Question

我在mySQL中遇到如下问题：

计算mySQL数据库中字符串字段中子字符串的实例
按子字符串（DESC）的出现次数排序结果

我从未做过除基本查询以外的任何事情。我无法在其他地方找到解决方案。

Answer 1

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

是的，看起来很臃肿，但是没有任何其他可能的解决方案。

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)

Answer 2

DELIMITER //
DROP FUNCTION IF EXISTS `subStringCount`//
CREATE FUNCTION `subStringCount` (sequence VARCHAR(1000), word VARCHAR(100)) RETURNS INT(4)
DETERMINISTIC
CONTAINS SQL
BEGIN
    DECLARE counter SMALLINT UNSIGNED DEFAULT 0;
    DECLARE word_length SMALLINT UNSIGNED;

    SET word_length = CHAR_LENGTH(word);

    WHILE (INSTR(sequence,word) != 0) DO
        SET counter = counter+1;
        SET sequence = SUBSTR(sequence, INSTR(sequence,word)+word_length);
    END WHILE; 

    RETURN counter;
END //
DELIMITER ;

可以通过以下方式执行：

SELECT sum(subStringCount(fieldName,'subString')) FROM  `table` WHERE 1

Answer 3

存储上述程序后，这样可以计算表格中的字符串频率...

SELECT sum(subStringCount(fieldName,'subString')) FROM  `table` WHERE 1

Mysql计算子串的实例，然后按顺序排序

3 个答案: