明确设置选择的键和值

Question

如何在散列中显式设置键和值，然后在选择中使用它们？我有一个链接模型，可以保存外部链接。链接有标题和网址。我想使用标题作为键，将url作为值。

---------------------更新-------------------------

这项工作..

links = Link.all

link_array = []

links.each do |link| 
  link_array << [link.title,link.url]
end

但是，现在这就是问题。我想将此数组连接到另一个数组，以便可以从单个表单选择中选择两个模型。像这样......

a = PagesController.action_methods
   # this grabs each action from the pages controller that will later be used as a route

b = a.select {|s| s.include? "callback"}
c = a - b
   # b and c removes any position in the array that includes the word 'callback' so that only actions defined in the controller are returned

links = Link.all

link_array = []

links.each do |link| 
  link_array << [link.title,link.url]
end

@all_links = c + link_array
   # desired result is an array used in a single form select containing both external links and internal links

Answer 1

我不确定你的目的是什么，但看起来你正在寻找这样的东西：

Link.where(title: url)

...其中url是一些局部变量。这将生成类似于以下内容的SQL语句：

SELECT * FROM links WHERE title='yoururl'

Answer 2

这就是我想要的。也许有更好的方法。

 a = PagesController.action_methods
 b = a.select {|s| s.include? "callback"}
 c = a - b
 @c_array = []
 c.each do |p|
   @c_array << [p, "#{p}_path"]
 end

 links = Link.all
 @link_array = []
 links.each do |link| 
   @link_array << [link.title, link.url]
 end

 @all_links = @c_array + @link_array

包含内部路由（到页面操作）和指向URL的外部链接的单个数组。现在，在第三个模型中，我可以有一个标题为“pathto”的字符串列，并存储路径（作为字符串）或url。然后，在帮助器中执行此操作：

    def send_route(pathto)
      if pathto.include? "http"
        pathto # just go to the url
      else
        self.send( pathto ) # create a route from the string stored in the db .. i.e. "home_path"
      end
    end