创建表时设置表名“引用”时发生错误
您不能提供表名称引用,因为它是保留关键字。使用反引号将其包装,例如“引用”。
让我们首先创建一个表-
mysql> create table `references`(Subject text);
使用插入命令在表中插入一些记录-
mysql> insert into `references` values('Introduction To MySQL'); mysql> insert into `references` values('Introduction To MongoDB'); mysql> insert into `references` values('Introduction To Spring and Hibernate'); mysql> insert into `references` values('Introduction To Java');
使用select语句显示表中的所有记录-
mysql> select *from `references`;
这将产生以下输出-
+--------------------------------------+ | Subject | +--------------------------------------+ | Introduction To MySQL | | Introduction To MongoDB | | Introduction To Spring and Hibernate | | Introduction To Java | +--------------------------------------+ 4 rows in set (0.00 sec)