DROP TABLE gab;
CREATE TABLE gab (usr VARCHAR2(10) NOT NULL, val NUMBER NOT NULL --, CONSTRAINT gab$uq
UNIQUE (usr, val)
);
INSERT INTO gab VALUES('GAP', 1);
INSERT INTO gab VALUES('GAP', 5);
INSERT INTO gab VALUES('GAP', 7);
INSERT INTO gab VALUES('JKL', 8);
INSERT INTO gab VALUES('JKL', 5);
COMMIT;SELECT usr
FROM (
SELECT DISTINCT usr, val FROM gab
)
WHERE val IN (1, 5, 7)
GROUP BY
usr
HAVING COUNT(*) = 3 -- number of elements in the list
/Depending on the existence of the constraint, here gab$uq, you can either use inline view of run it against original table. -- Vladimir Begun The statements and opinions expressed here are my own and do not necessarily represent those of Oracle Corporation.
Gabriel Aragon wrote:
I have a table with like this:
Usr val ---------- GAP 1 GAP 5 GAP 7 JKL 8 JKL 5
I need a query that returns the user (GAP o JKL) that has ALL the values in a list. Example: Having the list: 1,5,7 the result will be GAP, but with the values 1,5 or 1,5,7,8 there will be no result.
select distinct usr from xxx where val = All (1,3,5)
I was trying the ALL operator but it works with part of the list, I need the user that has (exactly) all the values in the list. Any idea?
Maybe it's a simple solution, but after several hours I feel blocked.
TIA Gabriel
-- Please see the official ORACLE-L FAQ: http://www.orafaq.net -- Author: Vladimir Begun INET: [EMAIL PROTECTED]
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