JPA查詢--使用原生sql 並且把查詢結果轉為實體對象

這篇文章主要記錄 使用原生sql查詢 並且把查詢結果轉為實體對象, 注意：這兒使用的數據庫 是oracle數據庫

這兒記錄了三種查詢：精確查詢，模糊查詢，分頁查詢。

1.把原生sql查詢的結果轉為實體類對象用的是 Query query = em.createNativeQuery(sql, Student.class);

 2.實體類 要使用@Entity 和 @Id 注解，否則會報錯

新建實體類

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. 精確查詢   where t.riskunit_id = :riskunit_id

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 ";
		// 執行查詢 並把結果專為實體類
		Query query = em.createNativeQuery(sql, Student.class);
		// 參數值
		query.setParameter("class_no", 400002);
		// 獲取查詢結果
		List<Student> students = query.getResultList();
		System.out.println("students===" + students.toString());
	}
}

2. 模糊查詢    where t.riskunit_id like '%'||:riskunit_id||'%' 

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||'%' ";
		// 執行查詢 並把結果專為實體類
		Query query = em.createNativeQuery(sql, Student.class);
		// 參數值
		query.setParameter("class_no", 400002);
		// 獲取查詢結果
		List<Student> students = query.getResultList();
		System.out.println("students===" + students.toString());
	}
}

3. 原生sql查詢並分頁  這里參考oracle sql 3.oracle sql 分頁實現

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. 有幾種可能hibernate或者sql的報錯：

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

 查詢sql的字段名與數據庫關鍵字沖突了，仔細查看sql 是因為where t.like 寫錯了

2.javax.persistence.PersistenceException: org.hibernate.MappingException: Unknown entity: com.mx.JpaDemo.entity.RiskUnit
需要在實體類RiskUnit上加上@Entity注解 

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

需要在實體類加上@Id注解