将 MySQL LIKE 实现为 MySQL IN 的查询?
mysqlmysqli database
要实现类似 MySQL IN() 的查询，您需要使用 COUNT()、IF() 以及 LIKE 运算符。让我们首先创建一个表 −
mysql> create table DemoTable -> ( -> Subject varchar(80) -> ); Query OK, 0 rows affected (0.58 sec)
使用 insert 命令在表中插入一些记录 −
mysql> insert into DemoTable values('MySQLMongoDB');
Query OK, 1 row affected (0.86 sec)
mysql> insert into DemoTable values('MySQL');
Query OK, 1 row affected (0.12 sec)
mysql> insert into DemoTable values('JavaMySQL');
Query OK, 1 row affected (0.18 sec)
mysql> insert into DemoTable values('MongoDB');
Query OK, 1 row affected (0.20 sec)
mysql> insert into DemoTable values('Java');
Query OK, 1 row affected (0.19 sec)
使用 select 语句显示表中的所有记录 −
mysql> select *from DemoTable;
这将产生以下输出 −
+--------------+ | Subject | +--------------+ | MySQLMongoDB | | MySQL | | JavaMySQL | | MongoDB | | Java | +--------------+ 5 rows in set (0.00 sec)
以下是将 LIKE 实现为 IN() 的查询 −
mysql> select -> count(if(Subject like '%MySQL%',1,NULL)) as MySQLCount, -> count(if(Subject like '%Java%',1,NULL)) as JavaCount, -> count(if(Subject like '%MongoDB%',1,NULL)) as MongoDBCount -> from DemoTable;
这将产生以下输出 −
+------------+-----------+--------------+ | MySQLCount | JavaCount | MongoDBCount | +------------+-----------+--------------+ | 3 | 2 | 2 | +------------+-----------+--------------+ 1 row in set (0.00 sec)
