MySQL的FIND_IN_SET如何在YashanDB改写
前言
MySQL应用使用的SQL的Where条件存在FIND_IN_SET，无法在YashanDB直接执行，需要改写。本文探讨不兼容的原因，并给出改写手段。
问题
SQL示例
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create table t1(c1 int, c2 varchar(10)); insert into t1 values(1,'b'); insert into t1 values(2,'d'); select c1 from t1 where find_in_set(c2,'a,b,c'); |
MySQL执行成功
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mysql> select c1 from t1 where find_in_set(c2,'a,b,c'); +------+ | c1 | +------+ | 1 | +------+ 1 row in set (0.00 sec) |
YashanDB报错
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SQL> select c1 from t1 where find_in_set(c2,'a,b,c'); [1:25]YAS-04336 boolean expression expected |
原因
MySQL的Boolean类型实际上是tinyint类型的同义词，0值为false，非0值为true。这意味着在MySQL的数值类型可以条件运算
参考：https://dev.mysql.com/doc/refman/8.4/en/numeric-type-syntax.html
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mysql> select find_in_set('b','a,b,c') from dual; +--------------------------+ | find_in_set('b','a,b,c') | +--------------------------+ | 2 | +--------------------------+ 1 row in set (0.00 sec)
mysql> select find_in_set('d','a,b,c') from dual; +--------------------------+ | find_in_set('d','a,b,c') | +--------------------------+ | 0 | +--------------------------+ 1 row in set (0.00 sec)
mysql> select 1 from dual where 2; +---+ | 1 | +---+ | 1 | +---+ 1 row in set (0.00 sec)
mysql> select 1 from dual where 0; Empty set (0.00 sec) |
但是，YashanDB的tinyint类型不能参与条件运算
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SQL> select find_in_set('b','a,b,c') from dual; FIND_IN_SET('B','A,B -------------------- 2
1 row fetched. SQL> select find_in_set('d','a,b,c') from dual; FIND_IN_SET('D','A,B -------------------- 0
1 row fetched. SQL> select 1 from dual where 2; [1:26]YAS-04336 boolean expression expected SQL> select 1 from dual where 0; [1:26]YAS-04336 boolean expression expected |
改写方案
按照find_in_set的逻辑，可以对所有参与where条件的find_in_set进行文本替换，例如示例中的替换规则就是 where find_in_set( 替换 where 0 < find_in_set(
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SQL> select c1 from t1 where 0 < find_in_set(c2,'a,b,c'); C1 ------------ 1 1 row fetched. |
备注：一般而言，单个项目的源代码由于开发人员的习惯，对find_in_set的写法基本类似，所以往往可以用少数文本替换规则，就能对所有不兼容的find_in_set进行改写
