尽管没有通配符，查询返回“LIKE”结果？

Question

我真的很困惑。运行以下查询：

SELECT * FROM `articles` WHERE `form` = 'Depotplåster' AND `size` = 5

返回也以“5”开头的行，尽管我既不使用LIKE也不使用%通配符运算符。怎么样？

size字段的类型为VARCHAR。

Answer 1

这是因为您正在使用数字和varchar数据之间的比较。 MySQL会隐式将您的列转换为double，从而生成5。看到这个简单的测试数据：

mysql> select * from test;
+-----------------+
| name            |
+-----------------+
| 5               |
| 5 and some crap |
+-----------------+
2 rows in set (0.00 sec)

现在，“好”方式：比较字符串：

mysql> select * from test where name = '5';
+------+
| name |
+------+
| 5    |
+------+
1 row in set (0.00 sec)

和“坏”方式：比较整数：

mysql> select * from test where name = 5;
+-----------------+
| name            |
+-----------------+
| 5               |
| 5 and some crap |
+-----------------+
2 rows in set, 1 warning (0.05 sec)

- 这是你的理由：

+---------+------+-----------------------------------------------------+
| Level   | Code | Message                                             |
+---------+------+-----------------------------------------------------+
| Warning | 1292 | Truncated incorrect DOUBLE value: '5 and some crap' |
+---------+------+-----------------------------------------------------+
1 row in set (0.00 sec)

最后，要明白，为什么会如此：

SELECT 
  CAST('5' AS DECIMAL) AS 5d, 
  CAST('5 and some crap' AS DECIMAL) AS 5sd, 
  CAST('5' AS DECIMAL) = CAST('5 and some crap' AS DECIMAL) AS areEqual;

将导致：

+----+-----+----------+
| 5d | 5sd | areEqual |
+----+-----+----------+
|  5 |   5 |        1 |
+----+-----+----------+
1 row in set (0.00 sec)

- 你可以看到，非重要部分被截断（如上面的警告信息中所述）

Answer 2

SELECT * FROM `articles` WHERE `form` = 'Depotplåster' AND `size` = '5'
-- this will compare the string 'size' with the string '5'


SELECT * FROM `articles` WHERE `form` = 'Depotplåster' AND `size` = 5
-- this will convert string 'size' to integer and then compare with the integer 5

将字符串转换为整数在字符串的开头查找整数，并取最大整数，直到第一个非数字字符。

select '5s4'=5, 's5'=5, '5'=5 -- =>1,0,1

Answer 3

SELECT *
FROM `articles`
WHERE `form` = 'Depotplåster'
AND `size` = '5'

你应该引用5，因为MySQL将表中的字符串转换为int而没有引号。