How to optimize this SELECT with sub query Oracle - oracle

Question

Here is my query,
SELECT ID As Col1,
(
SELECT VID FROM TABLE2 t
WHERE (a.ID=t.ID or a.ID=t.ID2)
AND t.STARTDTE =
(
SELECT MAX(tt.STARTDTE)
FROM TABLE2 tt
WHERE (a.ID=tt.ID or a.ID=tt.ID2) AND tt.STARTDTE < SYSDATE
)
) As Col2
FROM TABLE1 a
Table1 has 48850 records and Table2 has 15944098 records.
I have separate indexes in TABLE2 on ID,ID & STARTDTE, STARTDTE, ID, ID2 & STARTDTE.
The query is still too slow. How can this be improved? Please help.

Answer 1

I'm guessing that the OR in inner queries is messing up with the optimizer's ability to use indexes. Also I wouldn't recommend a solution that would scan all of TABLE2 given its size.
This is why in this case I would suggest using a function that will efficiently retrieve the information you are looking for (2 index scan per call):
CREATE OR REPLACE FUNCTION getvid(p_id table1.id%TYPE)
RETURN table2.vid%TYPE IS
l_result table2.vid%TYPE;
BEGIN
SELECT vid
INTO l_result
FROM (SELECT vid, startdte
FROM (SELECT vid, startdte
FROM table2 t
WHERE t.id = p_id
AND t.startdte < SYSDATE
ORDER BY t.startdte DESC)
WHERE rownum = 1
UNION ALL
SELECT vid, startdte
FROM (SELECT vid, startdte
FROM table2 t
WHERE t.id2 = p_id
AND t.startdte < SYSDATE
ORDER BY t.startdte DESC)
WHERE rownum = 1
ORDER BY startdte DESC)
WHERE rownum = 1;
RETURN l_result;
END;
Your SQL would become:
SELECT ID As Col1,
getvid(a.id) vid
FROM TABLE1 a
Make sure you have indexes on both table2(id, startdte DESC) and table2(id2, startdte DESC). The order of the index is very important.

Answer 2

Possibly try the following, though untested.
WITH max_times AS
(SELECT a.ID, MAX(t.STARTDTE) AS Startdte
FROM TABLE1 a, TABLE2 t
WHERE (a.ID=t.ID OR a.ID=t.ID2)
AND t.STARTDTE < SYSDATE
GROUP BY a.ID)
SELECT b.ID As Col1, tt.VID
FROM TABLE1 b
LEFT OUTER JOIN max_times mt
ON (b.ID = mt.ID)
LEFT OUTER JOIN TABLE2 tt
ON ((mt.ID=tt.ID OR mt.ID=tt.ID2)
AND mt.startdte = tt.startdte)

Answer 3

You can look at analytic functions to avoid having to hit the second table twice. Something like this might work:
SELECT id AS col1, vid
FROM (
SELECT t1.id, t2.vid, RANK() OVER (PARTITION BY t1.id ORDER BY
CASE WHEN t2.startdte < TRUNC(SYSDATE) THEN t2.startdte ELSE null END
NULLS LAST) AS rn
FROM table1 t1
JOIN table2 t2 ON t2.id IN (t1.ID, t1.ID2)
)
WHERE rn = 1;
The inner select gets the id and vid values from the two tables with a simple join on id or id2. The rank function calculates a ranking for each matching row in the second table based on the startdte. It's complicated a bit by you wanting to filter on that date, so I've used a case to effectively ignore any dates today or later by changing the evaluated value to null, and in this instance that means the order by in the over clause needs nulls last so they're ignored.
I'd suggest you run the inner select on its own first - maybe with just a couple of id values for brevity - to see what its doing, and what ranks are being allocated.
The outer query is then just picking the top-ranked result for each id.
You may still get duplicates though; if table2 has more than one row for an id with the same startdte they'll get the same rank, but then you may have had that situation before. You may need to add more fields to the order by to break ties in a way that makes sens to you.
But this is largely speculation without being able to see where your existing query is actually slow.