Removing duplicate rows from table in Oracle

Use the rowid pseudocolumn.

DELETE FROM your_table
WHERE rowid not in
(SELECT MIN(rowid)
FROM your_table
GROUP BY column1, column2, column3);

Where column1, column2, and column3 make up the identifying key for each record. You might list all your columns.

From DevX.com:

DELETE FROM our_table
WHERE rowid not in
(SELECT MIN(rowid)
FROM our_table
GROUP BY column1, column2, column3...) ;

Where column1, column2, etc. is the key you want to use.

You should do a small pl/sql block using a cursor for loop and delete the rows you don't want to keep. For instance:

declare
prev_var my_table.var1%TYPE;

begin

for t in (select var1 from my_table order by var 1) LOOP

-- if previous var equal current var, delete the row, else keep on going.
end loop;

end;

This blog post was really helpful for general cases:

If the rows are fully duplicated (all values in all columns can have copies) there are no columns to use! But to keep one you still need a unique identifier for each row in each group. Fortunately, Oracle already has something you can use. The rowid. All rows in Oracle have a rowid. This is a physical locator. That is, it states where on disk Oracle stores the row. This unique to each row. So you can use this value to identify and remove copies. To do this, replace min() with min(rowid) in the uncorrelated delete:

delete films
where  rowid not in (
  select min(rowid)
  from   films
  group  by title, uk_release_date
)

I didn't see any answers that use common table expressions and window functions. This is what I find easiest to work with.

DELETE FROM
 YourTable
WHERE
 ROWID IN
    (WITH Duplicates
          AS (SELECT
               ROWID RID, 
               ROW_NUMBER() 
               OVER(
               PARTITION BY First_Name, Last_Name, Birth_Date)
                  AS RN
               SUM(1)
               OVER(
               PARTITION BY First_Name, Last_Name, Birth_Date
               ORDER BY ROWID ROWS BETWEEN UNBOUNDED PRECEDING 
                                       AND UNBOUNDED FOLLOWING)
                   AS CNT
              FROM
               YourTable
              WHERE
               Load_Date IS NULL)
     SELECT
      RID
     FROM
      duplicates
     WHERE
      RN > 1);

Somethings to note:

1) We are only checking for duplication on the fields in the partition clause.

2) If you have some reason to pick one duplicate over others you can use an order by clause to make that row will have row_number() = 1

3) You can change the number duplicate preserved by changing the final where clause to "Where RN > N" with N >= 1 (I was thinking N = 0 would delete all rows that have duplicates, but it would just delete all rows).

4) Added the Sum partition field the CTE query which will tag each row with the number rows in the group. So to select rows with duplicates, including the first item use "WHERE cnt > 1".

solution :

delete from emp where rowid in
(
    select rid from
    (
        select rowid rid,
        row_number() over(partition by empno order by empno) rn
        from emp
    )
    where rn > 1
);

create or replace procedure delete_duplicate_enq as
    cursor c1 is
    select *
    from enquiry;
begin
    for z in c1 loop
        delete enquiry
        where enquiry.enquiryno = z.enquiryno
        and rowid > any
        (select rowid
        from enquiry
        where enquiry.enquiryno = z.enquiryno);
    end loop;
 end delete_duplicate_enq;

This is similar to the top answer but gives me a much better explain plan:

delete from your_table
 where rowid in (
        select max(rowid)
          from your_table
         group by column1, column2, column3
        having count(*) > 1
       );