java.sql.SQLException 无效的对象名称

Question

当我尝试从数据库中读取时出现以下错误，createQuery但如果我createNativeQuery在我的entityManager. 所以我想我的entityManager作品但我的 Hibernate 配置有问题？

应用程序属性：

security.user.password=<password>
logging.level.org.springframework.security=DEBUG
spring.datasource.url=jdbc:jtds:sqlserver://localhost:60830;databaseName=<db>
spring.datasource.username=<user>
spring.datasource.password=<password>
spring.datasource.driver-class-name=net.sourceforge.jtds.jdbc.Driver
spring.jpa.show-sql=true
spring.jpa.hibernate.ddl-auto=none
spring.jpa.properties.hibernate.id.new_generator_mappings=false
spring.jpa.properties.hibernate.format_sql=true
spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.SQLServerDialect
spring.jpa.database-platform=org.hibernate.dialect.SQLServerDialect

道：

package demo.dao;

import demo.entities.StampCard;
import org.springframework.stereotype.Repository;
import org.springframework.transaction.annotation.Transactional;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;

@Transactional
@Repository
public class StampCardDAO implements IStampCardDAO {

    @PersistenceContext
    private EntityManager entityManager;

    @Override
    public StampCard getStampCard(int id) {
        //These rows works fine
        //Query query = entityManager.createNativeQuery("select userId from StampCard where StampCardID = '1'");
        //String userId = (String) query.getSingleResult();
        Query query = entityManager.createQuery("SELECT s FROM StampCard s WHERE s.stampCardID = '1'");
        StampCard stampCard = (StampCard) query.getSingleResult();
        return stampCard;
    }
}

DAO 接口：

package demo.dao;

import demo.entities.StampCard;

public interface IStampCardDAO {

    StampCard getStampCard(int id);
}

实体：

package demo.entities;

import lombok.Getter;
import lombok.Setter;
import org.apache.commons.lang3.builder.EqualsBuilder;
import org.apache.commons.lang3.builder.HashCodeBuilder;

import javax.persistence.Entity;
import javax.persistence.Id;
import java.io.Serializable;

@Entity
public class StampCard implements Serializable {

    private static final long serialVersionUID = 6602888822739626415L;

    private int stampCardID;
    @Getter @Setter private String createdDate;
    @Getter @Setter private String userID;
    @Getter @Setter private int numberOfStamps;

    @Id
    public int getStampCardID() {
        return stampCardID;
    }

    public void setStampCardID(int stampCardID) {
        this.stampCardID = stampCardID;
    }

    @Override 
    public boolean equals(Object obj) {
        if (obj == null) {
            return false;
        }
        if (obj == this) {
            return true;
        }
        if (obj.getClass() != getClass()) {
            return false;
        }
        StampCard rhs = (StampCard) obj;
        return new EqualsBuilder().append(this.stampCardID, rhs.stampCardID).append(this.createdDate, rhs.createdDate).append(this.userID, rhs.userID)
                .append(this.numberOfStamps, rhs.numberOfStamps).isEquals();
    }

    @Override 
    public int hashCode() {
        return new HashCodeBuilder().append(stampCardID).append(createdDate).append(userID).append(numberOfStamps).toHashCode();
    }
}

数据库表：

CREATE TABLE [dbo].[StampCard](
    [StampCardID] [int] NOT NULL,
    [CreatedDate] [varchar](8) NOT NULL,
    [UserID] [varchar](255) NOT NULL,
    [NumberOfStamps] [int] NOT NULL,
 CONSTRAINT [PK_CardID] PRIMARY KEY CLUSTERED 
(
    [StampCardID] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]

数据库表内容：

StampCardID CreatedDate UserID  NumberOfStamps
1           20171208    USer    1

错误信息：

2017-12-10 18:04:09.837 WARN 37080 --- [restartedMain] ohengine.jdbc.spi.SqlExceptionHelper：SQL 错误：208，SQLState：S0002 2017-12-10 18:04:09.8700 [ERROR] restartedMain] ohengine.jdbc.spi.SqlExceptionHelper：对象名称“stamp_card”无效。

引起：java.sql.SQLException：无效的对象名称“stamp_card”。

我很感激我能得到的所有帮助。

Answer 1

@Entity 使用 StampCard 类名，hibernate 搜索具有名称的表 Stamp_Card

因此，如果您的表名只是StampCard使用如下表注释来指向您的表名。

@Entity
@Table(name = "STAMPCARD")