Question

I’m generating JSON as follows:

def suggest_gems
  @allgems = RubyGem.where("name like ?", "%#{params[:q]}%")
  respond_to do |format|
  format.json { render :json => @allgems.to_json }
  end
end 

However, the RubyGem table is quite large, with values for descriptions, links and created, updated at. How can I only output the name and id in the generated JSON to decrease loading times?
Thanks.

Answer 1

So I figured how to do that without changing the as_json method or using the respond_with statement, by adding a few lines of code. Here’s what I did:

def suggest_gems (result = [])
   @allgems = RubyGem.where("name like ?", "%#{params[:q]}%")
   @allgems.each { |g|
     result << { :id => g.id, :name => g.name }
     } 
   respond_to do |format|
   format.json { render :json => result.to_json }
   end
end 

I basically defined an array and associated the name and id attributes accordingly, then formatted the array as JSON. Shaved around 6 seconds off the loading time 🙂

Adding this answer here so that others might find it helpful!