I dont get it. A project status is a perfect field for a DB.
The query would even group the project in the right way.
select status , projectname
from todolist
group by statusRank
;
then you could have some code that goes through the objects
print project { arraycontainer.projectname }
If (arraycontainer[i+1].status <> arraycontainer[i].status) { print new_cat() };
sorry for the bad pseudocode but i hope you get the idea.my c++ is not good and
i want to go to bed.
Peter
> Am 17.04.2015 um 12:24 schrieb François K. <daithe...@free.fr>:
>
> Hi Lucien,
>
> Thanks a lot for your blazingly fast answer :D
>
>> Can't you simply make the hard-coded entries dynamic too. Just add a
>> column in the db saying that these entries are the hardcoded ones ?
>
> Sadly, nope, I can't hard-code them in the database because they are entry
> point to a different view on the same data.
>
> To make things more clear, I'm building a to-do app. The first screen is a
> list where the first items are : "Today", "Overdue", "Completed", "Trash", ...
> And then, I want to have the list of "Projects" (which are gathered from the
> database entries).
>
> "Today", "Overdue", etc... can't be added as Projects in the database, it
> makes no sense :(
>
>
>> And when doing db stuff, don't forget to use transactions. It can
>> help speeding up stuff a lot.
>
> Thanks for this advice. I'll do my best to respect this and use transactions
> everywhere.
>
>
> Cheers,
>
> --
> François
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