Hibernate不使用JPA实体创建PostgreSQL模式对象

问题描述 投票：0回答：1

我正在尝试使用JPA实体来创建数据库架构。但是，hibernate并未创建尝试在其下创建表的指定架构。以下是我的代码

package io.spring.models;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "brands", schema = "production")
public class Brand implements Serializable {

    private static final long serialVersionUID = -7234173905789565680L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "brand_id")
    private Integer brandId;

    @Column(name = "brandName")
    private String brandName;


    public Integer getBrandId() {
        return brandId;
    }

    public void setBrandId(Integer brandId) {
        this.brandId = brandId;
    }

    public String getBrandName() {
        return brandName;
    }

    public void setBrandName(String brandName) {
        this.brandName = brandName;
    }

    @Override
    public String toString() {
        return "Brand [brandId=" + brandId + ", brandName=" + brandName + "]";
    }

    public Brand() {
        super();
    }

}

我收到以下错误：

org.hibernate.tool.schema.spi.CommandAcceptanceException: Error executing DDL "create table production.brands (brand_id int4 not null, brand_name varchar(255), primary key (brand_id))" via JDBC Statement
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:67) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlString(SchemaCreatorImpl.java:439) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlStrings(SchemaCreatorImpl.java:423) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.createFromMetadata(SchemaCreatorImpl.java:314) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.performCreation(SchemaCreatorImpl.java:166) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:135) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:121) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:156) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:73) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:314) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:468) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1249) ~[hibernate-core-5.4.15.Final.jar:5.4.15.Final]
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:58) ~[spring-orm-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365) ~[spring-orm-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:391) ~[spring-orm-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:378) ~[spring-orm-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:341) ~[spring-orm-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1855) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1792) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:595) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:517) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:323) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:226) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:321) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:202) ~[spring-beans-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:1108) ~[spring-context-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:868) ~[spring-context-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:550) ~[spring-context-5.2.6.RELEASE.jar:5.2.6.RELEASE]
    at org.springframework.boot.web.servlet.context.ServletWebServerApplicationContext.refresh(ServletWebServerApplicationContext.java:141) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:747) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:397) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:315) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1226) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1215) ~[spring-boot-2.2.7.RELEASE.jar:2.2.7.RELEASE]
    at io.spring.SpringDataJpaApplication.main(SpringDataJpaApplication.java:10) ~[classes/:na]
    at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0(Native Method) ~[na:na]
    at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) ~[na:na]
    at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) ~[na:na]
    at java.base/java.lang.reflect.Method.invoke(Method.java:567) ~[na:na]
    at org.springframework.boot.devtools.restart.RestartLauncher.run(RestartLauncher.java:49) ~[spring-boot-devtools-2.2.7.RELEASE.jar:2.2.7.RELEASE]
Caused by: org.postgresql.util.PSQLException: ERROR: schema "production" does not exist

我的application.properties文件

spring.jpa.database=POSTGRESQL
spring.datasource.platform=postgres
spring.datasource.url=jdbc:postgresql://localhost:5432/BikeStore
spring.datasource.username=****
spring.datasource.password=****
spring.jpa.show-sql=true
spring.jpa.generate-ddl=true
spring.jpa.hibernate.ddl-auto=create-drop
hibernate.validator.apply_to_ddl = true

有什么方法可以使用Hibernate创建PostgreSQL模式对象以及表。我们可以手动创建模式，但是有一种方法可以仅使用休眠模式来创建模式。有人可以帮助我哪里出错了或丢失了。预先感谢。

hibernate jpa spring-data hibernate-mapping postgresql-9.4
1个回答
1
投票

错误消息很清楚，您没有名为“ production”]]的架构。在PostgreSQL中，数据库中可以有多个模式。

因此，请在BikeStore

数据库中创建一个称为生产的架构，或者在schema = "production"批注中从JPA实体中删除@Table。

如果您希望Hibernate（JPA）创建架构，则可以将javax.persistence.create-database-schemas属性设置为true

。如果您使用的是Spring，请按以下步骤进行设置：
spring.jpa.properties.javax.persistence.create-database-schemas = true

请注意，并非所有的DB方言都支持此功能。 Postgres方言支持它。 （它使用默认的Hibernate方言创建架构）

有关更多信息，请参见Hibernate Documentation。

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