With ActiveRecord you can pass a field and an array into WHERE like so:
Product.joins(:category).where('category.id' => [x,y,z])
(in this case Product has a many to many relationship with Category)
This uses the IN operator to find all products in categories with an ID of x, y, or z
What I would like to do is find all products in categories with an ID of x, y, AND z. I know you can produce this sort of result like so:
Product.joins(:category).where('category.id' => x).where('category.id' => y).where('category.id' => z)
In other words, I want to find products that have all of the categories supplied.
I feel like I could do be doing something much simpler here. Any ideas?
Update: I believe this question is relevant, still having trouble getting it to work. Still think there might be another way to do this.
Havent tried... but something like below should give you hint to get started -
Product.joins(:categories).select("products.*, GROUP_CONCAT('categories.id') as category_ids").group('products.id').having('category_ids = ?', [1,2,3])
How about this:
Product.joins(:category)
.where('category.id = ? AND category.id = ? AND category.id = ?',x,y,z)
Related
Update merge
Cama::PostType.first.posts.joins(:custom_field_values)
.where("cama_custom_fields_relationships.custom_field_slug = ? AND
cama_custom_fields_relationships.value LIKE ?","localization",
"%Paris%").merge(Cama::PostType.first.posts.joins(:custom_field_values)
.where("cama_custom_fields_relationships.custom_field_slug = ? AND
cama_custom_fields_relationships.value = ?","type-localization", "2"))
This also doesnt work. When executed seperatelty, it returns same AssociationRelation. I guess it only works for ActiveRecord:Relation
Update
I think Im looking for INTERSECT but don't know how to use it with where
There is another topic that I created and still can't find answer how to optimize it.
It goes likes this
I need to find "posts" by "other_model" values. Other model has relationship with posts throught another table but lets keep it simple. When i do
Foo.joins(:other_model).where("other_model.value = ? AND other_model.value = ?", "one", "two")
This of course won't find me any result because it contradicts itself.
When I do with OR instead of AND
Foo.joins(:other_model).where("other_model.value = ? OR other_model.value = ?", "one", "two")
It finds posts for me but... either it has one value or either has second value and...
I want to find posts based on other_model.value = one and other_model.value = two
Which means it looks for 2 seperate results and then I need to just return ids that covers each other... Does it make sense ?
I think you are looking for a query like:
Foo
.joins(:other_model)
.where('other_model.value = ? OR other_model.value = ?', 'one', 'two')
.group('foos.id')
.having('COUNT(other_models.id) >= 2')
In my RoR application, I've got a database lookup similar to this one:
Client.joins(:products).where({'product.id' => [1,2,3]})
Unfortunately this will return all clients that have bought product 1, 2 or 3 but I only want to get back the clients, that bought all of the three products. In other words, I'd like to write a query that matches for n elements in a given set.
Are there any elegant solutions for this?
This is not really elegant. But it should translate into the needed SQL.
Client.joins(:products).
where({'products.id' => [1,2,3]}).
group('users.id').
having('COUNT(DISTINCT products.id) >= 3')
Same answer with more dynamic way
ids = [1,2,3]
Client.joins(:products).
where({'products.id' => ids}).
group('users.id').
having('COUNT(DISTINCT products.id) >= ?', ids.size)
I have 2 models - Restaurant and Feature. They are connected via has_and_belongs_to_many relationship. The gist of it is that you have restaurants with many features like delivery, pizza, sandwiches, salad bar, vegetarian option,… So now when the user wants to filter the restaurants and lets say he checks pizza and delivery, I want to display all the restaurants that have both features; pizza, delivery and maybe some more, but it HAS TO HAVE pizza AND delivery.
If I do a simple .where('features IN (?)', params[:features]) I (of course) get the restaurants that have either - so or pizza or delivery or both - which is not at all what I want.
My SQL/Rails knowledge is kinda limited since I'm new to this but I asked a friend and now I have this huuuge SQL that gets the job done:
Restaurant.find_by_sql(['SELECT restaurant_id FROM (
SELECT features_restaurants.*, ROW_NUMBER() OVER(PARTITION BY restaurants.id ORDER BY features.id) AS rn FROM restaurants
JOIN features_restaurants ON restaurants.id = features_restaurants.restaurant_id
JOIN features ON features_restaurants.feature_id = features.id
WHERE features.id in (?)
) t
WHERE rn = ?', params[:features], params[:features].count])
So my question is: is there a better - more Rails even - way of doing this? How would you do it?
Oh BTW I'm using Rails 4 on Heroku so it's a Postgres DB.
This is an example of a set-iwthin-sets query. I advocate solving these with group by and having, because this provides a general framework.
Here is how this works in your case:
select fr.restaurant_id
from features_restaurants fr join
features f
on fr.feature_id = f.feature_id
group by fr.restaurant_id
having sum(case when f.feature_name = 'pizza' then 1 else 0 end) > 0 and
sum(case when f.feature_name = 'delivery' then 1 else 0 end) > 0
Each condition in the having clause is counting for the presence of one of the features -- "pizza" and "delivery". If both features are present, then you get the restaurant_id.
How much data is in your features table? Is it just a table of ids and names?
If so, and you're willing to do a little denormalization, you can do this much more easily by encoding the features as a text array on restaurant.
