To make this short, simple and interesting let’s try playing the game.

If there was “What Geek Wants to be a Millionaire” then this could go for the last question. You know the pressure and all could take couple of minutes to figure the answer out :)

Questions

So, here it goes. We have three simple MySQLv5 queries executed on default MySQL configuration. You have to give the result for these three simple queries.

1:

SELECT ('mysql' IN (0)) AS '@result';

2:

SELECT ('mysql' IN (1)) AS 'result';

3:

SELECT ('mysql' IN ('sql')) AS result;

If you know the answer so you could find the trick very quickly and you’re the winner of “Who Wants to be a Millionaire” :)

Answers

The answers are quite simple as queries. Here it goes:

1. The problem with the first query is that ‘mysql’ is of string type and 0 is numeric and there has to be type casting. So string ‘mysql’ is casted into numeric 0 and now we have query where we want to find 0 in set of only one value and that value is 0:

SELECT (0 IN (0))

So the result will be 1:

mysql> SELECT ('mysql' IN (0)) AS '@result';
+---------+
| @result |
+---------+
|       1 |
+---------+
1 row in set, 1 warning (0.01 sec)

mysql> SHOW WARNINGS;
+---------+------+-------------------------------------------+
| Level   | Code | Message                                   |
+---------+------+-------------------------------------------+
| Warning | 1292 | Truncated incorrect DOUBLE value: 'mysql' |
+---------+------+-------------------------------------------+
1 row in set (0.00 sec)

2. As I explained in the first example the query could be translated into:

SELECT (0 IN (1))

So the result will be 0.

3. This query is just for those newbies that don’t know MySQL’s IN() :) The answer unquestionably is 0.

So the first query was with the trick and I hope that there would be couple of people that will save ton of time trying to debug some similar query to first example.