One to many query for 3 tables

Rails
1

Hello,
I am looking for a join query between 3 tables that will give me the
results I’m looking for.

I have a properties tables that has many layouts and many photos as
well. The layouts table can also, have many layouts. When I run the
query, I want to show a list of properties w/ one main photo from the
photos table, and a list of layouts per property. Right now, the query
I have will show a list of properties, but properties will be displayed
multiple times if there are more than one photos. Here’s my current
query in rails form:

    @properties = Property.paginate(:page => params[:page],
                                      :select => ["properties.*,
layouts.*, photos.*"],
                                        :joins => ["INNER JOIN layouts
ON layouts.property_id = properties.property_id LEFT JOIN photos ON
photos.property_id = properties.property_id"],
                                          :per_page => 20)

or, in raw sql:

SELECT properties.*, layouts.*, photos.photo_file_name FROM `properties`
INNER JOIN layouts ON layouts.property_id = properties.property_id LEFT
JOIN photos ON photos.property_id = properties.property_id WHERE
(properties.property_status='available') ORDER BY layouts.rent LIMIT 20
OFFSET 0

Any help would be greatly appreciated!

2

On Mon, Jan 23, 2012 at 8:39 PM, Clem R. [email protected] wrote:

Hello,
I am looking for a join query between 3 tables that will give me the
results I’m looking for.

I have a properties tables that has many layouts and many photos as
well. The layouts table can also, have many layouts.

Is this a correct statement or a typo? “The layouts table can also, have
many layouts.”

layouts., photos."],
INNER JOIN layouts ON layouts.property_id = properties.property_id LEFT
JOIN photos ON photos.property_id = properties.property_id WHERE
(properties.property_status=‘available’) ORDER BY layouts.rent LIMIT 20
OFFSET 0
[/code]

Which paginator are you using?
Which version of Rails?

Taking abstraction of the effect the paginator may have …

Look in Active Record Query Interface — Ruby on Rails Guides

Start with the simplest cases in rails console and build up your
query.

Probably a style of

@properties = Property.includes(:photos).
includes(:layouts).

where(“properties.property_status=‘available’”).
order(“layouts.rent”).
all

will get you close (or maybe .joins(layouts) if you really need the
INNER
JOIN
on layouts).

Try a solution where you do not set the select manually first.

By the design of SQL querying language, the raw SQL will always return
the property for each photo (if it is all executed in 1 SQL query).
But Activerecord will consolidate the different returned records in 1
record
in the @properties list (which then will :have_many photos).

HTH,

Peter

3

Peter - big thanks for your help. You were right - I meant to say
“the layouts table can have many photos”

Looks like this was the winner for me:

@properties = Property.paginate(:page => params[:page], :per_page =>
20).includes(:photos).includes(:layouts).order(“properties.modify_date,
layouts.modify_date”).all

Cheers!

Peter V. wrote in post #1042203:

On Mon, Jan 23, 2012 at 8:39 PM, Clem R. [email protected] wrote:

Hello,
I am looking for a join query between 3 tables that will give me the
results I’m looking for.

I have a properties tables that has many layouts and many photos as
well. The layouts table can also, have many layouts.

Is this a correct statement or a typo? “The layouts table can also, have
many layouts.”

layouts., photos."],
INNER JOIN layouts ON layouts.property_id = properties.property_id LEFT
JOIN photos ON photos.property_id = properties.property_id WHERE
(properties.property_status=‘available’) ORDER BY layouts.rent LIMIT 20
OFFSET 0
[/code]

Which paginator are you using?
Which version of Rails?

Taking abstraction of the effect the paginator may have …

Look in Active Record Query Interface — Ruby on Rails Guides

Start with the simplest cases in rails console and build up your
query.

Probably a style of

@properties = Property.includes(:photos).
includes(:layouts).

where(“properties.property_status=‘available’”).
order(“layouts.rent”).
all

will get you close (or maybe .joins(layouts) if you really need the
INNER
JOIN
on layouts).

Try a solution where you do not set the select manually first.

By the design of SQL querying language, the raw SQL will always return
the property for each photo (if it is all executed in 1 SQL query).
But Activerecord will consolidate the different returned records in 1
record
in the @properties list (which then will :have_many photos).

HTH,

Peter

4

On Wed, Jan 25, 2012 at 1:35 PM, Clem R. [email protected] wrote:

Peter - big thanks for your help. You were right - I meant to say
“the layouts table can have many photos”

Looks like this was the winner for me:

@properties = Property.paginate(:page => params[:page], :per_page =>
20).includes(:photos).includes(:layouts).order(“properties.modify_date,
layouts.modify_date”).all

TL;DR : remove the ‘.all’ at the end

In my example, I terminated with ‘.all’ so that you would get a simple
Array
as the result.

Once you have settled on a certain formula for the @properties, it might
be
better to not longer write the ‘.all’ at the end of the chain. Without
the
.all’ ,
you will get an ActiveRecord::Relation on which you can continue to
chain
conditions etc. The real query will then only be done in a later stage
of
the
calculation when the description of the ActiveRecord::Relation is
complete.

If you ever test this in rails console, it will seem that the query is
always done
immediately. That is because irb calls .inspect on the last result of
the
line.

One trick to make this work in rails console is to add ; nil at the end:

1.9.3-p0 :030 > p1 = Parent.order(“parents.name”) ; nil
=> nil
1.9.3-p0 :031 > p2 = p1.where(“NOT parents.name IS NULL”) ; nil
=> nil
1.9.3-p0 :032 > p3 = p2.includes(:child) ; nil
=> nil
1.9.3-p0 :033 > p3.all
Parent Load (0.6ms) SELECT “parents”.* FROM “parents” WHERE (NOT
parents.name IS NULL) ORDER BY parents.name
Child Load (0.4ms) SELECT “children”.* FROM “children” WHERE
“children”.“parent_id” IN (1, 2, 3)
=> [#<Parent id: 1, name: “dad”, created_at: “2012-01-24 10:05:43”,
updated_at: “2012-01-24 10:05:43”>, #<Parent id: 2, name: “dad”,
created_at: “2012-01-24 10:06:59”, updated_at: “2012-01-24 10:06:59”>,
#<Parent id: 3, name: “mom”, created_at: “2012-01-24 10:07:42”,
updated_at:
“2012-01-24 10:07:42”>]
1.9.3-p0 :034 > p4 = p3.order(“children.created_at”) ; nil
=> nil
1.9.3-p0 :035 > p4.all
SQL (0.7ms) SELECT “parents”.“id” AS t0_r0, “parents”.“name” AS
t0_r1,
“parents”.“created_at” AS t0_r2, “parents”.“updated_at” AS t0_r3,
“children”.“id” AS t1_r0, “children”.“name” AS t1_r1,
“children”.“parent_id” AS t1_r2, “children”.“created_at” AS t1_r3,
“children”.“updated_at” AS t1_r4 FROM “parents” LEFT OUTER JOIN
“children”
ON “children”.“parent_id” = “parents”.“id” WHERE (NOT parents.name IS
NULL)
ORDER BY parents.name, children.created_at
=> [#<Parent id: 1, name: “dad”, created_at: “2012-01-24 10:05:43”,
updated_at: “2012-01-24 10:05:43”>, #<Parent id: 2, name: “dad”,
created_at: “2012-01-24 10:06:59”, updated_at: “2012-01-24 10:06:59”>,
#<Parent id: 3, name: “mom”, created_at: “2012-01-24 10:07:42”,
updated_at:
“2012-01-24 10:07:42”>]

HTH,

Peter