英文:
ManyToOne mapping not working when inside an Embeddable class
问题
我正在使用Spring Boot（2.2.6.RELEASE）和JPA开发一个项目，但我遇到一个问题让我夜不能寐...
我有一个弱实体，其主键是来自父表的复合主键组成的。
+-----------------+| BAR |+-----------------+| BAR_ID PK || NAME |+-----------------+|| +-----------------+| | FOO |+------<+-----------------++------<| BAR_ID PK FK || | BAZ_ID PK FK || +-----------------+|+-----------------+| BAZ |+-----------------+| BAZ_ID PK || NAME |+-----------------+
所以我有以下的映射...
@Entity@Table(name = "FOO")public class Foo implements Serializable {@Data@Embeddablepublic static class ID implements Serializable {@ManyToOne@JoinColumn(name="BAR_ID")private Bar bar;@ManyToOne@JoinColumn(name="BAZ_ID")private Baz baz;}@EmbeddedIdprivate ID id;}@Entity@Table(name = "BAR")@SequenceGenerator(name = "SEQ_BAR", sequenceName = "SEQ_BAR")public class Bar implements Serializable {@Id@GeneratedValue(strategy=GenerationType.SEQUENCE, generator = "SEQ_BAR")@Column(name = "BAR_ID")private Long id;@Column(name = "NAME")private String name;}@Entity@Table(name = "BAZ")@SequenceGenerator(name = "SEQ_BAZ", sequenceName = "SEQ_BAZ")public class Baz implements Serializable {@Id@GeneratedValue(strategy=GenerationType.SEQUENCE, generator = "SEQ_BAZ")@Column(name = "BAZ_ID")private Long id;@Column(name = "NAME")private String name;}
以及以下的Repository接口：
@Repositorypublic interface FooRepository extends JpaRepository<Foo, Foo.ID> {}
然后我尝试使用findById
fooRepository.findById(new Foo.ID(new Bar(1), new Baz(1)));
但出现了问题：
当我调用fooRepository.findById()时，尽管从数据库中检索到了行，但是Bar和Baz的ManyToOne映射却无法正常工作。这意味着无论我是执行 foo.getId().getBar().getName() 还是 foo.getId().getBaz().getName()，Bar和Baz对象的属性 name 都是null（当然，数据库中的属性不是null）。
所以我的问题是：
这应该能够工作，对吗？
我做错了什么？
我已经尝试将fetch类型更改为EAGER，但完全没有用。
我真的非常感谢任何回复。
英文:
I am working on a project with Spring Boot (2.2.6.RELEASE) and JPA, and I am facing an issue that is bringing me nightmares...
I have a weak entity whose PK is a composite key of primary keys from parent tables.
+-----------------+| BAR |+-----------------+| BAR_ID PK || NAME |+-----------------+|| +-----------------+| | FOO |+------<+-----------------++------<| BAR_ID PK FK || | BAZ_ID PK FK || +-----------------+|+-----------------+| BAZ |+-----------------+| BAZ_ID PK || NAME |+-----------------+
So I have the following mapping...
@Entity@Table(name = "FOO")public class Foo implements Serializable {@Data@Embeddablepublic static class ID implements Serializable {@ManyToOne@JoinColumn(name="BAR_ID")private Bar bar;@ManyToOne@JoinColumn(name="BAZ_ID")private Baz baz;}@EmbeddedIdprivate ID id;}@Entity@Table(name = "BAR")@SequenceGenerator(name = "SEQ_BAR", sequenceName = "SEQ_BAR")public class Bar implements Serializable {@Id@GeneratedValue(strategy=GenerationType.SEQUENCE, generator = "SEQ_BAR")@Column(name = "BAR_ID")private Long id;@Column(name = "NAME")private String name;}@Entity@Table(name = "BAZ")@SequenceGenerator(name = "SEQ_BAZ", sequenceName = "SEQ_BAZ")public class Baz implements Serializable {@Id@GeneratedValue(strategy=GenerationType.SEQUENCE, generator = "SEQ_BAZ")@Column(name = "BAZ_ID")private Long id;@Column(name = "NAME")private String name;}
and the following Repository interface:
@Repositorypublic interface FooRepository extends JpaRepository<Foo, Foo.ID> {}
then I try doing findById
fooRepository.findById(new Foo.ID(new Bar(1), new Baz(1));
and got the problem:
When I call fooRepository.findById(), although the row is fetched from database, ManyToOne mappings for Bar e Baz are not working properly. That means, whether I do foo.getId().getBar().getName() or foo.getId().getBaz().getName(), attribute name is null for both Bar and Baz objects (of course, the atrributes are not null in the database).
So my questions are:
This should work, right?<br/>
What am I doing wrong?<br/>
I have already tried changing fetch type to EAGER, but that didn't work at all.
I really appreciate any replies.
答案1
得分: 0
当我打印数据库执行语句时，语句为Hibernate: select foo0_.bar_id as bar_id2_2_0_，foo0_.baz_id as baz_id1_2_0_ from foo foo0_ where foo0_.bar_id=? and foo0_.baz_id=?。显然，JPA没有查询Bar和Baz表，因此foo.getBar().getName()肯定为null。
您可以使用以下方法实现相同的效果。
@Entity@Table(name = "FOO")@Datapublic class Foo implements Serializable {@Id@GeneratedValue(strategy = GenerationType.AUTO)Long id;@ManyToOne@JoinColumn(name="BAR_ID")private Bar bar;@ManyToOne@JoinColumn(name="BAZ_ID")private Baz baz;}@Repositorypublic interface FooRepository extends JpaRepository<Foo, Long> {Foo findByBarAndBaz(Bar bar, Baz baz);}
像这样使用它：
Foo byId = fooRepository.findByBarAndBaz(new Bar(1L), new Baz(1L));Bar bar = byId.getBar();System.out.println(bar.toString());
英文:
When I print database execution statement,the statement is Hibernate: select foo0_.bar_id as bar_id2_2_0_, foo0_.baz_id as baz_id1_2_0_ from foo foo0_ where foo0_.bar_id=? and foo0_.baz_id=?.Obviously the JPA does not query the Bar and Baz tables,so foo.getBar().getName() certainly will be null.
You can use follow way to achieve the same effect.
@Entity@Table(name = "FOO")@Datapublic class Foo implements Serializable {@Id@GeneratedValue(strategy = GenerationType.AUTO)Long id;@ManyToOne@JoinColumn(name="BAR_ID")private Bar bar;@ManyToOne@JoinColumn(name="BAZ_ID")private Baz baz;}@Repositorypublic interface FooRepository extends JpaRepository<Foo, Long> {Foo findByBarAndBaz(Bar bar, Baz baz);}
use it like this:
Foo byId = fooRepository.findByBarAndBaz(new Bar(1L), new Baz(1L));Bar bar = byId.getBar();System.out.println(bar.toString());
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