ActiveRecord HABTM finds with "AND"

Howdy railsters,

I have a problem which at first seemed to be very simple, but now I’m
quite stumped. I’d appreciate your help with this if possible. I’ve
scoured the web and the mailing list for several months now, and I
haven’t found an acceptable solution. Perhaps you’ve run into this?

I have two models, associated with a HABTM (actually using
has_many :through on both ends, along with a join table). I need to
retrieve all ModelA’s that is associated with BOTH of two ModelB’s. I
do NOT want all ModelA’s for ModelB_1 concatenated with all ModelA’s
for ModelB_2. I literally want all ModelA’s that are associated with
BOTH ModelB_1 and ModelB_2. It is not limited to only 2 ModelB’s, it
may be up to 50 ModelB’s, so this must scale.

I can describe the problem using a variety of analogies, that I think
better describes my problem than the previous paragraph:

  • Find all books that were written by all 3 authors together.
  • Find all movies that had the following 4 actors in them.
  • Find all blog posts that belonged to BOTH the Rails and Ruby
    categories for each post.
  • Find all users that had all 5 of the following tags: funny,
    thirsty, smart, thoughtful, and quick. (silly example!)
  • Find all people that have worked in both San Francisco AND San Jose
    AND New York AND Paris in their lifetimes.

I’ve thought of a variety of ways to accomplish this, but they’re
grossly inefficient and very frowned upon.

Taking an analogy above, say the last one, you could do something like
query for all the people in each city, then find items in each array
that exist across each array. That’s a minimum of 5 queries, all the
data of those queries transfered back to the app, then the app has to
intensively compare all 5 arrays to each other (loops galore!). That’s
nasty, right?

Another possible solution would be to chain the finds on top of each
other, which would essentially do the same as above, but won’t
eliminate the multiple queries and processing. Also, how would you
dynamicize the chain if you had user submitted checkboxes or values
that could be as high as 50 options? Seems dirty. You’d need a loop.
And again, that would intensify the search duration.

Obviously, if possible, we’d like to have the database perform this
for us, so, people have suggested to me that I simply put multiple
conditions in. Unfortunately, you can only do an OR with HABTM
typically.

Another solution I’ve run across is to use a search engine, like
sphinx or UltraSphinx. For my particular situation, I feel this is
overkill, and I’d rather avoid it. I still feel there should be a
solution that will let a user craft a query for an arbitrary number of
ModelB’s and find all ModelA’s.

How would you solve this problem?

Thanks a bunch,
Kevin

(I’ve subsequently cross-posted this to StackOverflow’s website since
they have a target audience that encompasses more than Rails, and is
still valid in those areas —


)

I’ll give you what advice I do know. I’m not sure it will fully help
you with your situation but it may help you to rethink your strategies.

First, make sure your tables are normalized before assigning
associations to them. If you are going to work with HABTM then 3NF or
greater…

The larger the query the better. Smaller queries are worse than one
enormously large query because rails caches that query for use and
doesn’t have to go out and do another… and another…

It will be less of a problem to process the data once you have it so I
wouldn’t worry about data processing at this point. It’s better to just
get the design and associations going.

Without seeing your models, it’s more difficult to guess what may be
right or wrong from a design point. You might want to state exactly how
many models you have, what tables and relationships you currently have
associated which will help tie into your original topic.

Thanks for the pointers. I’ve been designing databases for close to 15
years, and have been working with Rails for over 4 years. I’ve
certainly used HABTM and has_many :through for many projects without
issue. I already have the data constructed correctly, but I do
appreciate the hand holding effort anyway. This is a very particular
processing need. Should there be a wildly useful way to restructure
the database without over denormalizing/normalizing what I already
have, I’m open to it, but I doubt very much that’s where the problem
lies.

For sake of the conversation here are some fictitious models that
match my scenario identically:

TABLES

features
id:integer
name:string

schools
id:integer
name:string

features_schools
feature_id:integer
school_id:integer

MODELS

class School < ActiveRecord::Base
has_and_belongs_to_many :features
end

class Feature < ActiveRecord::Base
has_and_belongs_to_many :schools
end

How would I pull out all the schools that have ALL of the following
fictitious features: ‘Wheelchair Access’, ‘Playground’, ‘Sandbark’,
‘Library’, ‘Computer Lab’, and ‘Testing Center’? Note that it would be
unacceptable to return schools that do not have one of those
associations.

-Kevin

On Jul 7, 2009, at 8:18 PM, Älphä Blüë wrote:

Hi
Try this

Feature.find(:all,:conditions => [‘name in (?,?,?,?,?,?)’,‘Wheelchair
Access’,‘Playground’,‘Sandbark’,‘Library’,‘Computer Lab’,‘Testing
Center’]).each {|f| arr = arr && f.schools}

Sijo

Hi

arr = []
Feature.find(:all,:conditions => [‘name in (?,?,?,?,?,?)’,‘Wheelchair
Access’,‘Playground’,‘Sandbark’,‘Library’,‘Computer Lab’,‘Testing
Center’]).each {|f| arr = arr && f.schools}

Now arr have the required value.Is that you want?

Sijo

On Wed, Jul 8, 2009 at 1:21 AM, Sijo
Kg[email protected] wrote:

Hi

arr = []
Feature.find(:all,:conditions => [‘name in (?,?,?,?,?,?)’,‘Wheelchair
Access’,‘Playground’,‘Sandbark’,‘Library’,‘Computer Lab’,‘Testing
Center’]).each {|f| arr = arr && f.schools}

Now arr have the required value.Is that you want?

First off, I don’t think that works. arr = arr && f.schools will end
up with arr being set only to the set of schools associated with the
last feature instance found. I think you meant to use Array#& which
performs an array intersection, instead of the && logical operator.

But just changing that will end up with an empty arr since [] & [1] =>
[] i.e the intersection of an empty set with anything is the empty
set.

I believe that something like this is closer to the right meaning:

arr = Schools.find(:all)
Feature.find(:all, :conditions => [‘name in (?,?,?,?,?,?)’,‘Wheelchair
Access’,‘Playground’,‘Sandbark’,‘Library’,‘Computer Lab’,‘Testing
Center’]).each {|f| arr = arr & f.schools}

That’s going to do 2+n queries though, 1 to find the schools, 1 to
find the n features with one of the names, and then a query to find
each of those features schools.

Now I think that adding :include => :schools to the Feature.find will
get this down to 2 queries. But I’m not sure that this will turn out
to be the most efficient way to do it, since it potentially
instantiates lots of duplicate School objects and doing the Set
intersection in Ruby won’t be as efficient as letting SQL do it.

Not that I can come up with a way to let SQL do it, without thinking
harder than I want to this morning.


Rick DeNatale

Blog: http://talklikeaduck.denhaven2.com/
Twitter: http://twitter.com/RickDeNatale
WWR: http://www.workingwithrails.com/person/9021-rick-denatale
LinkedIn: http://www.linkedin.com/in/rickdenatale

I’m not sure that doing it in SQL is always going to be faster -
there’s got to be some performance penalty for doing a join with the
same table that many times.

One favorite trick that can sometimes make things like this more
efficient is to specifically request only the ids of the relevant
objects, do the intersection, and then pull in full objects only for
the resulting set. So something like this (things is an array of
feature names):

record_ids = Feature.find_by_name(things[0]).school_ids
things[1…-1].each do |thing|
record_ids &= Feature.find_by_name(thing).school_ids
end
@schools = School.find(record_ids)

Note that if you want eliminate some queries, you could pre-collect
the Feature objects with a call to Feature.find(:all, :conditions =>
{ :name => things }), which will find them all in one go.

You’ll find other example of this kind of code in the Rails
association preload logic, and several other places from what I
recall.

–Matt J.

On Jul 8, 2009, at 11:24 AM, Matt J. wrote:

record_ids = Feature.find_by_name(things[0]).school_ids
things[1…-1].each do |thing|
record_ids &= Feature.find_by_name(thing).school_ids
end
@schools = School.find(record_ids)

Note that if you want eliminate some queries, you could pre-collect
the Feature objects with a call to Feature.find(:all, :conditions =>
{ :name => things }), which will find them all in one go.

However, the problem with this solution is that you still need several
queries and a post-retrieval processing of whether or not the many
features share a list of schools in common. Across 40 features, this
would be too heavy and unusable in a production environment.

-Kevin

To further show an example of trying to do this, here is a ‘real
world’ example. However, in this particular case, I’m only showing two
features, but in real usage, a user might select as many as 50
features or so.

mysql> SELECT c.park_id, a.name, c.feature_id, b.name FROM parks a,
features b, features_parks c WHERE (c.park_id = a.id) AND
(c.feature_id = b.id) AND (c.feature_id = 21);
±--------±------------------------±-----------±-------+
| park_id | name | feature_id | name |
±--------±------------------------±-----------±-------+
| 3120 | Mitchell Park | 21 | Swings |
| 21385 | Boulware Park | 21 | Swings |
| 2351 | Bowden Park | 21 | Swings |
| 21561 | Cameron Park | 21 | Swings |
| 24791 | Ramos Park | 21 | Swings |
| 2585 | Eleanor Park | 21 | Swings |
| 2627 | Flood County Park | 21 | Swings |
| 2374 | Burgess Park | 21 | Swings |
| 22986 | Holbrook-Palmer Park | 21 | Swings |
| 2799 | Huddart Park | 21 | Swings |
| 3354 | Rinconada Park | 21 | Swings |
| 3693 | Werry Park | 21 | Swings |
| 26152 | Willow Oaks Park | 21 | Swings |
| 23559 | Live Oak Manor Park | 21 | Swings |
| 24295 | Oak Meadow Park | 21 | Swings |
| 25562 | Sunnyvale Baylands Park | 21 | Swings |
| 21354 | Cornelis Bol Park | 21 | Swings |
| 21459 | Bubb Park | 21 | Swings |
| 21910 | Cooper Park | 21 | Swings |
| 24859 | Rengstorff Park | 21 | Swings |
| 24043 | Monroe Mini Park | 21 | Swings |
| 24867 | Rex Manor Playground | 21 | Swings |
±--------±------------------------±-----------±-------+
22 rows in set (0.00 sec)

mysql> SELECT c.park_id, a.name, c.feature_id, b.name FROM parks a,
features b, features_parks c WHERE (c.park_id = a.id) AND
(c.feature_id = b.id) AND (c.feature_id = 51);
±--------±-------------------------±-----------±--------+
| park_id | name | feature_id | name |
±--------±-------------------------±-----------±--------+
| 3120 | Mitchell Park | 51 | Sandpit |
| 2351 | Bowden Park | 51 | Sandpit |
| 2374 | Burgess Park | 51 | Sandpit |
| 2474 | Coyote Point County Park | 51 | Sandpit |
| 24867 | Rex Manor Playground | 51 | Sandpit |
±--------±-------------------------±-----------±--------+
5 rows in set (0.00 sec)

So if I was trying to fetch all parks that have BOTH ‘Swings’ and a
‘Sandpit’, I’d expect to see:

Mitchell Park
Bowden Park
Burgess Park
Rex Manor Playground

If this was limited to only looking for 2 or 3 features at a time,
even on a hefty DB of 300,000 records, this wouldn’t be too difficult.
You would fetch all parks for each of the 2 or 3 features, then
eliminate all but the ones they share in common (exclusionary merging
of some sort). However, with 50 features, this becomes unusable, since
you’d have tons of redundancies when retrieving the data, and you’d
have to recursively iterate over 50 features.

Any way to combine the queries into one, and subquery it to eliminate
ones that don’t have all the features requested in the query?

-Kevin

Thanks to the guidance of an answer submitted by LucaM on the
StackOverflow post, I was able to come up with a SQL query and
ActiveRecord find query that does it.

SQL:

SELECT c.park_id, a.name FROM parks a, features b, features_parks c
WHERE (c.park_id = a.id) AND (c.feature_id = b.id) AND (c.feature_id
IN (21, 51, 15, 24)) GROUP BY a.id HAVING count(*)=4;

Rails:

   feature_ids = [21, 51, 15, 24]
   @parks = Park.all(
                 :joins => :features,
                 :group => "parks.id",
                 :select => "parks.*",
                 :conditions => ["features.id in (?)", feature_ids],
                 :having => "count(*)=#{feature_ids.size}"
            )

The key was grouping and using having to ensure that the items each
had the max number of features that we’re searching on.

I hope this helps someone.

-Kevin

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