Ruby on Rails - 搜索、排序、限制然后分页

Question

我想要一个数据搜索功能。搜索后，按列排序，然后将结果限制为 200。最后用 will_paginate 分页。

我的控制器

  def search
    title = params[:title]
    company = params[:company]
    location_id = params[:location_id]
    page = params[:page]
    @wages = Wage.search(title, company, location_id,page)
  end

我的模型

def self.search(title, company, location_id, page) 如果 location_id.present？

    paginate :conditions => ['title LIKE ? AND company LIKE ? AND location_id = ?', "%#{title}%", "%#{company}%", location_id],
                    :order => "total DESC",
                    :page => page,
                    :per_page => 20                       
else

    paginate :conditions => ['title LIKE ? AND company LIKE ?', "%#{title}%", "%#{company}%"],
                    :order => "total DESC",
                    :page => page,
                    :per_page => 20                                                                    
end

结束

我尝试更改为以下代码以限制结果：

paginate :conditions => ['title LIKE ? AND company LIKE ?', "%#{title}%", "%#{company}%"].limit(200)

但它不起作用。最好的方法是什么？

Answer 1

在你的模型中

class Wage < ActiveRecord::Base
  self.per_page = 10

  def self.search(title, company, location_id, page)

    if location_id     
      wages = Wage.where('title LIKE ? AND company LIKE ? AND location_id = ?', "%#{title}%", "%#{company}%", location_id).order('total DESC').limit(200).paginate(:page => page)
    else
      wages = Wage.where('title LIKE ? AND company LIKE ?', "%#{title}%", "%#{company}%").order('total DESC').limit(200).paginate(:page => page)
    end
    return wages
  end
end

或者，您可以将“total_entries”传递给您的分页方法，如下所示：

wages = Wage.where('title LIKE ? AND company LIKE ? AND location_id = ?', "%#{title}%", "%#{company}%", location_id).order('total DESC').paginate(:page => page, :total_entries => 200)

我希望这会有所帮助。

Answer 2

检查另一个问题will-paginate-limit-number-of-results 似乎你可以做这样的事情：

wages = Wage.limit(100).paginate(:per_page => 5)

在你的情况下（我没有测试过，但我认为它可以工作）：

@wages = Wage.where('title LIKE ? AND company LIKE ?', "%#{title}%", "%#{company}%").limit(200).paginate(:per_page => 20, :order => "total DESC", :page => page)

在这种情况下，@wages 是一个 will_paginate 对象：

@wages.total_pages
=> 20

或者您可以在视图中使用will_paginate 的html_helpers 进行渲染 (API-documentation#view-template-helpers)。

# controller
@wages = Wage.where('title LIKE ? AND company LIKE ?', "%#{title}%", "%#{company}%").limit(200)
# you should have @wages.count = 200

# view
<%= will_paginate @wages, :style => 'color:blue' %>