I need to get the previous and next active record objects with Rails. I did it, but don't know if it's the right way to do that.
What I've got:
Controller:
#product = Product.friendly.find(params[:id])
order_list = Product.select(:id).all.map(&:id)
current_position = order_list.index(#product.id)
#previous_product = #collection.products.find(order_list[current_position - 1]) if order_list[current_position - 1]
#next_product = #collection.products.find(order_list[current_position + 1]) if order_list[current_position + 1]
#previous_product ||= Product.last
#next_product ||= Product.first
product_model.rb
default_scope -> {order(:product_sub_group_id => :asc, :id => :asc)}
So, the problem here is that I need to go to my database and get all this ids to know who is the previous and the next.
Tried to use the gem order_query, but it did not work for me and I noted that it goes to the database and fetch all the records in that order, so, that's why I did the same but getting only the ids.
All the solutions that I found was with simple order querys. Order by id or something like a priority field.
Write these methods in your Product model:
class Product
def next
self.class.where("id > ?", id).first
end
def previous
self.class.where("id < ?", id).last
end
end
Now you can do in your controller:
#product = Product.friendly.find(params[:id])
#previous_product = #product.next
#next_product = #product.previous
Please try it, but its not tested.
Thanks
I think it would be faster to do it with only two SQL requests, that only select two rows (and not the entire table). Considering that your default order is sorted by id (otherwise, force the sorting by id) :
#previous_product = Product.where('id < ?', params[:id]).last
#next_product = Product.where('id > ?', params[:id]).first
If the product is the last, then #next_product will be nil, and if it is the first, then, #previous_product will be nil.
There's no easy out-of-the-box solution.
A little dirty, but working way is carefully sorting out what conditions are there for finding next and previous items. With id it's quite easy, since all ids are different, and Rails Guy's answer describes just that: in next for a known id pick a first entry with a larger id (if results are ordered by id, as per defaults). More than that - his answer hints to place next and previous into the model class. Do so.
If there are multiple order criteria, things get complicated. Say, we have a set of rows sorted by group parameter first (which can possibly have equal values on different rows) and then by id (which id different everywhere, guaranteed). Results are ordered by group and then by id (both ascending), so we can possibly encounter two situations of getting the next element, it's the first from the list that has elements, that (so many that):
have the same group and a larger id
have a larger group
Same with previous element: you need the last one from the list
have the same group and a smaller id
have a smaller group
Those fetch all next and previous entries respectively. If you need only one, use Rails' first and last (as suggested by Rails Guy) or limit(1) (and be wary of the asc/desc ordering).
This is what order_query does. Please try the latest version, I can help if it doesn't work for you:
class Product < ActiveRecord::Base
order_query :my_order,
[:product_sub_group_id, :asc],
[:id, :asc]
default_scope -> { my_order }
end
#product.my_order(#collection.products).next
#collection.products.my_order_at(#product).next
This runs one query loading only the next record. Read more on Github.
Related
I have a database query where I want to get an array of Users that are distinct for the set:
#range is a predefinded date range
#shift_list is a list of filtered shifts
def listing
Shift
.where(date: #range, shiftname: #shift_list)
.select(:user_id)
.distinct
.map { |id| User.find( id.user_id ) }
.sort
end
and I read somewhere that for readability, or isolating for testing, or code reuse, you could split this into seperate methods:
def listing
shiftlist
.select(:user_id)
.distinct
.map { |id| User.find( id.user_id ) }
.sort
end
def shift_list
Shift
.where(date: #range, shiftname: #shift_list)
end
So I rewrote this and some other code, and now the page takes 4 times as long to load.
My question is, does this type of method splitting cause the database to be hit twice? Or is it something that I did elsewhere?
And I'd love a suggestion to improve the efficiency of this code.
Further to the need to remove mapping from the code, this shift list is being created with the following code:
def _month_shift_list
Shift
.select(:shiftname)
.distinct
.where(date: #range)
.map {|x| x.shiftname }
end
My intention is to create an array of shiftnames as strings.
I am obviously missing some key understanding in database access, as this method is clearly creating part of the problem.
And I think I have found the solution to this with the following:
def month_shift_list
Shift.
.where(date: #range)
.pluck(:shiftname)
.uniq
end
Nope, the database will not be hit twice. The queries in both methods are lazy loaded. The issue you have with the slow page load times is because the map function now has to do multiple finds which translates to multiple SELECT from the DB. You can re-write your query to this:
def listing
User.
joins(:shift).
merge(Shift.where(date: #range, shiftname: #shift_list).
uniq.
sort
end
This has just one hit to the DB and will be much faster and should produce the same result as above.
