无法从实体 Spring 引导生成表

Question

Please help:!! I can't generated table from entities**:

// class Application
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.autoconfigure.domain.EntityScan;

@EntityScan("sb.entity")
@SpringBootApplication
public class Application {
    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }
}

// class Entity
package sb.entity;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Table;

@Entity
@Table(name = "category")
public class CategoryEntity {
    
    @Column(name = "name")
    private String name;
    
    @Column(name = "code")
    private String code;

}

// application.properties
server.port=8081

spring.datasource.url = jdbc:mysql://localhost:3306/springboot_database
spring.datasource.username = root
spring.datasource.password = hoa123456
spring.jpa.hibernate.ddl-auto = create-drop 

spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.MySQL5Dialect

// file pom.xml
<project xmlns="http://maven.apache.org/POM/4.0.0"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <groupId>sb</groupId>
    <artifactId>sb</artifactId>
    <version>1.0</version>

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>1.5.9.RELEASE</version>
    </parent>

    <properties>
        <mysql.version>8.0.13</mysql.version>
        <java.version>1.8</java.version>
    </properties>

    <dependencies>

        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>

        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-data-jpa</artifactId>
        </dependency>

    
        <dependency>
            <groupId>mysql</groupId>
            <artifactId>mysql-connector-java</artifactId>
            <version>${mysql.version}</version>
        </dependency>

        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-core</artifactId>
            <version>4.1.4.Final</version>
        </dependency>
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-entitymanager</artifactId>
            <version>5.2.3.Final</version>
        </dependency>

    </dependencies>

    <build>
        <plugins>
            <plugin>
                <groupId>org.springframework.boot</groupId>
                <artifactId>spring-boot-maven-plugin</artifactId>
            </plugin>
        </plugins>
    </build>

</project>

This is console: I seach on overstack but i just add spring-boot-starter-data-jpa already.

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'jpaMappingContext': Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: At least one JPA metamodel must be present. at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory:java.1628) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory:java.555) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory:java.483) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory:java.306) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.facto ry.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry:java.230) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory:java.302) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory:java.197) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.DefaultListableBeanFactory.preInstantiateSingletons(DefaultListableBeanFactory:java.742) ~[spring-beans-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationConte xt:java.867) ~[spring-context-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext:java.543) ~[spring-context-4.3.13.RELEASE:jar.4.3.13.RELEASE] at org.springframework.boot.context.embedded.EmbeddedWebApplicationContext.refresh(EmbeddedWebApplicationContext:java.122) ~[spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at org.springframework.boot.SpringApplication.refresh(SpringApplication:java.693) [spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at org.springframework.boot.SpringApplication.refreshContext(SpringApplication:java.360) [spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at org.springframework.boot.Spring Application.run(SpringApplication:java.303) [spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at org.springframework.boot.SpringApplication.run(SpringApplication:java.1118) [spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at org.springframework.boot.SpringApplication.run(SpringApplication:java.1107) [spring-boot-1.5.9.RELEASE:jar.1.5.9.RELEASE] at sb.api.Application.main(Application:java:11) [classes/:na] Caused by. java.lang:IllegalArgumentException. At least one JPA metamodel must be present. at org.springframework.util.Assert:notEmpty(Assert.java.277) ~[spring-core-4.3.13:RELEASE.jar.4.3.13.RELEASE] at org.springframework.data.jpa.mapping.JpaMetamodelMappingContext:(JpaMetamodelMappingContext.java.52) ~[spring-data-jpa-1.11.9:RELEASE.jar.na] at org.springframework.data.jpa.repository.config.JpaMetamodelMappingContextFactoryBean:createInstance(JpaMetamodelMappingContextFactoryBean.java.71) ~[spring-data-jpa-1.11.9:RELEASE.jar.na] at org.springframework.data.jpa.repository.config.JpaMetamodelMappingContextFactoryBean:createInstance(JpaMetamodelMappingContextFactoryBean.java.26) ~[spring-data-jpa-1.11.9:RELEASE.jar.na] at org.springframework.bea ns.factory.config.AbstractFactoryBean:afterPropertiesSet(AbstractFactoryBean.java.134) ~[spring-beans-4.3.13:RELEASE.jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory:invokeInitMethods(AbstractAutowireCapableBeanFactory.java.1687) ~[spring-beans-4.3.13:RELEASE.jar.4.3.13.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory:initializeBean(AbstractAutowireCapableBeanFactory.java.1624) ~[spring-beans-4.3.13:RELEASE.jar.4.3.13.RELEASE]... 16 common frames omitted

enter image description here **

Answer 1

Please ensure you have that you have the Application class package in correctly placed under package. For example if you have your Application.java( which has main) under com.sb, then Sb.java entity class should be under com.sb package.

Another example

com.sb --> Application.java

com.sb.enity --> CategoryEntity.java

com.notscanned.test --> NotScannedCategoryEntity.java

The hierarchy matters, anything under com.sb.* will be autoscanned by spring.

Hope this solves your problem.

Answer 2

When you create an Entity, you need to provide @Id annotation. This is mandatory. I am assuming you are getting error because of the same. Can you try creating entity as

package sb.entity;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Table;

@Entity
@Table(name = "category")
public class CategoryEntity {
    
@Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    Long id;
    @Column(name = "name")
    private String name;
    
    @Column(name = "code")
    private String code;

}