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 > > _______________________________________________ sqlite-users mailing list sqlite-users@mailinglists.sqlite.org http://mailinglists.sqlite.org/cgi-bin/mailman/listinfo/sqlite-users