EntityManager执行原生sql结果集与bean的映射

Posted 2023-04-29

createNativeQuery(sql,XXX.class)，如果XXX没有建立和表的关系的话，执行会报错。
但是要查询多张表的数据，所以想建这么一个XXX，存放多张表的字段，也要配置和数据库的关系才能自动把值注入到XXX的对象去么？
求代码。不要那种笨方法哦，我知道可以用object[]来获取，但是要把object[]一个个的赋值给XXX的属性，很麻烦。。

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JPA EntityManager查询--使用原生sql 并且把查询结果转为实体对象

这篇文章主要记录 使用原生sql查询 并且把查询结果转为实体对象, 注意：这儿使用的数据库 是oracle数据库

这儿记录了三种查询：精确查询，模糊查询，分页查询。

1.把原生sql查询的结果转为实体类对象用的是 Query query = em.createNativeQuery(sql, Student.class);

2.实体类 要使用@Entity 和 @Id 注解，否则会报错

新建实体类

import javax.persistence.Entity;
import javax.persistence.Id;
 
@Entity
public class Student 
    @Id
    private Long classNo;
    private Long studentNo;
    private String remark;
    //省略了getter setter方法
    @Override
    public String toString() 
        return "RiskUnit [classNo=" + classNo + ", studentNo=" + studentNo + ", remark=" + remark + "]";
       

1. 精确查询 where t.class_no = :class_no

import java.util.List;
 
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
  
public class JpaDemoTest 
    @PersistenceContext
    private EntityManager em;
 
    public void testJpaNativeQuery() 
 
        // sql
        String sql = "select t.class_no  classNo ,t.student_no studentNo ,t.remark remark  from t_student t where t.class_no = :class_no ";
        // 执行查询 并把结果专为实体类
        Query query = em.createNativeQuery(sql, Student.class);
        // 参数值
        query.setParameter("class_no", 400002);
        // 获取查询结果
        List<Student> students = query.getResultList();
        System.out.println("students===" + students.toString());
    

2. 模糊查询    where t.class_no like '%'||:class_no||'%' "

import java.util.List;
 
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
  
public class JpaDemoTest 
    @PersistenceContext
    private EntityManager em;
 
    public void testJpaNativeQuery() 
 
        // sql
        String sql = "select t.class_no  classNo ,t.student_no studentNo ,t.remark remark  from t_student t where t.class_no like '%'||:class_no||'%' ";
        // 执行查询 并把结果专为实体类
        Query query = em.createNativeQuery(sql, Student.class);
        // 参数值
        query.setParameter("class_no", 400002);
        // 获取查询结果
        List<Student> students = query.getResultList();
        System.out.println("students===" + students.toString());
    

3. 原生sql查询并分页  这里参考oracle sql 3.oracle sql 分页实现

import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
 
public class JpaDemoTest 
    @PersistenceContext
    private EntityManager em;
 
    public void testJpaNativeQuery() 
        Integer pageNo = 2;
        Integer pageSize = 10;
        String sql = "select t.class_no  classNo ,t.student_no studentNo ,t.remark remark  from t_student t where t.class_no like '%' || :class_no || '%' order by t.class_no ";
        sql = " select * from ( select a.*, rownum as a_rownum from (" + sql + ") a where rownum <= "
                + (pageNo + 1) * pageSize + ") b where b.a_rownum > " + pageNo * pageSize;
        Query query = em.createNativeQuery(sql, Student.class);
        query.setParameter("class_no", 40000);
        List<Student> students = query.getResultList();
        System.out.println("student===" + students.toString());
    

4. 有几种可能hibernate或者sql的报错：

1.Caused by: java.sql.SQLSyntaxErrorException: ORA-01747: invalid user.table.column, table.column, or column specification

 查询sql的字段名与数据库关键字冲突了，仔细查看sql 是因为where t.like 写错了

2.javax.persistence.PersistenceException: org.hibernate.MappingException: Unknown entity: com.mx.JpaDemo.entity.RiskUnit
需要在实体类RiskUnit上加上@Entity注解 

3.Caused by: org.hibernate.AnnotationException: No identifier specified for entity: com.mx.JpaDemo.entity.RiskUnit

需要在实体类加上@Id注解