Ruby on Rails: custom rendering in controller

Question

So basically I have this for my index page:

def index
  @articles = Article.paginate(:per_page => 15, :page => params[:page])
end

I implemented soft deletion, which basically works like this: there is an additional boolean column in db that is called is_active and is true by default. Deleting is rewritten to just change that to false instead of destruction, and made a page to view soft-deleted entries.

The issue: current workaround I found for paginate is to simply add

<% if article.is_active %>

in my index.html.erb. The flaw is: when I delete something it still considered there by paginate, so instead of say 15 entries I will see 14. Even worse, on undeleting page it shows same amount of blank pages, and deleted entries are on their would-be-appropriate pages (so for example first entry may end up being on page 14 instead of 1). It is not critical flaw, but I'd like to know if I can fix it without rewriting too much.

Maybe I can change something in controller so it doesn't send any entries that have true or false in that field depending on what I want to output?

Eyeslandic · Accepted Answer · 2017-03-27 23:22:40Z

2

Can't you just filter by this field?

@articles = Article.where(is_active: true).paginate(per_page: 15, page: params[:page])

answered Mar 27, 2017 at 23:22

Eyeslandic

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2 Comments

user3735111 Over a year ago

Worked like a charm! Thanks! I'm just getting hang of ruby on rails, but not completely new to programming, so I suspected there is more than just .all or .paginate, etc., just wasn't sure how to google it properly.

Eyeslandic Over a year ago

guides.rubyonrails.org/active_record_querying.html This guide is the best place to start, good luck

archana · Accepted Answer · 2017-03-27 23:23:27Z

0

You can use a scope or a where clause to get only active articles:

def index
  @articles = Article.where("is_active = ?", true).paginate(:per_page => 15, :page => params[:page])
end

answered Mar 27, 2017 at 23:23

archana

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