Re: Count child objects for each line of a table: LEFT JOIN, LATERAL JOIN or subqueries ?

Hi Frank,

Thanks for your answer !

It seems it would perform better to aggregate as soon as possible, like you
illustrated in your example.
I will rewrite the query with "WITH" clauses to improve readability.

Thanks also for the Coalesce idea. It is better to see 0 instead of NULL.

Michaël

Le lun. 23 mai 2022 à 16:15, kimaidou <kimaidou(at)gmail(dot)com> a écrit :

> So you
>
> Le lun. 23 mai 2022 à 15:14, Frank Streitzig <fstreitzig(at)gmx(dot)net> a
> écrit :
>
>> Am Mon, May 23, 2022 at 01:55:07PM +0200 schrieb kimaidou:
>> > Hi list,
>> >
>> > I have a basic need, often encountered in spatial analysis: I have a
>> list
>> > of cities, parks, childcare centres, schools. I need to count the
>> number of
>> > items for each city (0 if no item exists for this city)
>> >
>> > I have tested 3 different SQL queries to achieve this goal:
>> >
>> > * one with several LEFT JOINS: http://sqlfiddle.com/#!17/fe902/3
>> > * one with sub-queries: http://sqlfiddle.com/#!17/fe902/4
>> > * one with several LATERAL JOINS: http://sqlfiddle.com/#!17/fe902/6
>>
>> Hello,
>>
>> Cost of queries see link "View Execution Plan" in fiddle
>>
>> query 1: 134.62
>> query 2: 8522.32
>> query 3: 134.62
>>
>> query 1 and 3 have wrong count in result (columns nb_school,
>> nb_childcare, nb_park)
>>
>> My try has cost of 81.83
>>
>> select c.*
>> , coalesce(s.cnt,0) as cnt_school
>> , s.schools
>> , coalesce(cc.cnt,0) as cnt_childcare
>> , cc.childcares
>> , coalesce(p.cnt,0) as cnt_park
>> , p.parks
>> from city c
>> left outer join
>> (select fk_id_city, count(*) as cnt
>> ,string_agg(name, ', ') AS schools
>> from school
>> group by fk_id_city) s
>> on s.fk_id_city = c.id
>> left outer join
>> (select fk_id_city, count(*) as cnt
>> ,string_agg(name, ', ') AS childcares
>> from childcare
>> group by fk_id_city) cc
>> on cc.fk_id_city = c.id
>> left outer join
>> (select fk_id_city, count(*) as cnt
>> ,string_agg(name, ', ') AS parks
>> from park
>> group by fk_id_city) p
>> on p.fk_id_city = c.id
>> order by c.id
>> ;
>>
>> IMHO, but without a where clause, the cost will increase with the amount
>> of data.
>>
>> Regards,
>> Frank
>>
>>