一个迁移项目遇到的,MySQL的instr函数只能查找子串是否在父串中,没法按照出现的次数进行查找。
这里我自己写了一个,以便迁移。当然我这里仅仅针对的是迁移,可能没有完全实现原有函数的细节。
Oracle 里用了几次如下的调用,
SQL> select instr(‘This is belong to you, but not to me.‘,‘to‘,1,1) as pos from dual; POS -------------------- 16 已用时间: 00: 00: 00.00 SQL> select instr(‘This is belong to you, but not to me.‘,‘to‘,1,2) as pos from dual; POS -------------------- 32 已用时间: 00: 00: 00.00 SQL> select instr(‘This is belong to you, but not to me.‘,‘belong‘,-1,1) as pos from dual; POS -------------------- 9 已用时间: 00: 00: 00.00 SQL> select instr(‘This is belong to you, but not to me.‘,‘belong‘,-1,2) as pos from dual; POS -------------------- 0 已用时间: 00: 00: 00.00
MySQL里效果如下,
mysql> select func_instr_oracle(‘This is belong to you, but not to me.‘,‘to‘,1,1) as pos; +------+ | pos | +------+ | 16 | +------+ 1 row in set (0.00 sec) mysql> select func_instr_oracle(‘This is belong to you, but not to me.‘,‘to‘,1,2) as pos; +------+ | pos | +------+ | 32 | +------+ 1 row in set (0.00 sec) mysql> select func_instr_oracle(‘This is belong to you, but not to me.‘,‘belong‘,-1,1) as pos; +------+ | pos | +------+ | 9 | +------+ 1 row in set (0.00 sec) mysql> select func_instr_oracle(‘This is belong to you, but not to me.‘,‘belong‘,-1,2) as pos; +------+ | pos | +------+ | 0 | +------+ 1 row in set (0.00 sec)
附上函数func_instr_oracle的代码:
DELIMITER $$ USE `oracle12c`$$ DROP FUNCTION IF EXISTS `func_instr_oracle`$$ CREATE DEFINER=`root`@`localhost` FUNCTION `func_instr_oracle`( f_str VARCHAR(1000), -- Parameter 1 f_substr VARCHAR(100), -- Parameter 2 f_str_pos INT, -- Postion f_count INT UNSIGNED -- Times ) RETURNS INT(10) UNSIGNED BEGIN -- Created by ytt. Simulating Oracle instr function. -- Date 2015/12/5. DECLARE i INT DEFAULT 0; -- Postion iterator DECLARE j INT DEFAULT 0; -- Times compare. DECLARE v_substr_len INT UNSIGNED DEFAULT 0; -- Length for Parameter 1. DECLARE v_str_len INT UNSIGNED DEFAULT 0; -- Length for Parameter 2. SET v_str_len = LENGTH(f_str); SET v_substr_len = LENGTH(f_substr); -- Unsigned. IF f_str_pos > 0 THEN SET i = f_str_pos; SET j = 0; WHILE i <= v_str_len DO IF INSTR(LEFT(SUBSTR(f_str,i),v_substr_len),f_substr) > 0 THEN SET j = j + 1; IF j = f_count THEN RETURN i; END IF; END IF; SET i = i + 1; END WHILE; -- Signed. ELSEIF f_str_pos <0 THEN SET i = v_str_len + f_str_pos+1; SET j = 0; WHILE i <= v_str_len AND i > 0 DO IF INSTR(RIGHT(SUBSTR(f_str,1,i),v_substr_len),f_substr) > 0 THEN SET j = j + 1; IF j = f_count THEN RETURN i - v_substr_len + 1; END IF; END IF; SET i = i - 1; END WHILE; -- Equal to 0. ELSE RETURN 0; END IF; RETURN 0; END$$ DELIMITER ;
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原文地址:http://yueliangdao0608.blog.51cto.com/397025/1719867