To create the table, you need to insert the following lines into application.properties -
spring.jpa.hibernate.ddl-auto=update
Here, Hibernate will automatically create the table demo88. The application.properties code is as follows-
spring.datasource.platform=mysql spring.jpa.database-platform=org.hibernate.dialect.MySQL5InnoDBDialect spring.jpa.hibernate.ddl-auto=update server.port=8191 spring.datasource.driverClassName=com.mysql.jdbc.Driver spring.datasource.url=jdbc:mysql://localhost:3306/sampledatabase spring.datasource.username=root spring.datasource.password=123456
demo88 entity class creation table column is as follows-
package com.automaticallytablecreation; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.Id; import javax.persistence.Table; @Entity @Table public class demo88 { @Id private int id; @Column(name="name") private String name; }
The main class code is as follows-
package com.automaticallytablecreation; import org.springframework.boot.SpringApplication; import org.springframework.boot.autoconfigure.SpringBootApplication; @SpringBootApplication public class AutomaticTableApplication { public static void main(String[] args) { SpringApplication.run(AutomaticTableApplication.class, args); } }
Now right click on the main class and click on the "Run as Java Application" menu. The application will run and create the table.
This will produce the following output-
For verification, let us display the table-
mysql> show create table demo88;
This will produce the following output-
+--------+------------------------------------------------------------+| Table | Create Table | +--------+------------------------------------------------------------+| demo88 | CREATE TABLE `demo88` ( `id` int NOT NULL, `name` varchar(255) DEFAULT NULL, PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci | +--------+-------------------------------------------------------------+1 row in set (0.03 sec)
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