verwende ich Federdaten JPA in meiner Web-app, ich Entität Benutzer habenSpring Data JPA-Abfrage funktioniert nicht, Spalte existiert nicht
@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
}
Und ich habe UserRepository Schnittstelle, die CrudRepository erstreckt. Wenn ich Methode findAll in meinem Controller-anrufe, bekomme ich diesen Fehler
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
Meine Feder-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>
Mein Tisch in postgresql
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)
)
Ich denke, die Spalte in der Tabelle vorhanden ist? –
Ja, und ich weiß nicht, wo das Präfix "user0_." ist genommen. – Birthright
Es ist nur der Alias, der von Hibernate verwendet wird, um Ihre Benutzertabelle zu benennen –