orderby = ~person_cnt
суббота, 14 ноября 2020 г. в 19:23:11 UTC+3, [email protected]:
> Assuming sqlite:
> person_cnt =
> db.PERSON_PROJECT_PRIORITY.person_fk.count().with_alias('person_cnt')
>
>
> prioritySet = db((db.PERSON.id <http://db.person.id/> ==
> PERSON-PROJECT-PRIORITY.person_fk) & (PROJECT.id ==
> PERSON-PROJECT-PRIORITY.project_fk)).select(
> ...,
> person_cnt,
> 'GROUP_CONCAT(person.person_name,",") AS person_list'
> groupby = db.PERSON-PROJECT-PRIORITY.project_fk,
> )
>
> суббота, 14 ноября 2020 г. в 09:40:17 UTC+3, [email protected]:
>
>> How to write a statement that counts which projects are the priority of
>> most people
>>
>> PROJECT
>> id
>> project_name
>>
>> PERSON
>> id
>> person_name
>>
>> PERSON_PROJECT_PRIORITY
>> person_fk
>> project_fk
>>
>> prioritySet = db((db.PERSON.id == PERSON-PROJECT-PRIORITY.person_fk) &
>> (PROJECT.id == PERSON-PROJECT-PRIORITY.project_fk)).select()
>>
>> How to sort by count of projects which have priority in order of the most
>> persons' priority?
>>
>> Output looks like this:
>>
>> Paint-the-house (10) [Means is the top priority for 10 people]
>> Plant-a-garden (5)
>> Clean-out-garage (2)
>>
>> If you have additional time, how to write so output looks like:
>>
>> Paint-the-house (10) Tom, Sue, Tony, Ted, Mary, Fred, Sal, Chris, Ed,
>> Sally
>> Plant-a-garden (5) Harry, George, Joanne, Tony, Janet
>> Clean-out-garage (2) Clyde, Jane
>>
>> Thanks,
>>
>> Alex Glaros
>>
>
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