With this scheme your queries boil down to
select * from restaurants where restaurants.features #> ARRAY['pizza', 'delivery']
If you want to maintain your features table because it contains useful data, you can store the array of feature ids on the restaurant and do a query like this:
select * from restaurants where restaurants.feature_ids #> ARRAY[5, 17]
If you don't know the ids up front, and want it all in one query, you should be able to do something along these lines:
select * from restaurants where restaurants.feature_ids #> (
select id from features where name in ('pizza', 'delivery')
) as matched_features
That last query might need some more consideration...
Anyways, I've actually got a pretty detailed article written up about Tagging in Postgres and ActiveRecord if you want some more details.
This is not "copy and paste" solution but if you consider following steps you will have fast working query.
index feature_name column (I'm assuming that column feature_id is indexed on both tables)
place each feature_name param in exists():
select fr.restaurant_id
from
features_restaurants fr
where
exists(select true from features f where fr.feature_id = f.feature_id and f.feature_name = 'pizza')
and
exists(select true from features f where fr.feature_id = f.feature_id and f.feature_name = 'delivery')
group by
fr.restaurant_id
Maybe you're looking at it backwards?
Maybe try merging the restaurants returned by each feature.
Simplified:
pizza_restaurants = Feature.find_by_name('pizza').restaurants
delivery_restaurants = Feature.find_by_name('delivery').restaurants
pizza_delivery_restaurants = pizza_restaurants & delivery_restaurants
Obviously, this is a single instance solution. But it illustrates the idea.
UPDATE
Here's a dynamic method to pull in all filters without writing SQL (i.e. the "Railsy" way)
def get_restaurants_by_feature_names(features)
# accepts an array of feature names
restaurants = Restaurant.all
features.each do |f|
feature_restaurants = Feature.find_by_name(f).restaurants
restaurants = feature_restaurants & restaurants
end
return restaurants
end
Since its an AND condition (the OR conditions get dicey with AREL). I reread your stated problem and ignoring the SQL. I think this is what you want.
# in Restaurant
has_many :features
# in Feature
has_many :restaurants
# this is a contrived example. you may be doing something like
# where(name: 'pizza'). I'm just making this condition up. You
# could also make this more DRY by just passing in the name if
# that's what you're doing.
def self.pizza
where(pizza: true)
end
def self.delivery
where(delivery: true)
end
# query
Restaurant.features.pizza.delivery
Basically you call the association with ".features" and then you use the self methods defined on features. Hopefully I didn't misunderstand the original problem.
Cheers!
Restaurant
.joins(:features)
.where(features: {name: ['pizza','delivery']})
.group(:id)
.having('count(features.name) = ?', 2)
This seems to work for me. I tried it with SQLite though.
I have a model called Shops with an attribute called brands, brands is a text field and contains multiple brands. What i would like to do is select all unique brands and display them sorted in alphabetic order
#brands = Shop.all(:select => 'distinct(brands)')
What to do from here?
If Shop#brands can hold multiple values like for example: "rony, hoke, fike", then I can reluctantly suggest doing something like this:
#brands = Shop.all(:select => 'brands').each { |s|
s.brands.split(',').map { |b|
b.strip.downcase
}
}.flatten.uniq.sort
BUT, you should really think about your data model here to prevent such hackery. You couuld break out the brands into it's own table + model and do a many to many relationship with Shop.
ActiveRecord objects of the class 'Location' (representing the db-table Locations) have the attributes 'url', 'lat' (latitude) and 'lng' (longitude).
Lat-lng-combinations on this model should be unique. The problem is, that there are a lot of Location-objects in the database having duplicate lat-lng-combinations.
I need help in doing the following
Find objects that share the same
lat-lng-combination.
If the 'url' attribute of the object
isn't empty, keep this object and delete the
other duplicates. Otherwise just choose the
oldest object (by checking the attribute
'created_at') and delete the other duplicates.
As this is a one-time-operation, solutions in SQL (MySQL 5.1 compatible) are welcome too.
If it's a one time thing then I'd just do it in Ruby and not worry too much about efficiency. I haven't tested this thoroughly, check the sorting and such to make sure it'll do exactly what you want before running this on your db :)
keep = []
locations = Location.find(:all)
locations.each do |loc|
# get all Locations's with the same coords as this one
same_coords = locations.select { |l| l.lat == loc.lat and \
l.lng == loc.lng }
with_urls = same_coords.select { |l| !l.url.empty? }
# decide which list to use depending if there were any urls
same_coords = with_urls.any? ? with_urls : same_coords
# pick the best one
keep << same_coords.sort { |a,b| b.created_at <=> a.created_at }.first.id
end
# only keep unique ids
keep.uniq!
# now we just delete all the rows we didn't decide to keep
locations.each do |loc|
loc.destroy unless keep.include?( loc.id )
end
Now like I said, this is definitely poor, poor code. But sometimes just hacking out the thing that works is worth the time saved in thinking up something 'better', especially if it's just a one-off.
If you have 2 MySQL columns, you can use the CONCAT function.
SELECT * FROM table1 GROUP BY CONCAT(column_lat, column_lng)
If you need to know the total
SELECT COUNT(*) AS total FROM table1 GROUP BY CONCAT(column_lat, column_lng)
Or, you can combine both
SELECT COUNT(*) AS total, table1.* FROM table1
GROUP BY CONCAT(column_lat, column_lng)
But if you can explain more on your question, perhaps we can have more relevant answers.