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如何从Ruby中的表返回唯一的随机行?

[英]How do I return a unique random row from a table in Ruby?

模型

class Answer < ActiveRecord::Base
def self.energy(v, w)
  a = self.where('energy_id = ? AND weight = ?', v, w)
  a.offset(rand(a.count)).first.name
end

视图

<%= form_for(@answer) do |f| %>
  <%= f.submit "#{Answer.energy(3, 1)}", name: "answer", class: "btn" %>
  <%= f.submit "#{Answer.energy(4, 1)}", name: "answer", class: "btn" %>
<% end %>

我有,并且它正确返回一个随机值。 我虽然打电话36次(18对2),但我不希望同一个值返回一次以上。 我尝试了各种.pop变体,但每次都失败。

谢谢您的帮助!

对于呼叫,我使用了这些表单按钮,因此我选择了:

<% names = [] %>

<div id='one' class='center'>
  <%= form_for(@answer) do |f| %>
    <%= f.submit "#{record = Answer.energy(3, 1, names)}", name: "answer", class: "btn btn-large btn-primary" %>
    <% names << record %>
    <%= f.submit "#{record = Answer.energy(4, 1, names)}", name: "answer", class: "btn btn-large btn-primary" %>
    <% names << record %>
  <% end %>
</div>
~~ 17 more times ~~

奇迹般有效! 谢谢jvnill的帮助!

您可以将.order("RAND()")用于mysql,将.order("RANDOM()")用于postgre。

更新:无重复。

def self.energy(v, w)
  where('energy_id = ? AND weight = ?', v, w).limit(16).uniq.pluck(:name)
end

更新:16次,没有重复

这将返回第一个与能量和体重匹配的随机记录,并且不包含在exception_names中

# model
def self.energy(v, w, except_names = [])
  klass = where('energy_id = ? AND weight = ?', v, w)
  klass = klass.where('name NOT IN (?)', except_names) if except_names.any?
  klass.order('RAND()').first.name
end

这将调用该方法16次,每次在名称中添加一个uniq名称

# how to call
names = []
16.times do
  if record = Model.energe(1, 1, names)
    names << record
  end
end
def self.energy(v, w)
  c = []
  a = self.where('energy_id = ? AND weight = ?', v, w)
  b = a.offset(rand(a.count)).first.name
  unless c.include?(b) do
    puts b
    c << b
    end
end

如果您调用它16次,可以将已经可用的行作为参数传递吗? 如果可以的话,将可以工作吗?

def self.energy(v, w, records = [])
  a = self.where("energy_id = ? AND weight = ? AND id NOT IN ('-1',?)", v, w, Array(records).map(&:to_param))
  a.offset(rand(a.count)).first.name
end

用法:

records = []
16.times do
 records << Answer.energy(10, 5, records)
end

您还可以传递单个记录而不是数组:

other_record = Answer.energy(10,5)
Answer.energy(10, 5, other_record)

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