Avoiding default date value to be used when only time is provided to a datetime field on Redshift - amazon-redshift

Question

I created a table with a datetime field "dt". Using COPY command to load data. The corresponding value for the field from the file is just the hour information, i.e., say, 14:50:00. So, the value being stored is 1900-01-01 14:50:00. I don't need the date part. How to do that.
Or may be an alternate datatype which can store only time.

Answer 1

Amazon Redshift supports only date(year month day) and timestamp(year month day hour minute second) format, and it doesn't support time(hour minute second) format of Postgresql.
In my idea, there are two ways to work around.
As #Damien_The_Unbeliever mentioned, ignore the date part of the timestamp format.
create table date_test(id int, timestamp timestamp);
insert into date_test2 values (1, '1900-01-01 14:50:00');
insert into date_test2 values (2, '1900-01-01 17:20:00');
select * from date_test2 where timestamp > '1900-01-01 14:50:00';
select * from date_test where date_test.timestamp > '1900-01-01 14:50:00';
id | timestamp
----+---------------------
2 | 1900-01-01 17:20:00
(1 row)
Use char or varchar type to store the time value.
create table date_test2(id int, timestamp char(8));
insert into date_test2 values (1, '14:50:00');
insert into date_test2 values (2, '17:20:00');
select * from date_test2 where timestamp > '14:50:00';
id | timestamp
----+-----------
2 | 17:20:00
(1 row)
The second solution looks easier, but it is worse performance as Redshift doc says. If you store a large amount of data, you should consider of the first one.
Here are the related links to the document about date/time column.
http://docs.aws.amazon.com/redshift/latest/dg/c_best-practices-timestamp-date-columns.html
http://docs.aws.amazon.com/redshift/latest/dg/r_Datetime_types.html