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java - Spring Data JPA 查询不起作用,列不存在

转载 作者:行者123 更新时间:2023-11-30 06:53:52 25 4
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我在我的网络应用程序中使用 spring data jpa,我有实体用户

@Entity
public class User implements Serializable {
private static final long serialVersionUID = 1L;

private Long id;
private String password;
private String email;
private Boolean enabled;
private String name;
private String lastname;
private String userRole;

public User() {
}

public User(String password, String email, Boolean enabled, String name, String lastname, String userRole) {
this.password = password;
this.email = email;
this.enabled = enabled;
this.name = name;
this.lastname = lastname;
this.userRole = userRole;
}

@Id
@GeneratedValue(strategy = GenerationType.AUTO, generator = "users_id_seq")
@SequenceGenerator(name="users_id_seq", sequenceName="users_id_seq", allocationSize = 1)
@Column(name = "id", nullable = false)
public Long getId() {
return id;
}

//Other columns
}

我有扩展 CrudRepository 的 UserRepository 接口(interface)。当我在我的 Controller 中调用方法 findAll 时,出现此错误

01-May-2016 22:45:58.674 WARN [http-nio-8080-exec-3] org.hibernate.engine.jdbc.spi.SqlExceptionHelper.logExceptions SQL Error: 0, SQLState: 42703
01-May-2016 22:45:58.675 ERROR [http-nio-8080-exec-3] org.hibernate.engine.jdbc.spi.SqlExceptionHelper.logExceptions Error: column user0_.id does not exist
Position: 8

我的spring-config.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:jpa="http://www.springframework.org/schema/data/jpa"

xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx.xsd http://www.springframework.org/schema/data/jpa http://www.springframework.org/schema/data/jpa/spring-jpa.xsd">

<jpa:repositories base-package="com.birthright.repository"/>

<bean id="myEmf" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource"/>
<property name="packagesToScan" value="com.birthright.entity"/>
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter"/>
</property>
<property name="jpaProperties">
<props>
<prop key="hibernate.dialect">org.hibernate.dialect.PostgreSQL9Dialect</prop>
<prop key="hibernate.hbm2ddl.auto">update</prop>
</props>
</property>
</bean>

<tx:annotation-driven/>

<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="org.postgresql.Driver"/>
<property name="url" value="jdbc:postgresql://localhost:5432/AutoService"/>
<property name="username" value="postgres"/>
<property name="password" value="root"/>
</bean>

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="myEmf"/>
</bean>

</beans>

我在 postgresql 中的表

CREATE TABLE public."user"
(
id integer NOT NULL DEFAULT nextval('users_id_seq'::regclass),
password character varying,
email character varying,
enabled boolean,
name character varying,
lastname character varying,
user_role character varying,
CONSTRAINT users_pkey PRIMARY KEY (id)
)

最佳答案

User 是 PostgreSQL 中的保留关键字。使用默认命名策略,您可能有 User 表名。

不知道为什么 @Table(name = "user", schema = "public") 有效。也许 PostgreSQL 不会将 public.user 视为与 User 相对的关键字。

请为表格使用复数名称。对表名使用系统或子系统前缀 (xxx_users) 也是一个好主意。

命名策略可以用于这种方法。引用这个例子:Hibernate5NamingStrategy

前缀示例: StrategyOptions

关于java - Spring Data JPA 查询不起作用,列不存在,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36971264/

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