sorry, did not mean direct SQL through db.executesql , meant through DAL

On Saturday, November 14, 2020 at 8:23:11 AM UTC-8 valq...@gmail.com wrote:

> 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, alexg...@gmail.com: 
>
>> 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
>>
>

-- 
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
--- 
You received this message because you are subscribed to the Google Groups 
"web2py-users" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to web2py+unsubscr...@googlegroups.com.
To view this discussion on the web visit 
https://groups.google.com/d/msgid/web2py/7c8aea14-a3d7-40c9-a7a2-c342f872329bn%40googlegroups.com.

Reply via email to