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，然后将结果限制在200以内。最后用will_pagenate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

我的型号

def self.search（标题、公司、位置id、页面）如果location_id.spresent？

    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

为了限制结果，我尝试更改为以下代码：

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

在您的模型中

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"传递给分页方法，如图所示：

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

我希望这能有所帮助。

Answer 2

用另一个问题检查结果的页码限制似乎可以这样做：

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

在你的情况下（我还没有测试过，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will_paginate对象：

@wages.total_pages
=> 20

也可以使用视图中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' %>