何在rails中创建递归函数,它将返回父级的所
发布时间:2020-12-17 03:03:27 所属栏目:百科 来源:网络整理
导读:我有一个方法如图所示: def all_pages_and_its_n_level_child @parent_pages = Page.where(:parent_id = params[:page_id]) #this will give me all child of this page for page in @parent_pages child_exists=Page.where(:parent_id=page.id) #this will
我有一个方法如图所示:
def all_pages_and_its_n_level_child @parent_pages = Page.where(:parent_id => params[:page_id]) #this will give me all child of this page for page in @parent_pages child_exists=Page.where(:parent_id=>page.id) #this will give all children of that page #*I want to make this loop further so that i can check to N-levels,#this is only for two levels* end end 解决方法
这是一些样本(它没有经过测试!但它会给你一些线索):
def all_children(children_array = []) children = Page.where(parent_id: self.id) children_array += children.all children.each do |child| child.all_children(children_array) end children_array end 所以这是递归,它将尝试从父级中查找所有子级(也嵌套).我知道这是非常丑陋和低效的,但我希望ii会给你一个关于找到嵌套元素的线索. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |