Question

共有5个级别。

每个级别都有一定的天数，在习惯升级到下一级别之前必须通过（按n_days分解）：

      case n_days     
          when 0..9
            1
          when 10..24
            2
          when 25..44
            3  #Level 3
          when 45..69
            4
          when 70..99
            5
          else
            "Mastery"
         end
      end

如何使用n_days这样的习惯指数从当前级别调用<%= habit.current_level.n_days.count_off_from_zero_to_show %>？

例如，如果我们在第3级专门处于50，则会在习惯指数中显示第5天。

habit.rb

class Habit < ActiveRecord::Base
    belongs_to :user
    has_many :comments, as: :commentable
    has_many :levels
    serialize :committed, Array
    validates :date_started, presence: true
    before_save :current_level
    acts_as_taggable
    scope :private_submit, -> { where(private_submit: true) }
    scope :public_submit, -> { where(private_submit: false) }

attr_accessor :missed_one, :missed_two, :missed_three

    def save_with_current_level
        self.levels.build
        self.levels.build
        self.levels.build
        self.levels.build
        self.levels.build
        self.save
    end

    def self.committed_for_today
    today_name = Date::DAYNAMES[Date.today.wday].downcase
    ids = all.select { |h| h.committed.include? today_name }.map(&:id)
    where(id: ids)
  end 

    def current_level_strike
      levels[current_level - 1] # remember arrays indexes start at 0
    end

    def current_level
            return 0 unless date_started
            committed_wdays = committed.map { |day| Date::DAYNAMES.index(day.titleize) }
            n_days = ((date_started.to_date)..Date.today).count { |date| committed_wdays.include? date.wday } - self.missed_days

      case n_days     
          when 0..9
            1
          when 10..24
            2
          when 25..44
            3
          when 45..69
            4
          when 70..99
            5
          else
            "Mastery"
        end
    end
end

level.rb

class Level < ActiveRecord::Base
  belongs_to :habit
end

模式

create_table "habits", force: true do |t|
  t.integer  "missed_days",    default: 0
  t.text     "committed"
  t.integer  "days_lost",   default: 0
  t.datetime "date_started"
  t.string   "trigger"
  t.string   "target"
  t.string   "reward"
  t.boolean  "private_submit"
  t.integer  "user_id"
  t.datetime "created_at",                 null: false
  t.datetime "updated_at",                 null: false
  t.integer  "order"
end

add_index "habits", ["user_id", "created_at"], name: "index_habits_on_user_id_and_created_at"
add_index "habits", ["user_id"], name: "index_habits_on_user_id"

create_table "levels", force: true do |t|
  t.integer  "habit_id"
  t.integer  "days_lost",   default: 0
  t.integer  "missed_days",   default: 0
  t.integer  "current_level"
  t.datetime "created_at",                null: false
  t.datetime "updated_at",                null: false
end

add_index "levels", ["habit_id"], name: "index_levels_on_habit_id"

要点：https://gist.github.com/RallyWithGalli/c66dee6dfb9ab5d338c2

Answer 1

在-lboost_regex方法中，您有两行逻辑，最终代表current_level的值。如果要在该方法或case语句之外重用该值，则应将值存储在实例变量中，或将逻辑移动到另一个方法中。如果逻辑只对这个类很重要，那么我可能会把它变成私有方法。

n_days

私有方法可以进一步完善，但希望这可以让您了解如何继续提取共享逻辑。

从current_level计算天数？

1 个答案: