ruby-on-rails - 使用Rails 5，如何使FriendlyId附加-"count+1"来重复slugs而不是UUID？

Question

显然，FriendlyId 已更改了之前将数字序列附加到重复 slugs 的默认方法(这正是我想要的)，现在使用 UUID:

Previous versions of FriendlyId appended a numeric sequence to make slugs unique, but this was removed to simplify using FriendlyId in concurrent code.

我目前对此功能不感兴趣，并且更喜欢使用原始方法来生成更清晰的 URL。我发现了一个类似的问题，其中 someone provided下面的代码覆盖了FriendlyId normalize_Friendly_id方法来获得我想要的功能，但是使用它会导致错误(参数数量错误(给定1，预期0)):

def normalize_friendly_id
  count = self.count "name = #{name}"
  super + "-" + count if name > 0
end

我试图将其“转换”为FriendlyId“候选人”，但我真的不知道我在做什么，并且以下内容不起作用。关于如何调整 name_candidate 方法来产生我想要的结果有什么想法吗？

class Folder < ApplicationRecord
  extend FriendlyId
  friendly_id :name_candidates, use: [ :slugged, :scoped ], scope: :account_id

  has_ancestry

  belongs_to :account
  has_many :notes, dependent: :destroy

  validates :name, presence: true

  # # https://stackoverflow.com/a/25380607/523051
  # # overrride friendlyId to append -number to duplicate folders instead of uuid's
  # def normalize_friendly_id
  #   count = self.count "name = #{name}"
  #   super + "-" + count if name > 0
  # end

  def name_candidates
    append_number = self.count "name = #{name}" if name > 0
    [
      :name,
      :name, append_number
    ]
  end
end

注意，我正在利用FriendlyId 的:scoped 功能，因此现有文件夹名称的检查应正确限定为:account_id。

Answer 1

Friendly_id 5 现在有一个 slug_candidates，可让您自定义 slug。

因此，要生成连续的 slug，您可以这样做:

friendly_id :slug_candidates, use: :slugged

def slug_candidates
  [:name, :name_and_sequence]
end

def name_and_sequence
  slug = normalize_friendly_id(name)
  sequence = Model.where("slug like '#{slug}--%'").count + 2
  "#{slug}--#{sequence}"
end

这将在以下问题中讨论: https://github.com/norman/friendly_id/issues/480

According to the author, sequential slugs are bad for performance .