String compare exact in query MySQL - mysql

Question

I created table like that in MySQL:
DROP TABLE IF EXISTS `barcode`;
CREATE TABLE `barcode` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`code` varchar(40) COLLATE utf8_bin DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;
INSERT INTO `barcode` VALUES ('1', 'abc');
INSERT INTO `barcode` VALUES ('2', 'abc ');
Then I query data from table barcode:
SELECT * FROM barcode WHERE `code` = 'abc ';
The result is:
+-----+-------+
| id | code |
+-----+-------+
| 1 | abc |
+-----+-------+
| 2 | abc |
+-----+-------+
But I want the result set is only 1 record. I workaround with:
SELECT * FROM barcode WHERE `code` = binary 'abc ';
The result is 1 record. But I'm using NHibernate with MySQL for generating query from mapping table. So that how to resolve this case?

Answer 1

There is no other fix for it. Either you specify a single comparison as being binary or you set the whole database connection to binary. (doing SET NAMES binary, which may have other side effects!)
Basically, that 'lazy' comparison is a feature of MySQL which is hard coded. To disable it (on demand!), you can use a binary compare, what you apparently already do. This is not a 'workaround' but the real fix.
from the MySQL Manual:
All MySQL collations are of type PADSPACE. This means that all CHAR and VARCHAR values in MySQL are compared without regard to any trailing spaces
Of course there are plenty of other possiblities to achieve the same result from a user's perspective, i.e.:
WHERE field = 'abc ' AND CHAR_LENGTH(field) = CHAR_LENGTH('abc ')
WHERE field REGEXP 'abc[[:space:]]'
The problem with these is that they effectively disable fast index lookups, so your query always results in a full table scan. With huge datasets that makes a big difference.
Again: PADSPACE is default for MySQLs [VAR]CHAR comparison. You can (and should) disable it by using BINARY. This is the indended way of doing this.

Answer 2

You can try with a regular expression matching :
SELECT * FROM barcode WHERE `code` REGEXP 'abc[[:space:]]'

Answer 3

i was just working on case just like that when using LIKE with wildcard (%) resulting in an unexpected result. While searching i also found STRCMP(text1, text2) under string comparison feature of mysql which compares two string. however using BINARY with LIKE solved the problem for me.
SELECT * FROM barcode WHERE `code` LIKE BINARY 'abc ';

Answer 4

You could do this:
SELECT * FROM barcode WHERE `code` = 'abc '
AND CHAR_LENGTH(`code`)=CHAR_LENGTH('abc ');

Answer 5

I am assuming you only want one result, you could use LIMIT
SELECT * FROM barcode WHERE `code` = 'abc ' LIMIT 1;
To do exact string matching you could use Collation
SELECT *
FROM barcode
WHERE code COLLATE utf8_bin = 'abc';

Answer 6

The sentence right after the one quoted by Kaii basically says "use LIKE" :
“Comparison” in this context does not include the LIKE pattern-matching operator, for which trailing spaces are significant
and the example below shows that 'Monty' = 'Monty ' is true, but not 'Monty' LIKE 'Monty '.
However, if you use LIKE, beware of literal strings containing the '%', '_' or '\' characters : '%' and '_' are wildcard characters, '\' is used to escape sequences.