select distinct on (Key) Key, Date, Value
from <<table name>>
order by Key, Date desc
MYDATABASE=> create table aaa (key varchar(1), date date, value text);
CREATE TABLE
Time: 1518.002 ms
MYDATABASE=> insert into aaa (key, date, value) values ('A',
'2008-05-01', 'foo');
INSERT 0 1
Time: 1.125 ms
MYDATABASE=> insert into aaa (key, date, value) values ('A',
'2008-04-01', 'bar');
INSERT 0 1
Time: 0.290 ms
MYDATABASE=> insert into aaa (key, date, value) values ('A',
'2008-03-01', 'foo');
INSERT 0 1
Time: 0.310 ms
MYDATABASE=> insert into aaa (key, date, value) values ('B',
'2008-03-01', 'baz');
INSERT 0 1
Time: 0.304 ms
MYDATABASE=> insert into aaa (key, date, value) values ('B',
'2008-02-01', 'bar');
INSERT 0 1
Time: 0.330 ms
MYDATABASE=> insert into aaa (key, date, value) values ('C',
'2008-06-03', 'foo');
INSERT 0 1
Time: 0.298 ms
MYDATABASE=> insert into aaa (key, date, value) values ('C',
'2008-04-04', 'baz');
INSERT 0 1
Time: 0.295 ms
MYDATABASE=> insert into aaa (key, date, value) values ('C',
'2008-03-04', 'bar');
INSERT 0 1
Time: 0.319 ms
MYDATABASE=> commit;
COMMIT
Time: 569.591 ms
MYDATABASE=> select * from aaa;
key | date | value
-----+------------+-------
A | 2008-05-01 | foo
A | 2008-04-01 | bar
A | 2008-03-01 | foo
B | 2008-03-01 | baz
B | 2008-02-01 | bar
C | 2008-06-03 | foo
C | 2008-04-04 | baz
C | 2008-03-04 | bar
(8 rows)
Time: 0.520 ms
MYDATABASE=> select distinct on (key) key, date, value from aaa order by
key, date desc;
key | date | value
-----+------------+-------
A | 2008-05-01 | foo
B | 2008-03-01 | baz
C | 2008-06-03 | foo
(3 rows)
Time: 0.524 ms
-Mark
On Sat, 2008-10-04 at 00:25 +0530, Raj Mathur wrote:
> I have some data of the form:
>
> Key | Date | Value
> A | 2008-05-01 | foo *
> A | 2008-04-01 | bar
> A | 2008-03-01 | foo *
> B | 2008-03-04 | baz
> B | 2008-02-04 | bar
> C | 2008-06-03 | foo *
> C | 2008-04-04 | baz
> C | 2008-03-04 | bar
>
> Is there any way to select only the rows marked with a (*) out of
> these
> without doing a join? I.e. I wish to find the row with the highest
> Date for each Key and use the Value from that.
>
> Regards,
>
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