创建表时设置表名“references”时,将出现错误


你无法给表名引用,因为它是一个保留关键字。用反引号将它包起来,例如,`references`。

让我们先创建一个表格 -

mysql> create table `references`(Subject text);
Query OK, 0 rows affected (0.44 sec)

使用 insert 命令在表中插入一些记录 -

mysql> insert into `references` values('Introduction To MySQL');
Query OK, 1 row affected (0.28 sec)
mysql> insert into `references` values('Introduction To MongoDB');
Query OK, 1 row affected (0.15 sec)
mysql> insert into `references` values('Introduction To Spring and Hibernate');
Query OK, 1 row affected (0.13 sec)
mysql> insert into `references` values('Introduction To Java');
Query OK, 1 row affected (0.18 sec)

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)

更新于: 2019-08-21

107 浏览

开启你的职业生涯

完成课程以获得认证

开始
广告
© . All rights reserved.