How to get correct year, month and day in firebird function datediff

Question

I have to ask another question about datediff in Firebird. I don`t know how to get correct result in this case: worker x has two contract of employment, first in the period 1988

Answer 1

You should sum days first then sum the result and then calculate Y, M, D

SELECT
KP3.id_contact 
, (KP3.D2-KP3.D1) / (12*31) AS Y
, ((KP3.D2-KP3.D1) - ((KP3.D2-KP3.D1) / (12*31)) * 12 * 31) / 31 AS M
, CAST(MOD((KP3.D2-KP3.D1) - (((KP3.D2-KP3.D1) / (12*31)) * 12 * 31), 31) AS INTEGER) AS D
FROM
(SELECT
KP2.id_contact, SUM(KP2.D1) AS D1, SUM(KP2.D2) AS D2
FROM
    (
    SELECT
    KP.id_contact, DATEDIFF(MONTH, KP.DATE_FROM, KP.DATE_TO) / 12 AS Y, CAST(MOD(DATEDIFF(MONTH, KP.DATE_FROM, KP.DATE_TO), 12) AS INTEGER) AS M 
    , EXTRACT(YEAR FROM KP.DATE_FROM)*12*31+EXTRACT(MONTH FROM KP.DATE_FROM)*31+EXTRACT(DAY FROM KP.DATE_FROM) D1
    , EXTRACT(YEAR FROM KP.DATE_TO)*12*31+EXTRACT(MONTH FROM KP.DATE_TO)*31+EXTRACT(DAY FROM KP.DATE_TO) D2 
    FROM
    KP  
    ) AS KP2
GROUP BY KP2.id_contact
) AS KP3