Mysql tæller forekomster af understreng, og bestil derefter efter

SELECT (CHAR_LENGTH(str) - CHAR_LENGTH(REPLACE(str, substr, ''))) / CHAR_LENGTH(substr) AS cnt
...
ORDER BY cnt DESC

Ja, det ser oppustet ud, men der er ikke nogen anden mulig løsning.

mysql> select (CHAR_LENGTH('asd') - CHAR_LENGTH(REPLACE('asd', 's', ''))) / CHAR_LENGTH('s');
+-----------------------------------------------------------------+
| (CHAR_LENGTH('asd') - CHAR_LENGTH(REPLACE('asd', 's', ''))) / CHAR_LENGTH('s') |
+-----------------------------------------------------------------+
|                                                          1.0000 |
+-----------------------------------------------------------------+
1 row in set (0.00 sec)



mysql> select host, (CHAR_LENGTH(host) - CHAR_LENGTH(REPLACE(host, 'l', ''))) / CHAR_LENGTH('l') AS cnt from user;
+-----------+--------+
| host      | cnt    |
+-----------+--------+
| 127.0.0.1 | 0.0000 |
| honeypot  | 0.0000 |
| honeypot  | 0.0000 |
| localhost | 2.0000 |
| localhost | 2.0000 |
+-----------+--------+
5 rows in set (0.00 sec)