CODEWITHSUNDEEP
ruby-on-railsrubypagination
Patrick Ng
Patrick Ng

Reputation: 203

Ruby on Rails - Search, Order, Limit then Paginate

I want to have a search function for the data. After search, then order by a column and then limit the results to 200 only. Finally paginate with will_paginate.

My Controller

  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

My model

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

end

I tried changing to the below code in order to limit the result: 

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

But it is not working. What is a best way to do it?

Upvotes: 0

Views: 2529

Answers (2)

damoiser
damoiser

Reputation: 6238

Checking with this other question will-paginate-limit-number-of-results seems that you can do something like this:

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

In your case (I haven't tested it, but I think that it can work):

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

In this case @wages is a will_paginate object:

@wages.total_pages
=> 20

Or you can render with the html_helpers of will_paginate in the view (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' %>

Upvotes: 0

Yosep Kim
Yosep Kim

Reputation: 2941

In your model

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

Alternatively, you can pass in "total_entries" to your paginate method, as shown:

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

I hope this helps.

Upvotes: 0

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