无法通过reflection设置器设置字段值

在使用hibernate和MySQL的spring mvc应用程序中，我收到一个错误，这似乎表明Name实体无法找到Patient实体的BaseEntity超类的id属性的setter。

我该如何解决这个错误？

这是错误消息：

 Caused by: org.hibernate.PropertyAccessException: could not set a field value by reflection setter of myapp.mypackage.Name.patient 

以下是触发错误的代码行：

 ArrayList names = (ArrayList) this.clinicService.findNamesByPatientID(patntId); 

这是BaseEntity ，它是Patient和Name的超类：

 @Entity @Inheritance(strategy = InheritanceType.TABLE_PER_CLASS) @DiscriminatorFormula("(CASE WHEN dtype IS NULL THEN 'BaseEntity' ELSE dtype END)") public class BaseEntity { @Transient private String dtype = this.getClass().getSimpleName(); @Id @GeneratedValue(strategy = GenerationType.TABLE) protected Integer id; public void setId(Integer id) {this.id = id;} public Integer getId() {return id;} public void setDtype(String dt){dtype=dt;} public String getDtype(){return dtype;} public boolean isNew() {return (this.id == null);} } 

这是Patient实体：

 @Entity @Table(name = "patient") public class Patient extends BaseEntity{ @OneToMany(mappedBy = "patient") private Set names; protected void setNamesInternal(Set nms) {this.names = nms;} protected Set getNamesInternal() { if (this.names == null) {this.names = new HashSet();} return this.names; } public List getNames() { List sortedNames = new ArrayList(getNamesInternal()); PropertyComparator.sort(sortedNames, new MutableSortDefinition("family", true, true)); return Collections.unmodifiableList(sortedNames); } public void addName(Name nm) { getNamesInternal().add(nm); nm.setPatient(this); } //other stuff } 

这是Name实体：

 @Entity @Table(name = "name") public class Name extends BaseEntity{ @ManyToOne @JoinColumn(name = "patient_id") private Patient patient; public Patient getPatient(){return patient;} public void setPatient(Patient ptnt){patient=ptnt;} //other stuff } 

可以在此链接中查看完整的堆栈跟踪。

Hibernate为上述查询生成的SQL是：

 select distinct hl7usname0_.id as id1_0_0_, givennames1_.id as id1_45_1_, hl7usname0_.family as family1_44_0_, hl7usname0_.patient_id as patient3_44_0_, hl7usname0_.person_id as person4_44_0_, hl7usname0_.suffix as suffix2_44_0_, hl7usname0_.usecode as usecode5_44_0_, hl7usname0_.codesystem as codesyst6_44_0_, givennames1_.given as given2_45_1_, givennames1_.name_id as name3_45_1_, givennames1_.name_id as name3_0_0__, givennames1_.id as id1_45_0__ from hl7_usname hl7usname0_ left outer join hl7_usname_given givennames1_ on hl7usname0_.id=givennames1_.name_id where hl7usname0_.patient_id=1 

当我通过My​​SQL命令行客户端运行此查询时，它返回测试数据库表中的唯一记录。

Answer 1

这不是堆栈跟踪所说的。 堆栈跟踪并未说明无法设置ID。 它说：

引起：java.lang.IllegalArgumentException：无法将org.springframework.samples.knowledgemanager.model.HL7Patient字段org.springframework.samples.knowledgemanager.model.HL7USName.patient设置为org.springframework.samples.knowledgemanager.model.HL7USName

因此，您的HL7USName类有一个名为HL7Patient类型的HL7Patient ，并且无法使用HL7USName类型的值设置此字段。

这意味着您的数据库包含一个Name，该Name具有Name类型行的外键，而不是Patient类型的行。