Firebird Interview Question:

How do convert or display the date or time as string?

Submitted by: Administrator

Simply use CAST to appropriate CHAR or VARCHAR data type (big enough). Example:

CREATE TABLE t1 ( t time, d date, ts timestamp );
INSERT INTO t1 (t,d,ts) VALUES ('14:59:23', '2007-12-31', '2007-12-31 14:59');

SELECT CAST(t as varchar(13)), CAST(d as varchar(10)), CAST(ts as varchar(24))
FROM t1;

Firebird would output times in HH:MM:SS.mmmm format (hours, minutes, seconds, milliseconds), and dates in YYYY-MM-DD (year, month, day) format.

If you wish a different formatting you can either use SUBSTRING to extract the info from char column, or use EXTRACT to buld a different string:

SELECT extract(day from d)||'.'||extract(month from d)||'.'||extract(year from d)
FROM t1;
Submitted by: Administrator