对 INT 字段执行 MySQL LIKE 比较?
mysqlmysqli database
您需要使用 cast() 方法对 INT 字段进行比较。语法如下 −
SELECT yourColumnName1,yourColumnName2,......N yourTableName WHERE CAST(yourColumnName as CHAR) LIKE ‘%yourIntegerValue%’;
为了理解上述语法，让我们创建一个表。以下是创建用于对 INT 字段执行 LIKE 比较的表的查询 −
mysql> create table ComparisonOnIntField -> ( -> StudentId int NOT NULL, -> StudentName varchar(20), -> StudentAge int -> ); Query OK, 0 rows affected (1.00 sec)
在表中插入一些记录，对 INT 字段执行 MySQL LIKE 比较。插入记录的查询如下 −
mysql> insert into ComparisonOnIntField values(10,'Carol',24); Query OK, 1 row affected (0.17 sec) mysql> insert into ComparisonOnIntField values(12,'Bob',21); Query OK, 1 row affected (0.18 sec) mysql> insert into ComparisonOnIntField values(14,'Sam',23); Query OK, 1 row affected (0.10 sec) mysql> insert into ComparisonOnIntField values(16,'Mike',25); Query OK, 1 row affected (0.26 sec) mysql> insert into ComparisonOnIntField values(18,'John',27); Query OK, 1 row affected (0.14 sec) mysql> insert into ComparisonOnIntField values(20,'David',26); Query OK, 1 row affected (0.15 sec)
使用 select 语句显示表中的所有记录。查询如下 −
mysql> select *from ComparisonOnIntField;
以下是输出。
+-----------+-------------+------------+ | StudentId | StudentName | StudentAge | +-----------+-------------+------------+ | 10 | Carol | 24 | | 12 | Bob | 21 | | 14 | Sam | 23 | | 16 | Mike | 25 | | 18 | John | 27 | | 20 | David | 26 | +-----------+-------------+------------+ 6 rows in set (0.00 sec)
这是对 INT 字段执行 MySQL LIKE 比较的查询 −
mysql> select StudentName,StudentAge from ComparisonOnIntField -> where cast(StudentId as CHAR) Like '%18%';
以下是输出 −
+-------------+------------+ | StudentName | StudentAge | +-------------+------------+ | John | 27 | +-------------+------------+ 1 row in set (0.05 sec)
