Обсуждение: subquery question
Hi,
I have a table: (date timestamp, id integer, value integer)
What Iam trying to do is to get a result that looks like this:
day sum_odd sum_even
2009-01-01 6565 78867
2009-01-02 876785 87667
basically a need to combine these two queries into one:
SELECT date_trunc('day',date) AS day, sum(value) AS sum_odd FROM xyz WHERE id % 2 = 1 GROUP BY date_trunc('day',date)
SELECT date_trunc('day',date) AS day, sum(value) AS sum_even FROM xyz WHERE id % 2 = 0 GROUP BY date_trunc('day',date)
I found various ways to do this via unions or joins, but none of them seem efficient, what is the best way to do that ?
thank you very much
Sebastian
Does this help
Here is my test table data.
ID;DATE;VALUE
1;"2009-03-13";5
2;"2009-03-13";2
3;"2009-03-11";1
4;"2009-03-11";2
5;"2009-03-11";3
SELECT mydate AS day, SUM(CASE WHEN id % 2 = 1 THEN value END) AS sum_odd, SUM(CASE WHEN id % 2 = 0 THEN
valueEND) AS sum_even
FROM xyz
GROUP BY mydate;
DATE;SUM_ODD;SUM_EVEN
"2009-03-11";4;2
"2009-03-13";5;2
Check the plans generated to see if one query actually appears better
than another.
Bob
On Thu, Mar 12, 2009 at 9:06 PM, Sebastian Böhm <seb@exse.net> wrote:
> Hi,
> I have a table: (date timestamp, id integer, value integer)
> What Iam trying to do is to get a result that looks like this:
> day sum_odd sum_even
> 2009-01-01 6565 78867
> 2009-01-02 876785 87667
>
> basically a need to combine these two queries into one:
> SELECT date_trunc('day',date) AS day, sum(value) AS sum_odd FROM
> xyz WHERE id % 2 = 1 GROUP BY date_trunc('day',date)
> SELECT date_trunc('day',date) AS day, sum(value) AS sum_even FROM
> xyz WHERE id % 2 = 0 GROUP BY date_trunc('day',date)
> I found various ways to do this via unions or joins, but none of them seem
> efficient, what is the best way to do that ?
>
> thank you very much
> Sebastian