Question

I am struggling to find a working query in SQLite, that will return my records in their upcoming birthday along with the number of days it will happen

CREATE TABLE contact_birthday ('contact_id' varchar,'data1' varchar,'display_name' varchar)

data1 holds the birthday in YYYY-MM-DD format

Something like:

"1986-06-28","Angel","0"
"1979-06-29","Bea","1"
"1984-07-02","John","4"
"1984-06-26","Mark","364"

I’ve found some other results but they are working for MySQL, or SQL Server not in SQLite.

How would be the query in SQLite without user defined functions?

Answer 1

I got it to work using julianday (the only way to get a ‘day’ count AFAIK)

select name,
julianday(strftime('%Y', 'now')||strftime('-%m-%d', data1))-julianday('now') as birthday
from foo
where birthday between -1 and 30;

My result:

Angel|-0.479438499547541

Bea|0.520561488810927

John|3.52056147716939

Quick explanation:

julianday(strftime('%Y', 'now') <- this pulls the current year so that julianday for data1 doesn’t consider years from the 80’s

||strftime('-%m-%d', data1)) <- this concatenates the month/day for the actual birthday to the current year

where birthday between -1 and 30; <- the -1 ensures you will see results all the way up through ‘today’. The 30 is the maximum amount of days you want to see in advance.

Edit 1:

Here’s a modification that uses seconds since 1970 epoch rather than julian day:

select name,
(strftime('%s',strftime('%Y', 'now')||strftime('-%m-%d', data1))-strftime('%s','now'))/86400.0 as birthday
from foo
where birthday between -1 and 30
order by birthday;

Result:

Angel|-0.487118055555556

Bea|0.512881944444445

John|3.51288194444444

Edit 2: [Pentium10]

includes proper localtime

select data1,display_name,
(strftime('%s',strftime('%Y', 'now','localtime')||strftime('-%m-%d', data1))-strftime('%s','now','localtime'))/86400.0 as birthday
from contact_birthday
order by birthday asc

Edit 3: [Pentium10]

includes the order by if from today to next year this time are exactly 365 days
** still needs the dynamic calculation of 365 day, as that can be 366 for leap years

select data1,display_name,
((strftime('%s',strftime('%Y', 'now','localtime')||strftime('-%m-%d', data1))-strftime('%s','now','localtime'))/86400.0+1+365) % 365 as birthday
from contact_birthday
order by birthday asc

Edit 4: [Matt]

This actually might work for you:

select data1,name,
365-(strftime('%s',strftime('%Y', 'now', '+1 year', 'localtime')||strftime('-%m-%d', data1))-strftime('%s','now','localtime'))/86400.0 as birthday
from foo
order by birthday asc

Edit 5: [Pentium10]
THE ANSWER

select data1,display_name,
((strftime('%s',strftime('%Y', 'now','localtime')||strftime('-%m-%d', data1))-strftime('%s','now','localtime'))/86400.0+1+((strftime('%s','now', 'localtime','+1 year')-strftime('%s','now',  'localtime'))/86400.0)) % ((strftime('%s','now', 'localtime','+1 year')-strftime('%s','now',  'localtime'))/86400.0) as indays
from contact_birthday
order by indays asc

Result:

1984-07-02|John|-3.65773148148151

1979-06-29|Bea|-0.657731481481505

1986-06-28|Angel|0.342268518518495

1984-06-26|Mark|2.34226851851849

1987-02-16|Matt|132.342268518519

So birthdays that are coming up in the current year will be negative (e.g. John’s is 3 days in the FUTURE), birthdays that have already passed this year will be positive.

This should allow for the correct ordering.