Hello Stephen,
Are the nice date/time features at https://sqlite.org/lang_datefunc.html
enough?
For example:
select date ('2016-08-01', '+1 year'); -- gives 2017-08-01
Regarding: "- using today as the base date, a date of Aug 1, 2016 with a
weekly
recurring date, I'd like to get Sept 2, 2017."
I'm not sure I understand. If I run:
select julianday( '2017-09-02') - julianday('2016-08-01');
I get a difference of 397 days, which equals 56 weeks and 5 days -- not
evenly divisible by 7.
select 397.0 / 7; -- gives 56.714...
select 397.0 % 7; -- gives 5
>
>
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