How to do a subquery in group_concat - mysql

Question

I have the following data model:
`title`
- id
- name
`version`
- id
- name
- title_id
`version_price`
- id
- version_id
- store
- price
And here is an example of the data:
`title`
- id=1, name=titanic
- id=2, name=avatar
`version`
- id=1, name="titanic (dubbed in spanish)", title_id=1
- id=2, name="avatar directors cut", title_id=2
- id=3, name="avatar theatrical", title_id=2
`version_price`
- id=1, version_id=1, store=itunes, price=$4.99
- id=1, version_id=1, store=google, price=$4.99
- id=1, version_id=2, store=itunes, price=$5.99
- id=1, version_id=3, store=itunes, price=$5.99
I want to construct a query that will give me all titles that have a version_price on iTunes but not on Google. How would I do this? Here is what I have so far:
select
title.id, title.name, group_concat(distinct store order by store)
from
version inner join title on version.title_id=title.id inner join version_price on version_price.version_id=version.id
group by
title_id
This gives me a group_concat which shows me what I have:
id name group_concat(distinct store order by store)
1 Titanic Google,iTunes
2 Avatar iTunes
But how would I construct a query to include whether the item is on Google (using a case statement or whatever's needed)
id name group_concat(distinct store order by store) on_google
1 Titanic Google,iTunes true
2 Avatar iTunes false
It would basically be doing a group_concat LIKE '%google%' instead of a normal where clause.
Here's a link for a SQL fiddle of the current query I have: http://sqlfiddle.com/#!9/e52b53/1/0

Answer 1

Use conditional aggregation to determine if the title is in a specified store.
select title.id, title.name, group_concat(distinct version_price.store order by store),
if(count(case when store = 'google' then 1 end) >= 1,'true','false') as on_google
from version
inner join title on version.title_id=title.id
inner join version_price on version_price.version_id=version.id
group by title.id, title.name
count(case when store = 'google' then 1 end) >= 1 counts all the rows for a given title after assigning 1 to the rows which have google in them. (Or else they would be assigned null and the count ignores nulls.) Thereafter, the if checks for the countand classifies a title if it has atleast one google store on it.

Answer 2

http://sqlfiddle.com/#!9/b8706/2
you can just:
SELECT
title.id,
title.name,
group_concat(distinct version_price.store),
MAX(IF(version_price.store='google',1,0)) on_google
FROM version
INNER JOIN title
ON version.title_id=title.id
INNER JOIN version_price
ON version_price.version_id=version.id
GROUP BY title_id;
and add HAVING to the query if need records to be filtered:
SELECT
title.id,
title.name,
group_concat(distinct version_price.store),
MAX(IF(version_price.store='google',1,0)) on_google
FROM version
INNER JOIN title
ON version.title_id=title.id
INNER JOIN version_price
ON version_price.version_id=version.id
GROUP BY title_id
HAVING on_google;

Answer 3

This will give you the number of version prices not on google, and the number on google. (COUNT does not count null values.)
SELECT t.id, t.name
, COUNT(DISTINCT vpNotG.id) > 0 AS onOtherThanGoogle
, COUNT(DISTINCT vpG.id) > 0 AS onGoogle
FROM title AS t
INNER JOIN version AS v ON t.id=v.title_id
LEFT JOIN version_price AS vpNotG
ON v.id=vpNotG.version_id
AND vpNotG.store <> 'Google'
LEFT JOIN version_price AS vpG
ON v.id=vpG.version_id
AND vpG.store = 'Google'
GROUP BY t.id
or for another solution similar to vkp's:
SELECT t.id, t.name
, COUNT(DISTINCT CASE WHEN store = 'Google' THEN vp.id ELSE NULL END) AS googlePriceCount
, COUNT(DISTINCT CASE WHEN store = 'iTunes' THEN vp.id ELSE NULL END) AS iTunesPriceCount
, COUNT(DISTINCT CASE WHEN store <> 'Google' THEN vp.id ELSE NULL END) AS nonGooglePriceCount
FROM title AS t
INNER JOIN version AS v ON t.id = v.title_id
INNER JOIN version_price AS vp ON v.id = vp.version_id
GROUP BY t.id
Note: The ELSE NULL can be omitted, because if no ELSE is provided it is implied; but I included for clarity.

Answer 4

I would do it like below
SELECT
*
FROM
title t
INNER JOIN
version v ON
v.title_id = t.id
CROSS APPLY (
SELECT
*
FROM
version_price vp
WHERE
vp.store <> 'google'
) c ON c.version_id == v.id
Syntax may be just a little off as I can't test it right now, but I believe this is the spirit of what you would want. Cross apply is also a very efficient join which is always helpful!

Answer 5

This might be the most inefficient of the above answers, but the following subquery would work, using a %like% condition:
select *, case when stores like '%google%' then 1 else 0 end on_google
from (select title.id, title.name, group_concat(distinct store order by store) stores
from version inner join title on version.title_id=title.id inner join version_price
on version_price.version_id=version.id group by title_id) x
id name stores on_google
1 Titanic Google,iTunes 1
2 Avatar iTunes 0