Swapping column values in MySQL

I just had to deal with the same and I'll summarize my findings.

1. The UPDATE table SET X=Y, Y=X approach obviously doesn't work, as it'll just set both values to Y.

2. Here's a method that uses a temporary variable. Thanks to Antony from the comments of http://beerpla.net/2009/02/17/swapping-column-values-in-mysql/ for the "IS NOT NULL" tweak. Without it, the query works unpredictably. See the table schema at the end of the post. This method doesn't swap the values if one of them is NULL. Use method #3 that doesn't have this limitation.

   UPDATE swap_test SET x=y, y=@temp WHERE (@temp:=x) IS NOT NULL;

3. This method was offered by Dipin in, yet again, the comments of http://beerpla.net/2009/02/17/swapping-column-values-in-mysql/. I think it’s the most elegant and clean solution. It works with both NULL and non-NULL values.

   UPDATE swap_test SET x=(@temp:=x), x = y, y = @temp;

4. Another approach I came up with that seems to work:

   UPDATE swap_test s1, swap_test s2 SET s1.x=s1.y, s1.y=s2.x WHERE s1.id=s2.id;

Essentially, the 1st table is the one getting updated and the 2nd one is used to pull the old data from.
Note that this approach requires a primary key to be present.

This is my test schema:

CREATE TABLE `swap_test` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `x` varchar(255) DEFAULT NULL,
  `y` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB;

INSERT INTO `swap_test` VALUES ('1', 'a', '10');
INSERT INTO `swap_test` VALUES ('2', NULL, '20');
INSERT INTO `swap_test` VALUES ('3', 'c', NULL);

You could take the sum and subtract the opposing value using X and Y

UPDATE swaptest SET X=X+Y,Y=X-Y,X=X-Y;

Here is a sample test (and it works with negative numbers)

mysql> use test
Database changed
mysql> drop table if exists swaptest;
Query OK, 0 rows affected (0.03 sec)

mysql> create table swaptest (X int,Y int);
Query OK, 0 rows affected (0.12 sec)

mysql> INSERT INTO swaptest VALUES (1,2),(3,4),(-5,-8),(-13,27);
Query OK, 4 rows affected (0.08 sec)
Records: 4  Duplicates: 0  Warnings: 0

mysql> SELECT * FROM swaptest;
+------+------+
| X    | Y    |
+------+------+
|    1 |    2 |
|    3 |    4 |
|   -5 |   -8 |
|  -13 |   27 |
+------+------+
4 rows in set (0.00 sec)

mysql>

Here is the swap being performed

mysql> UPDATE swaptest SET X=X+Y,Y=X-Y,X=X-Y;
Query OK, 4 rows affected (0.07 sec)
Rows matched: 4  Changed: 4  Warnings: 0

mysql> SELECT * FROM swaptest;
+------+------+
| X    | Y    |
+------+------+
|    2 |    1 |
|    4 |    3 |
|   -8 |   -5 |
|   27 |  -13 |
+------+------+
4 rows in set (0.00 sec)

mysql>

Give it a Try !!!

The following code works for all scenarios in my quick testing:

UPDATE swap_test
   SET x=(@temp:=x), x = y, y = @temp

UPDATE table SET X=Y, Y=X will do precisely what you want (edit: in PostgreSQL, not MySQL, see below). The values are taken from the old row and assigned to a new copy of the same row, then the old row is replaced. You do not have to resort to using a temporary table, a temporary column, or other swap tricks.

@D4V360: I see. That is shocking and unexpected. I use PostgreSQL and my answer works correctly there (I tried it). See the PostgreSQL UPDATE docs (under Parameters, expression), where it mentions that expressions on the right hand side of SET clauses explicitly use the old values of columns. I see that the corresponding MySQL UPDATE docs contain the statement "Single-table UPDATE assignments are generally evaluated from left to right" which implies the behaviour you describe.

Good to know.