在使用 hibernate 和 MySQL 的 spring mvc 应用程序中，我收到一个错误，该错误似乎表明 Name 实体找不到 id 属性的 setter Patient 实体的 BaseEntity 父类(super class)。

我该如何解决这个错误？

这是错误信息:

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

这是触发错误的代码行:

ArrayList<Name> names = (ArrayList<Name>) this.clinicService.findNamesByPatientID(patntId);

这里是 BaseEntity，它是 Patient 和 Name 的父类(super class):

@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<Name> names;

    protected void setNamesInternal(Set<Name> nms) {this.names = nms;}

    protected Set<Name> getNamesInternal() {
        if (this.names == null) {this.names = new HashSet<Name>();}
        return this.names;
    }

    public List<Name> getNames() {
        List<Name> sortedNames = new ArrayList<Name>(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

}

可以查看完整的堆栈跟踪at this link .

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 MySQL 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。

最佳答案

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:

Caused by: java.lang.IllegalArgumentException: Can not set org.springframework.samples.knowledgemanager.model.HL7Patient field org.springframework.samples.knowledgemanager.model.HL7USName.patient to org.springframework.samples.knowledgemanager.model.HL7USName

因此，您的 HL7USName 类有一个名为 patient 且类型为 HL7Patient 的字段，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名称，该名称具有指向 Name 类型行而不是 Patient 类型行的外键。

关于java - 无法通过反射 setter 设置字段值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25295695/