计算两个时间戳之间的小时差？

Question

I'm working on a PHP project with Firebird, about working hours. Basically, every worker has a card, which he stamps on entering and leaving the building. Each stamp is recorded into the ARCHIVE table, with date ( DATUM ), id ( NREC ) and user_id ( NCODE ). What I have to do is get the last and first time of each day, and calculate hours. So to get the first and last, I used this:

SELECT MAX(NREC)
FROM ARCHIVE
WHERE DATUM BETWEEN '$day.$month.$year' AND '$day+1.$month.$year';

SELECT MIN(NREC)
FROM ARCHIVE
WHERE DATUM BETWEEN '$day.$month.$year' AND '$day+1.$month.$year';

I save both max and max in a variable, then I select the both dates and calculate. But that are 4 queries. Is there any simpler/easier way? Also, DATUM is a timestamp

Answer 1

Could try:

SELECT DATUM FROM ARCHIVE JOIN (
    SELECT MIN(NREC) AS minid, MAX(NREC) as maxid FROM ARCHIVE 
    WHERE DATUM  DATUM BETWEEN '$day.$month.$year' AND '$day+1.$month.$year' 
    AND NCODE = user_id) 
as mm ON mm.maxid = ARCHIVE.NREC OR mm.minid=ARCHIVE.NREC 

If Firebird works anything like MySQL this should give you two timestamps.

Answer 2

What about

SELECT CAST(DATUM AS DATE), MAX(NREC), MIN(NREC)
FROM ARCHIVE
GROUP BY 1

Optionally combined with a condition for the right date range?