mysql互换表中两列数据方法
1.创建表及记录用于测试
CREATE TABLE `product` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT COMMENT '产品id',
`name` varchar(50) NOT NULL COMMENT '产品名称',
`original_price` decimal(5,2) unsigned NOT NULL COMMENT '原价',
`price` decimal(5,2) unsigned NOT NULL COMMENT '现价',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `product` (`id`, `name`, `original_price`, `price`) VALUES
(NULL, '雪糕', '5', '3.5'),
(NULL, '鲜花', '18', '15'),
(NULL, '甜点', '25', '12.5'),
(NULL, '玩具', '55', '45'),
(NULL, '钱包', '285', '195');
mysql> select * from product;
+----+--------+----------------+--------+
| id | name | original_price | price |
+----+--------+----------------+--------+
| 1 | 雪糕 | 5.00 | 3.50 |
| 2 | 鲜花 | 18.00 | 15.00 |
| 3 | 甜点 | 25.00 | 12.50 |
| 4 | 玩具 | 55.00 | 45.00 |
| 5 | 钱包 | 285.00 | 195.00 |
+----+--------+----------------+--------+
5 rows in set (0.00 sec)
2.互换original_price与price的值
新手可能会使用以下方法进行互换
update product set original_price=price,price=original_price;- 1
但这样执行的结果只会使original_price与price的值都是price的值，因为update有顺序的，
先执行original_price=price , original_price的值已经更新为price，
然后执行price=original_price，这里相当于没有更新。
执行结果：
mysql> select * from product;
+----+--------+----------------+--------+
| id | name | original_price | price |
+----+--------+----------------+--------+
| 1 | 雪糕 | 5.00 | 3.50 |
| 2 | 鲜花 | 18.00 | 15.00 |
| 3 | 甜点 | 25.00 | 12.50 |
| 4 | 玩具 | 55.00 | 45.00 |
| 5 | 钱包 | 285.00 | 195.00 |
+----+--------+----------------+--------+
5 rows in set (0.00 sec)
mysql> update product set original_price=price,price=original_price;
Query OK, 5 rows affected (0.00 sec)
Rows matched: 5 Changed: 5 Warnings: 0
mysql> select * from product;
+----+--------+----------------+--------+
| id | name | original_price | price |
+----+--------+----------------+--------+
| 1 | 雪糕 | 3.50 | 3.50 |
| 2 | 鲜花 | 15.00 | 15.00 |
| 3 | 甜点 | 12.50 | 12.50 |
| 4 | 玩具 | 45.00 | 45.00 |
| 5 | 钱包 | 195.00 | 195.00 |
+----+--------+----------------+--------+
5 rows in set (0.00 sec)
正确的互换方法如下：
update product as a, product as b set a.original_price=b.price, a.price=b.original_price where a.id=b.id;- 1
执行结果：
mysql> select * from product;
+----+--------+----------------+--------+
| id | name | original_price | price |
+----+--------+----------------+--------+
| 1 | 雪糕 | 5.00 | 3.50 |
| 2 | 鲜花 | 18.00 | 15.00 |
| 3 | 甜点 | 25.00 | 12.50 |
| 4 | 玩具 | 55.00 | 45.00 |
| 5 | 钱包 | 285.00 | 195.00 |
+----+--------+----------------+--------+
5 rows in set (0.00 sec)
mysql> update product as a, product as b set a.original_price=b.price, a.price=b.original_price where a.id=b.id;
Query OK, 5 rows affected (0.01 sec)
Rows matched: 5 Changed: 5 Warnings: 0
mysql> select * from product;
+----+--------+----------------+--------+
| id | name | original_price | price |
+----+--------+----------------+--------+
| 1 | 雪糕 | 3.50 | 5.00 |
| 2 | 鲜花 | 15.00 | 18.00 |
| 3 | 甜点 | 12.50 | 25.00 |
| 4 | 玩具 | 45.00 | 55.00 |
| 5 | 钱包 | 195.00 | 285.00 |
+----+--------+----------------+--------+
5 rows in set (0.00 sec)
最后我用update product as a, product as b set a.original_price=b.price, a.price=b.original_price where a.id=b.id;http://blog.csdn.net/fdipzone/article/details/50864196