The assumption here is that there is a has_one/has_many relationship on the User model for Shifts
class User < ActiveRecord::Base
has_one :shift
end
If you don't want to establish the has_one/has_many relationship on User, you can re-write it to:
def listing
User.
joins("INNER JOIN shifts on shifts.user_id = users.id").
merge(Shift.where(date: #range, shiftname: #shift_list).
uniq.
sort
end
ALTERNATIVE:
You can use 2 queries if you experience issues with using ActiveRecord#merge.
def listing
user_ids = Shift.where(date: #range, shiftname: #shift_list).uniq.pluck(:user_id).sort
User.find(user_ids)
end
I have a simple association like
class Slot < ActiveRecord::Base
has_many :media_items, dependent: :destroy
end
class MediaItem < ActiveRecord::Base
belongs_to :slot
end
The MediaItems are ordered per Slot and have a field called ordering.
And want to avoid n+1 querying but nothing I tried works. I had read several blogposts, railscasts etc but hmm.. they never operate on a single model and so on...
What I do is:
def update
#slot = Slot.find(params.require(:id))
media_items = #slot.media_items
par = params[:ordering_media]
# TODO: IMP remove n+1 query
par.each do |item|
item_id = item[:media_item_id]
item_order = item[:ordering]
media_items.find(item_id).update(ordering: item_order)
end
#slot.save
end
params[:ordering_media] is a json array with media_item_id and an integer for ordering
I tried things like
#slot = Slot.includes(:media_items).find(params.require(:id)) # still n+1
#slot = Slot.find(params.require(:id)).includes(:media_items) # not working at all b/c is a Slot already
media_items = #slot.media_items.to_a # looks good but then in the array of MediaItems it is difficult to retrieve the right instance in my loop
This seems like a common thing to do, so I think there is a simple approach to solve this. Would be great to learn about it.
First at all, at this line media_items.find(item_id).update(ordering: item_order) you don't have an n + 1 issue, you have a 2 * n issue. Because for each media_item you make 2 queries: one for find, one for update. To fix you can do this:
params[:ordering_media].each do |item|
MediaItem.update_all({ordering: item[:ordering]}, {id: item[:media_item_id]})
end
Here you have n queries. That is the best we can do, there's no way to update a column on n records with n distinct values, with less than n queries.
Now you can remove the lines #slot = Slot.find(params.require(:id)) and #slot.save, because #slot was not modified or used at the update action.
With this refactor, we see a problem: the action SlotsController#update don't update slot at all. A better place for this code could be MediaItemsController#sort or SortMediaItemsController#update (more RESTful).
At the last #slot = Slot.includes(:media_items).find(params.require(:id)) this is not n + 1 query, this is 2 SQL statements query, because you retrieve n media_items and 1 slot with only 2 db calls. Also it's the best option.
I hope it helps.
Let's say you have an assocation in one of your models like this:
class User
has_many :articles
end
Now assume you need to get 3 arrays, one for the articles written yesterday, one of for the articles written in the last 7 days, and one of for the articles written in the last 30 days.
Of course you might do this:
articles_yesterday = user.articles.where("posted_at >= ?", Date.yesterday)
articles_last7d = user.articles.where("posted_at >= ?", 7.days.ago.to_date)
articles_last30d = user.articles.where("posted_at >= ?", 30.days.ago.to_date)
However, this will run 3 separate database queries. More efficiently, you could do this:
articles_last30d = user.articles.where("posted_at >= ?", 30.days.ago.to_date)
articles_yesterday = articles_last30d.select { |article|
article.posted_at >= Date.yesterday
}
articles_last7d = articles_last30d.select { |article|
article.posted_at >= 7.days.ago.to_date
}
Now of course this is a contrived example and there is no guarantee that the array select will actually be faster than a database query, but let's just assume that it is.
My question is: Is there any way (e.g. some gem) to write this code in a way which eliminates this problem by making sure that you simply specify the association conditions, and the application itself will decide whether it needs to perform another database query or not?
ActiveRecord itself does not seem to cover this problem appropriately. You are forced to decide between querying the database every time or treating the association as an array.
There are a couple of ways to handle this:
You can create separate associations for each level that you want by specifying a conditions hash on the association definition. Then you can simply eager load these associations for your User query, and you will be hitting the db 3x for the entire operation instead of 3x for each user.
class User
has_many articles_yesterday, class_name: Article, conditions: ['posted_at >= ?', Date.yesterday]
# other associations the same way
end
User.where(...).includes(:articles_yesterday, :articles_7days, :articles_30days)
You could do a group by.
