springboot 之jpa高级查询操作

springboot的jpa可以根据方法名自动解析sql 非常方便, 只需要在 dao接口中定义方法即可;

下面是一个 demo

package com.bus365.root.dao;
import java.io.Serializable;
import java.util.List;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.JpaSpecificationExecutor;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.query.Param;
import com.bus365.root.model.User;
public interface UserDao extends JpaRepository<User,Long>,JpaSpecificationExecutor<User>,Serializable {
 User findByName(String name);
 User findByNameAndAge(String name, Integer age);
 User findByNameOrAge(String name, Integer age); 

 /*@Query(value = "from User where name = :name")
 List<User> findbyname(@Param("name") String name);*/
}

下面展示service层调用:

@Override
 public User findByName(String name) {
 User user = userDao.findByName(name);
 return user;
 }

 @Override
 public User findByNameAndAge(String name, Integer age) {
 User user = userDao.findByNameAndAge(name,age);
 return user;
 }

 @Override
 public User findByNameOrAge(String name, Integer age) {
 User user = userDao.findByNameOrAge(name,age);
 return user;
 }

具体的关键字，使用方法和生产成SQL如下表所示

下面介绍使用java原生的jpa操作数据库,对jpa熟悉的朋友应该很快就能理解,springboot使用原生jpa的关键是引入entitymanger

看一下service层

package com.bus365.root.service.impl;
import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import org.springframework.stereotype.Service;
import com.bus365.root.model.Address;
import com.bus365.root.service.AddressService;

@Service
public class AddressServiceImpl implements AddressService {

	@PersistenceContext
	private EntityManager entityManager;

	public List<Address> listAddress(){
		List resultList = entityManager.createNativeQuery("select * from address ", Address.class).getResultList();
		return resultList;
	}
}

注意 @PersistenceContext

private EntityManager entityManager;

动态引入entitymanger , 之后就能正常使用了;

createNativeQuery是操作原生mysql方法;支持跨表查询;

jpa的事务 直接使用注解Transactional 参数rollbackon表示回滚条件, 这个注解一搬加在service层; 注意getSingleResult 如果查不到数据会报错;

@Transactional(rollbackOn= {Exception.class})
 public Address getAddressByid(Long id) {
 Address singleResult = null;
 try {
  singleResult = (Address) entityManager
   .createNativeQuery("select * from address where id = :id", Address.class).setParameter("id", id)
   .getSingleResult();
 } catch (Exception e) {
  e.printStackTrace();
 }
 return singleResult;
 }

jpa实现多表联查;

@Transactional
 public List<Object[]> getUserWithAddrByid(Long id) {
 List resultList = entityManager.createNativeQuery(
  "select u.id id,u.age age,u.name name,a.name aname,a.completeaddress addre from user u left join address a on u.addressid = a.id where u.id = :id")
  .setParameter("id", id).getResultList();
 return resultList;
 }

这是一个联查user 和address的例子, 返回的结果是个List<Object[]> 项目中一般封装成vo 类,或者List<Map<String,Object>> 的形式

github项目地址 https://github.com/Christain1993/SpringBootIntegration

补充：springBootJpa的复杂查询

分页

/**
   * 条件查询+分页
   * @param whereMap
   * @param page
   * @param size
   * @return
   */
  public Page<CaseManage> findSearch(Map whereMap, int page, int size,Integer createId) {
    Sort sort = new Sort(Sort.Direction.DESC,"id");
    Specification<CaseManage> specification = createSpecification(whereMap,createId);
    PageRequest pageRequest = new PageRequest(page,size,sort);
    return caseDao.findAll(specification, pageRequest);
  }

  /**
   * 条件查询
   * @param whereMap
   * @return
   */
  public List<CaseManage> findSearch(Map whereMap,Integer createId) {
    Specification<CaseManage> specification = createSpecification(whereMap, createId);
    return caseDao.findAll(specification);
  }

  /**
   * 动态条件构建
   * @param searchMap
   * @return
   */
  private Specification<CaseManage> createSpecification(Map searchMap,Integer createId) {
    return new Specification<CaseManage>() {
      @Override
      public Predicate toPredicate(Root<CaseManage> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
        List<Predicate> predicateList = new ArrayList<Predicate>();
        // 案件名称
        if (searchMap.get("case_name")!=null && !"".equals(searchMap.get("case_name"))) {
          predicateList.add(cb.like(root.get("case_name").as(String.class), "%"+(String)searchMap.get("case_name")+"%"));
        }
        // 案件编号uuid类型
        if (searchMap.get("case_uuid")!=null && !"".equals(searchMap.get("case_uuid"))) {
          predicateList.add(cb.equal(root.get("case_uuid").as(String.class), (String)searchMap.get("case_uuid")));
        }
        return cb.and( predicateList.toArray(new Predicate[predicateList.size()]));
      }
    };
  }

or查询

想实现这样的效果

where （state=1 or state=2）and name='zhangsan'

java代码

List<Predicate> predicateList = new ArrayList<Predicate>();
Predicate or = cb.or(cb.and(cb.equal(root.get("case_authority").as(String.class), "0")), cb.and(cb.equal(root.get("create_id").as(String.class), String.valueOf(createId))));
predicateList.add(or);

以上为个人经验，希望能给大家一个参考，也希望大家多多支持我们。如有错误或未考虑完全的地方，望不吝赐教。