Question

我通过Spring Data JPA使用Hibernate并尝试添加计算字段。一个简单的SELECT 1*1查询可以工作，但是当我添加真正的公式时，Hibernate会完全混淆并生成一个语法无效的查询。

父表：

@Entity
@Table(name = "szallitolevel")
public class Szallitolevel {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;

    @Min(1)
    private Long szam;

    @ManyToOne(cascade = CascadeType.REFRESH, fetch = FetchType.EAGER)
    @Fetch(FetchMode.JOIN)
    private Partner partner;

    @Formula("(select sum(xx.mennyiseg) from szallitolevel_sor xx where xx.szallitolevel = id)")
//    @Formula("(select 1*1)")
    private Long sumMennyiseg;

    @OneToMany(mappedBy="szallitolevel", cascade = CascadeType.ALL, fetch = FetchType.EAGER, orphanRemoval=true)
    @Fetch(FetchMode.SUBSELECT)
    @Valid
    private List<SzallitolevelSor> sorok = new AutoPopulatingList<SzallitolevelSor>(SzallitolevelSor.class);
}

子表：

@Entity
@Table(name = "szallitolevel_sor")
public class SzallitolevelSor {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;

    private String nev;

    @Min(0)
    private Long mennyiseg;

    @ManyToOne
    private Szallitolevel szallitolevel;
}

生成的查询中甚至没有sum（），并且公式以某种方式找到了WHERE部分：

select 
  sorok0_.szallitolevel as szallito4_2_1_, 
  sorok0_.id as id1_3_1_, 
  sorok0_.id as id1_3_0_,
  sorok0_.mennyiseg as mennyise2_3_0_, 
  sorok0_.nev as nev3_3_0_, 
  sorok0_.szallitolevel as szallito4_3_0_ 
from szallitolevel_sor sorok0_ 
where sorok0_.szallitolevel 
  in (select szallitole0_.id 
      from szallitolevel_sor xx 
      where xx.szallitolevel = szallitole0_.id) as formula0_0_,
        partner1_.nev as nev2_1_1_, 
        partner1_.penz as penz3_1_1_ 
from szallitolevel szallitole0_ 
  left outer join partner partner1_ 
    on szallitole0_.partner=partner1_.id) 

 {FAILED after 0 msec}
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'as formula0_0_, partner1_.nev as nev2_1_1_, partner1_.penz as penz3_1_1_ from sz' at line 1

更新：我正在使用JDBCTemplate进行相关查询，请参阅答案

Answer 1

所以最后我决定尝试将JPA保留为简单的CRUD并允许JDBC转义路径，如果需要超过10分钟才能使JPA工作。使用JDBC解决上述问题需要执行以下步骤：

在其中一个配置类中创建JDBCTemplate bean：

@Bean
public NamedParameterJdbcTemplate jdbcTemplate(DataSource dataSource) {
    return new NamedParameterJdbcTemplate(dataSource);
}

设置计算的实体字段瞬态，以便Hibernate忽略它。还要确保有一个吸气剂和一个二传手：

@Entity
@Table(name = "szallitolevel")
public class Szallitolevel {
    ...
    @Transient
    private Long sumMennyiseg = 0L;
    ...
}

自定义JPA存储库（您还可以通过向findAll提供自定义查询来查看此处针对Hibernate N + 1问题的修复）：

public interface SzallitolevelRepoCustom {
    List<Szallitolevel> customFindAll();
}


public interface SzallitolevelRepo extends CrudRepository<Szallitolevel, Long>, SzallitolevelRepoCustom {
    @Query("select s from Szallitolevel s left join fetch s.partner p")
    List<Szallitolevel> findAll();
}


public class SzallitolevelRepoImpl implements SzallitolevelRepoCustom {
    private static final String FIND_ALL = "SELECT sz.id, sum(sor.mennyiseg) as sumMennyiseg, partner.nev as 'partner.nev' "
            + "from szallitolevel sz "
            + "left join szallitolevel_sor sor on sor.szallitolevel = sz.id "
            + "left join partner on partner.id = sz.partner "
            + "group by sz.id ";

    @Autowired
    private NamedParameterJdbcTemplate jdbcTemplate;

    @Override
    public List<Szallitolevel> customFindAll() {
        List<Szallitolevel> result = jdbcTemplate.query(FIND_ALL, new NestedRowMapper<>(Szallitolevel.class));
        return result;
    }
}

注意SQL中的别名partner.nev。行映射器将使用它来创建一个空的Partner对象，并仅设置nev属性。（如果您需要填充整个对象并且不介意另一个数据库查询，您还可以编写一个Spring转换器，从Long到Partner，从数据库中检索整个对象。）

从here偷走NestedRowMapper。

如何混合JPA和JDBCTemplate因为Hibernate @Formula错误

1 个答案: