Oracle SQL How to find duplicate values in different columns? - sql

Question

I have a set of rows with many columns. For example,
ID | Col1 | Col2 | Col3 | Duplicate
------------------------------------
81 | 101 | 102 | 101 | YES
82 | 101 | 103 | 104 | NO
I need to calculate the "Duplicate" column. It is duplicate because it has the same value in Col1 and Col3. I know there is the LEAST function, which is similar to the MIN function but with columns. Does something similar to achieve this exists?
The approach I have in mind is to write all possible combinations in a case like this:
SELECT ID, col1, col2, col3,
CASE WHEN col1 = col2 or col1 = col3 or col2 = col3 then 1 else 0 end as Duplicate
FROM table
But, I wish to avoid that, since I have too many columns in some cases, and is very prone to errors.
What is the best way to solve this?

Answer 1

Hmmm. You are looking for within-row duplicates. This is painful. More recent versions of Oracle support lateral joins. But for just a handful of non-NULL columns, you can do:
select id, col1, col2, col3,
(case when col1 in (col2, col3) or col2 in (col3) then 1 else 0 end) as Duplicate
from t;
For each additional column, you need to add one more in comparison and update the other in-lists.

Answer 2

Something like this... note that in the lateral clause we still need to unpivot, but that is one row at a time - resulting in possibly much faster execution than simple unpivot and standard aggregation.
with
input_data ( id, col1, col2, col3 ) as (
select 81, 101, 102, 101 from dual union all
select 82, 101, 103, 104 from dual
)
-- End of simulated input data (for testing purposes only).
-- Solution (SQL query) begins BELOW THIS LINE.
select i.id, i.col1, i.col2, i.col3, l.duplicates
from input_data i,
lateral ( select case when count (distinct val) = count(val)
then 'NO' else 'YES'
end as duplicates
from input_data
unpivot ( val for col in ( col1, col2, col3 ) )
where id = i.id
) l
;
ID COL1 COL2 COL3 DUPLICATES
-- ---- ---- ---- ----------
81 101 102 101 YES
82 101 103 104 NO

Answer 3

You can do this by unpivoting and then counting the distinct values per id and checking if it equals the number of rows for that id. Equal means there are no duplicates. Then left join this result to the original table to caclulate the duplicate column.
SELECT t.*,
CASE WHEN x.id IS NOT NULL THEN 'Yes' ELSE 'No' END AS duplicate
FROM t
LEFT JOIN
(SELECT id
FROM
(SELECT *
FROM t
unpivot (val FOR col IN (col1,col2,col3)) u
) t
GROUP BY id
HAVING count(*)<>count(DISTINCT val)
) x ON x.id=t.id

Answer 4

The best way† is to avoid storing repeating groups of columns. If you have multiple columns that essentially store comparable data (i.e. a multi-valued attribute), move the data to a dependent table, and use one column.
CREATE TABLE child (
ref_id INT,
col INT
);
INSERT INTO child VALUES
(81, 101), (81, 102), (81, 101),
(82, 101), (82, 103), (82, 104);
Then it's easier to find cases where a value occurs more than once:
SELECT id, col, COUNT(*)
FROM child
GROUP BY id, col
HAVING COUNT(*) > 1;
If you can't change the structure of the table, you could simulate it using UNIONs:
SELECT id, col1, COUNT(*)
FROM (
SELECT id, col1 AS col FROM mytable
UNION ALL SELECT id, col2 FROM mytable
UNION ALL SELECT id, col3 FROM mytable
... for more columns ...
) t
GROUP BY id, col
HAVING COUNT(*) > 1;
† Best for the query you are trying to run. A denormalized storage strategy might be better for some other types of queries.

Answer 5

SELECT ID, col1, col2,
NVL2(NULLIF(col1, col2), 'Not duplicate', 'Duplicate')
FROM table;
If you want to compare more than 2 columns can implement same logic with COALESCE

Answer 6

I think you want to use fresh data that doesnot contains any duplicate values inside table if it right then use SELECT DISTINCT statement like
SELECT DISTINCT * FROM TABLE_NAME
It will conatins duplicate free data,
Note: It will also applicable for a particular column like
SELECT DISTINCT col1 FROM TABLE_NAME