I'd like to make a newsfeed for the homepage of a site i'm playing around with. There are two models: Articles, and Posts. If I wanted just one in the newsfeed it would be easy:
#newsfeed_items = Article.paginate(:page => params[:page])
But I would like for the two to be both paginated into the same feed, in reverse chronological order. The default scope for the article and post model are already in that order.
How do I get the articles and posts to be combined in to the newsfeed as such?
Thanks!
EDIT: What about using SQL in the users model?
Just wondering: maybe would it be possible define in User.rb:
def feed
#some sql like (SELECT * FROM articles....)
end
Would this work at all?
in my last project i stuck into a problem, i had to paginate multiple models with single pagination in my search functionality. it should work in a way that the first model should appear first when the results of the first model a second model should continue the results and the third and so on as one single search feed, just like facebook feeds. this is the function i created to do this functionality
def multi_paginate(models, page, per_page)
WillPaginate::Collection.create(page, per_page) do |pager|
# set total entries
pager.total_entries = 0
counts = [0]
offsets = []
for model in models
pager.total_entries += model.count
counts << model.count
offset = pager.offset-(offsets[-1] || 0)
offset = offset>model.count ? model.count : offset
offsets << (offset<0 ? 0 : offset)
end
result = []
for i in 0...models.count
result += models[i].limit(pager.per_page-result.length).offset(offsets[i]).to_a
end
pager.replace(result)
end
end
try it and let me know if you have any problem with it, i also posted it as an issue to will_paginate repository, if everyone confirmed that it works correctly i'll fork and commit it to the library. https://github.com/mislav/will_paginate/issues/351
for those interested, please check this question: Creating a "feed" from multiple rails models, efficiently?
Here, Victor Piousbox provides a good, efficient solution.
Look at paginate_by_sql method. You can write unione query to fetch both articles and posts:
select 'article' as type, id from articles
union
select 'post' as type, id from posts
You can paginate both if you use AJAX. Here is well explained how to paginate using AJAX with WillPaginate.
You can paginate an array using WillPaginate::Collection.create. So you'd need to use ActiveRecord to find both sets of data and then combine them in a single array.
Then take a look at https://github.com/mislav/will_paginate/blob/master/lib/will_paginate/collection.rb for documentation on how to use the Collection to paginate any array.
Related
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.
In my rails app, each user has a karma/score that i'm calculating through the user model as follows:
after_invitation_accepted :increment_score_of_inviter
def increment_score_of_inviter
invitation_by.increment!(:score, 10)
end
def comment_vote_count
Vote.find(comment_ids).count * 2
end
def calculated_vote_count
base_score + comment_vote_count
end
def recalculate_score!
update_attribute(:score, calculated_vote_count)
end
I'm trying to create paginated list of all the users, sorted by their scores. With thousands of users, how do I do this efficiently?
I was think of using:
User.all.sort_by(&:calculated_vote_count)
But, this would be pretty heavy.
Well...using User.all upon a table full of records will be a memory hog for your application. Instead you should try to accomplish what you want on the DB layer.
At this point I'm assuming base_score is one of the table columns (is base_score same as score?), so you'd have to do something like the following (using LEFT JOIN):
User.select("users.*, (COUNT(votes.id) * 2 + users.base_score) AS calculated_vote_count").joins("LEFT JOIN votes ON votes.user_id = user.id").order("calculated_vote_count DESC")
And then you can paginate the results the way you like.
I didn't test it, but it should work. Let me know if doesn't.
It's pretty straight forward:
User.order('score DESC').all
Obviously you'd have pagination, User.order('score DESC').page(params[:page]).per(20) with Kaminari.
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'll explain this as best as possible. I have a query on user posts:
#selected_posts = Posts.where(:category => "Baseball")
I would like to write the following statement. Here it is in pseudo terms:
User.where(user has a post in #selected_posts)
Keep in mind that I have a many to many relationship setup so post.user is usable.
Any ideas?
/EDIT
#posts_matches = User.includes(#selected_posts).map{ |user|
[user.company_name, user.posts.count, user.username]
}.sort
Basically, I need the above to work so that it uses the users that HAVE posts in selected_posts and not EVERY user we have in our database.
