Question

假设我们有两个这样的类：

人：

@Entity
@Table (name= "person")
public class Person
{
@Id
@GeneratedValue
@Column(name = "Id")
private Long id;

@Column(name = "name")
private String name;

@Column(name = "phone")
private String phone;

@OneToMany(mappedBy = "bank")
private List<Bank> banks = new ArrayList<>();
}

银行：

@Entity
@Table(name = "bank")
public class Bank {

@Id
@GeneratedValue
@Column(name = "id")
private Long id;

@Column(name = "bank")
private String name;

@Columt(name = "phone")
private String phone;

@ManyToOne
@JoinColumn(name = "client_id")
@NotFound(action = NotFoundAction.EXCEPTION)
private Person person;
}

我们只想获得某些字段：person.name，person.phone，person.bank.phone

我试过这样的事情：

List result = session.createCriteria( Person.class )
.setProjection( Projections.projectionList()
.add( Projections.property( "name" ) )
.add( Projections.property( "phone" ) )
.add( Projections.property( "banks.phone" ) )
).list();

但最后我得到了这个例外：

org.hibernate.QueryException: could not resolve property: banks.phone of: training.net5.Person
at org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:83)
at org.hibernate.persister.entity.AbstractPropertyMapping.toColumns(AbstractPropertyMapping.java:98)
at org.hibernate.persister.entity.BasicEntityPropertyMapping.toColumns(BasicEntityPropertyMapping.java:61)
at org.hibernate.persister.entity.AbstractEntityPersister.toColumns(AbstractEntityPersister.java:1964)
at org.hibernate.loader.criteria.CriteriaQueryTranslator.getColumns(CriteriaQueryTranslator.java:511)
at org.hibernate.criterion.PropertyProjection.toSqlString(PropertyProjection.java:67)
at org.hibernate.criterion.ProjectionList.toSqlString(ProjectionList.java:116)
at org.hibernate.loader.criteria.CriteriaQueryTranslator.getSelect(CriteriaQueryTranslator.java:379)
at org.hibernate.loader.criteria.CriteriaJoinWalker.<init>(CriteriaJoinWalker.java:110)
at org.hibernate.loader.criteria.CriteriaJoinWalker.<init>(CriteriaJoinWalker.java:92)
at org.hibernate.loader.criteria.CriteriaLoader.<init>(CriteriaLoader.java:97)
at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1651)
at org.hibernate.internal.CriteriaImpl.list(CriteriaImpl.java:380)
at training.net5.MainClass.main(MainClass.java:82)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:601)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:120)
could not resolve property: banks.phone of: training.net5.Person

基本上所需要的是一种仅通过相对于根实体的路径获取某些字段的方法。所以我想知道我是否可以使用投影，或者我应该寻找另一种方法。

P.S。对不起我的英语我不是母语人士

Answer 1

你可以尝试从另一个实体开始吗？

List result = session.createCriteria( Bank.class )
.setProjection( Projections.projectionList()
.add( Projections.property( "person.name" ) )
.add( Projections.property( "person.phone" ) )
.add( Projections.property( "phone" ) )
).list();

也许它应该是

@OneToMany(mappedBy = "person")
private List<Bank> banks = new ArrayList<>();

Answer 2

您想要获得的内容称为fetch groups。如果您有一个包含许多列的表，可能需要对其进行标准化，而不是使用此技术。在您的代码banks中有OneToMany关系。你有什么期待？

使用Hibernate - 仅获取指定的字段（按名称）

2 个答案: