34

I have an array @level1 which looks like this :

[[3.0, 4, 2], [2.0, 48, 3], [2.1, 56, 4], ............]

I want to apply pagination on this array such each page displays only a few rows at once. I tried this:

@temp1 = @level1.paginate(:page => params[:page])

But it throws the following error: 

undefined method `paginate' for [[3.0, 4, 2], [2.0, 48, 3], [2.1, 56, 4]]:Array

How can I perform pagination on this using will_paginate?

Pi Horse
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5 Answers5

82

See https://github.com/mislav/will_paginate/blob/master/lib/will_paginate/array.rb

Just require 'will_paginate/array' before you try it and it will work just fine. If it were me, I'd put that in say config/initalizers/will_paginate_extensions.rb so it's available right away and from everywhere.

Philip Hallstrom
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    This is a much better solution. Thanks! – Nathan Colgate Apr 04 '13 at 20:09
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    @NathanColgate If you have to use a lot, then maybe yes. But if you want to use this in just a few places, I would not pollute the standard classes with stuff that aren't their responsibility. A `PaginatedArray`, that inherits from `Array` and has this method might be the better way. But I'll go with Chris' solution. – iGEL Jul 22 '13 at 13:59
27

Sure, you can use WillPaginate::Collection to construct arbitrary pagination across collections. Assuming an array called values:

@values = WillPaginate::Collection.create(current_page, per_page, values.length) do |pager|
  pager.replace values
end

You can then treat @values just like any other WillPaginate collection, including passing it to the will_paginate view helper.

Chris Heald
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    This works great if you have an array of objects representing one page (and the total count) and want to use various WillPaginate view helpers - for example if you need to bypass ActiveRecord for performance reasons. –  Apr 15 '14 at 21:24
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    I tried this approach and was half successful. It generates the `will_paginate` view helper, but the array is not broken down: the full list is generated. Anything in particular needed in the view? – Jerome May 27 '14 at 06:53
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    @Jerome You need to populate the values array with only the values for the current page. You then pass in current_page/per_page/the total number of values in the whole collection manually, and it constructs the pagination helpers. – Chris Heald May 27 '14 at 07:02
  • That sounds logical. In other words , since I'm sorting the array, first sort array, then '@values.page(params[:page])` or `.pagination[...]` and finally: `WillPaginate::Collection.create(1, 10, @values.length)` . This however returns `undefined method `page' for # < Array`... Is this documented someplace? I've been searching... Specific posting http://stackoverflow.com/questions/23875093/rails-willpaginatecollection-not-paginating-array – Jerome May 27 '14 at 08:23
  • Wow, you have no idea how this answer helped me! Massive thumbs up – Uri Klar Jul 30 '14 at 15:31
  • The answer suggesting that you require will_paginate/array is a better solution – Aaron B. Russell Sep 20 '14 at 17:02
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    Assuming you don't mind extending Array. If you need it in one or two places, extending Array might be a code smell. will_paginate/array does exactly what this code does, except it extends it onto Array. – Chris Heald Sep 21 '14 at 04:14
5

Given a collection array in a Rails project

will_paginate ~>3.0.5

collection.paginate(page: params[:page], per_page: 10)

returns the pagination for the array values.

Do not forget require 'will_paginate/array' on top of the controller.

Francisco Quintero
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2

This is my example based in the answer of @chris-heald

products = %w(i love ruby on rails)
@products = WillPaginate::Collection.create(params[:page], params[:per_page], products.length) do |pager|
   pager.replace products[pager.offset, pager.per_page]
end
Alberto Camargo
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0

I wasn't really happy with the answers provided in this thread.

I came up with this solution.

# app/lib/paged_array.rb

require 'will_paginate/array'

class PagedArray
  delegate_missing_to :@paged_collection

  def initialize(collection, page: 1, per_page: 20)
    @paged_collection = collection.paginate(
      page: page,
      per_page: per_page
    )
  end
end

And you can use it like that:

superheroes = %w[spiderman batman superman ironman]

PagedArray.new(superheroes, page: 1, per_page: 2)

I like this solution since it's explicit and you are not polluting your standard classes.

That being said, you are dependent of rails since I use "delegate_missing_to".

Jeremie Ges
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