Try this:
user.posts.where("posts.category = ?", "Baseball")
Edit 1:
user.posts.where("posts.id IN (?)", #selected_posts)
Edit 2:
User.select("users.company_name, count(posts.id) userpost_count, user.username").
joins(:posts).
where("posts.id IN (?)", #selected_posts).
order("users.company_name, userpost_count, user.username")
Just use the following:
User.find(#selected_posts.map(&:user_id).uniq)
This takes the user ids from all the selected posts, turns them into an array, and removes any duplicates. Passing an array to user will just find all the users with matching ids. Problem solved.
To combine this with what you showed in your question, you could write:
#posts_matches = User.find(#selected_posts.map(&:user_id).uniq).map{ |user|
[user.company_name, user.posts.size, user.username]
}
Use size to count a relation instead of count because Rails caches the size method and automatically won't look it up more than once. This is better for performance.
Not sure what you were trying to accomplish with Array#sort at the end of your query, but you could always do something like:
#users_with_posts_in_selected = User.find(#selected_posts.map(&:user_id).uniq).order('username DESC')
I don't understand your question but you can pass an array to the where method like this:
where(:id => #selected_posts.map(&:id))
and it will create a SQL query like WHERE id IN (1,2,3,4)
By virtue of your associations your selected posts already have the users:
#selected_posts = Posts.where("posts.category =?", "Baseball")
#users = #selected_posts.collect(&:user);
You'll probably want to remove duplicate users from #users.
Pretty sure that I'm missing something really simple here:
I'm trying to display a series of pages that contain instances of two different models - Profiles and Groups. I need them ordering by their name attribute. I could select all of the instances for each model, then sort and paginate them, but this feels sloppy and inefficient.
I'm using mislav-will_paginate, and was wondering if there is any better way of achieving this? Something like:
[Profile, Group].paginate(...)
would be ideal!
Good question, I ran into the same problem a couple of times. Each time, I ended it up by writing my own sql query based on sql unions (it works fine with sqlite and mysql). Then, you may use will paginate by passing the results (http://www.pathf.com/blogs/2008/06/how-to-use-will_paginate-with-non-activerecord-collectionarray/). Do not forget to perform the query to count all the rows.
Some lines of code (not tested)
my_query = "(select posts.title from posts) UNIONS (select profiles.name from profiles)"
total_entries = ActiveRecord::Base.connection.execute("select count(*) as count from (#{my_query})").first['count'].to_i
results = ActiveRecord::Base.connection.select_rows("select * from (#{my_query}) limit #{limit} offset #{offset}")
Is it overkilled ? Maybe but you've got the minimal number of queries and results are consistent.
Hope it helps.
Note: If you get the offset value from a http param, you should use sanitize_sql_for_conditions (ie: sql injection ....)
You can get close doing something like:
#profiles, #groups = [Profile, Group].map do |clazz|
clazz.paginate(:page => params[clazz.to_s.downcase + "_page"], :order => 'name')
end
That will then paginate using page parameters profile_page and group_page. You can get the will_paginate call in the view to use the correct page using:
<%= will_paginate #profiles, :page_param => 'profile_page' %>
....
<%= will_paginate #groups, :page_param => 'group_page' %>
Still, I'm not sure there's a huge benefit over setting up #groups and #profiles individually.
in my last project i stuck into a problem, i had to paginate multiple models with single pagination in my search functionality.
it should work in a way that the first model should appear first when the results of the first model a second model should continue the results and the third and so on as one single search feed, just like facebook feeds.
this is the function i created to do this functionality
def multi_paginate(models, page, per_page)
WillPaginate::Collection.create(page, per_page) do |pager|
# set total entries
pager.total_entries = 0
counts = [0]
offsets = []
for model in models
pager.total_entries += model.count
counts << model.count
offset = pager.offset-(offsets[-1] || 0)
offset = offset>model.count ? model.count : offset
offsets << (offset<0 ? 0 : offset)
end
result = []
for i in 0...models.count
result += models[i].limit(pager.per_page-result.length).offset(offsets[i]).to_a
end
pager.replace(result)
end
end
try it and let me know if you have any problem with it, i also posted it as an issue to will_paginate repository, if everyone confirmed that it works correctly i'll fork and commit it to the library.
https://github.com/mislav/will_paginate/issues/351
Have you tried displaying two different sets of results with their own paginators and update them via AJAX? It is not exactly what you want, but the result is similar.