MySQL duplicate data removal with loop - mysql

Question

I have a table called Positions which has data like this:
Id PositionId
1 'a'
2 'a '
3 'b '
4 'b'
Some of them has spaces so my idea is to remove those spaces, this is not actual table just an example of a table which has much more data.
So i created procedure to iterate over PositionIds and compare them if trimed they match remove one of them:
CREATE PROCEDURE remove_double_positions()
BEGIN
DECLARE done INT DEFAULT 0;
DECLARE current VARCHAR(255);
DECLARE previous VARCHAR(255) DEFAULT NULL;
DECLARE positionCur CURSOR FOR SELECT PositionId FROM Positions ORDER BY PositionId;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
OPEN positionCur;
clean_duplicates: LOOP
FETCH positionCur INTO current;
IF done THEN
LEAVE clean_duplicates;
END IF;
IF previous LIKE current THEN
DELETE FROM Positions WHERE PositionId = current;
END IF;
SET previous = current;
END LOOP clean_duplicates;
CLOSE positionCur;
END
For some reason it shows that 2 rows were affected but actually deletes all 4 of them and i don't know the reason why, could you help me.

Answer 1

From the manual https://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html#operator_like under the like operator - Per the SQL standard, LIKE performs matching on a per-character basis, thus it can produce results different from the = comparison operator:...In particular, trailing spaces are significant, which is not true for CHAR or VARCHAR comparisons performed with the = operator:
mysql> SELECT 'a' = 'a ', 'a' LIKE 'a ';
+------------+---------------+
| 'a' = 'a ' | 'a' LIKE 'a ' |
+------------+---------------+
| 1 | 0 |
+------------+---------------+
1 row in set (0.00 sec)
This is true when = or like is used in where or case.
Your procedure would work as desired if you amended the delete bit to
IF trim(previous) = trim(current) THEN
DELETE FROM Positions WHERE PositionId like current;
END IF;

Answer 2

Just some other solution without cursor and procedure. I've check it on ORACLE. Hope it helps.
DELETE FROM positions
WHERE id IN ( SELECT t1.id
FROM positions t1,
positions t2
WHERE t1.positionId = TRIM(t2.positionId)
AND t1.positionId != t2.positionId
);
UPDATE
There are some crasy things are going on with mysql. Some problem with blank at the end of a strong and this error 1093 error.
Now my solution checked with MySQL 5.5.9
CREATE TABLE positions (
id INT NOT NULL,
positionid VARCHAR(2) NOT NULL
);
INSERT INTO positions VALUES
( 1, 'a'),
( 2, 'a '),
( 3, 'b'),
( 4, 'b ');
DELETE FROM positions
WHERE id IN ( SELECT t3.id FROM
(SELECT t2.id
FROM positions t1,
positions t2
WHERE t1.positionid = t2.positionid
AND LENGTH(t1.positionid) = 1
AND length(t2.positionid) = 2
) t3
);
mysql> SELECT * from positions;
+----+------------+
| id | positionid |
+----+------------+
| 1 | a |
| 3 | b |
+----+------------+
2 rows in set (0.00 sec)
mysql>
This "double" from delete SQL will fix this error 1093
Hope this helps.