What it comes down to is you need to profile your code and determine what's going to be fastest for your app (or if you should even bother with it at all)
You can get rid of the necessity of checking the query with something like the code below.
class User
has_many :articles
def article_30d
#articles_last30d ||= user.articles.where("posted_at >= ?", 30.days.ago.to_date)
end
def articles_last7d
#articles_last7d ||= articles_last30d.select { |article| article.posted_at >= 7.days.ago.to_date }
end
def articles_yesterday
#articles_yesterday ||= articles_last30d.select { |article| article.posted_at >= Date.yesterday }
end
end
What it does:
Makes only one query maximum, if any of the three is used
Calculates only the used array, and the 30d version in any case, but only once
It does not however simplifies the initial 30d query even if you do not use it. Is it enough, or you need something more?
I have a Search resource that returns posts based on a filter as described in Railscast111, and have the following code:
def filter_posts
posts = Post.order('created_at DESC')
posts = posts.where("name ilike ?", "%#{keywords}%")
posts = posts.where(... #numerous other filters
posts
end
The filter itself seems to work fine. However, the content is not always returned in order of 'created_at DESC'. How can I sort the final output so that it's always in order of 'created_at DESC'? Currently, there is no association between the Post and Search models. Do I need to build one? If so, how?
Have you tried chaining the two conditions together?
posts = Post.where("name like?", "%#{keywords}%").order('created_at DESC')
Depending on how many filters you end up calling, you'll need to keep updating your original result, with the updated scope (based on your filter), as each time you use where it creates a new scope, instead of adjusting the original one. So you seem to be on the right path, as your original code does this, e.g
posts = Post.where("filter1")
posts = posts.where("filter2")
Have you tried sorting after all of the filters have been applied, so something like
posts = posts.order('created_at DESC')
or
posts = posts.sort_by &:created_at
Also, I'm not really sure what you mean by a Search resource, when (at least in this case) it appears you could keep the search logic within the Post model itself. Can you clarify, or maybe post the model?
I'm using Ruby on Rails. I have a couple of models which fit the normal order/order lines arrangement, i.e.
class Order
has_many :order_lines
end
class OrderLines
belongs_to :order
belongs_to :product
end
class Product
has_many :order_lines
end
(greatly simplified from my real model!)
It's fairly straightforward to work out the most popular individual products via order line, but what magical ruby-fu could I use to calculate the most popular combination(s) of products ordered.
Cheers,
Graeme
My suggestion is to create an array a of Product.id numbers for each order and then do the equivalent of
h = Hash.new(0)
# for each a
h[a.sort.hash] += 1
You will naturally need to consider the scale of your operation and how much you are willing to approximate the results.
External Solution
Create a "Combination" model and index the table by the hash, then each order could increment a counter field. Another field would record exactly which combination that hash value referred to.
In-memory Solution
Look at the last 100 orders and recompute the order popularity in memory when you need it. Hash#sort will give you a sorted list of popularity hashes. You could either make a composite object that remembered what order combination was being counted, or just scan the original data looking for the hash value.
Thanks for the tip digitalross. I followed the external solution idea and did the following. It varies slightly from the suggestion as it keeps a record of individual order_combos, rather than storing a counter so it's possible to query by date as well e.g. most popular top 10 orders in the last week.
I created a method in my order which converts the list of order items to a comma separated string.
def to_s
order_lines.sort.map { |ol| ol.id }.join(",")
end
I then added a filter so the combo is created every time an order is placed.
after_save :create_order_combo
def create_order_combo
oc = OrderCombo.create(:user => user, :combo => self.to_s)
end
And finally my OrderCombo class looks something like below. I've also included a cached version of the method.
class OrderCombo
belongs_to :user
scope :by_user, lambda{ |user| where(:user_id => user.id) }
def self.top_n_orders_by_user(user,count=10)
OrderCombo.by_user(user).count(:group => :combo).sort { |a,b| a[1] <=> b[1] }.reverse[0..count-1]
end
def self.cached_top_orders_by_user(user,count=10)
Rails.cache.fetch("order_combo_#{user.id.to_s}_#{count.to_s}", :expiry => 10.minutes) { OrderCombo.top_n_orders_by_user(user, count) }
end
end
It's not perfect as it doesn't take into account increased popularity when someone orders more of one item in an order.