Hibernate 和 Java 泛型 - 无法使用 MappedSuperClass 创建会话工厂

Question

我有几个 classes 我们正在尝试扩展以允许重用代码，但 hibernate 有 none 它。以下是新的 classes 及其扩展：

超级声明class

@MappedSuperclass
public abstract class CoreStatement<S extends Approval>
    implements java.io.Serializable
{

    public abstract Long getId();

    public abstract void setId(Long id);

    public abstract Set<S> getApprovals();

    public abstract void setApprovals(Set<S> approvals);

}

基本语句 class - 这稍后会得到扩展，但通过单个 table 继承

@Entity
@Table(name="EXPNS_STTMNT")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
        name="CLASS_ID",
        discriminatorType = DiscriminatorType.INTEGER
)
public abstract class ExpenseStatement extends CoreStatement<ExpenseApproval>
{
    private Set<ExpenseApproval> approvals;

    @Override
    @Id
    @Column(name="ID", unique=true, nullable=false, precision=10, scale=0)
    public Long getId() {
        return this.id;
    }

    @Override
    public void setId(Long id) {
        this.id = id;
    }

    @Override
    @OneToMany(cascade=CascadeType.ALL, fetch=FetchType.LAZY, mappedBy="statement",
        targetEntity = ExpenseApproval.class)
    public Set<ExpenseApproval> getApprovals() {
        return approvals;
    }

    public void setApprovals(Set<ExpenseApproval> approvals) {
        this.approvals = approvals;
    }
}

认可超级class

@MappedSuperclass
public abstract class Approval<T extends CoreStatement> implements java.io.Serializable  {
    public abstract Long getId();

    public abstract void setId(Long id);
    public abstract T getStatement();

    public abstract void setStatement(T statement);
}

认可度class

@Entity
@Table(name="APPRVL")
public class ExpenseApproval extends Approval<ExpenseStatement>{
    private Long id;
    private ExpenseStatement statement;
    @Id
    @Column(name="ID", unique=true, nullable=false, precision=10, scale=0)
    public Long getId() {
        return this.id;
    }

    @Override
    public void setId(Long id) {
        this.id = id;
    }
    @Override
    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="EXPENSE_STATEMENT_ID", nullable=true)
    public ExpenseStatement getStatement() {
        return statement;
    }

    @Override
    public void setStatement(ExpenseStatement statement) {
        this.statement= statement;
    }
}

当运行通过单元测试时，我们得到错误：

java.lang.ExceptionInInitializerError Caused by: org.hibernate.MappingException: Could not determine type for: java.util.Set, at table: EXPNS_STTMNT, for columns: [org.hibernate.mapping.Column(approvals)] at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:314) at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:292) ....

这似乎是某种映射错误，但我无法缩小范围。许多之前发布过该问题的人都遇到过他们的注释位于私有属性和 getters 之上的问题，即他们混合并匹配了他们的注释位置，但这里似乎不是这种情况。其他人对可能导致问题的原因有任何建议吗？

Answer 1

Hibernate 及其错误消息通常很糟糕，但我设法解决了这个问题。这与集合无关，而与注释的位置有关。注释通常应该在 superclass 而不是 base class.

所以这个例子是：

@MappedSuperclass
public abstract class Approval<T extends CoreStatement> implements java.io.Serializable  {

    private Long id;
    private T statement


    @Id
    @Column(name="ID", unique=true, nullable=false)
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="STATEMENT_ID", nullable=true)
    public T getStatement() {
        return statement;
    }

    public void setStatement(T statement) {
        this.statement = statement;
    }
}

然后要设置适当的列或覆盖注释的属性，您可以在 class 命名约定之上使用 @AssociationOverrides。

@Entity
@Table(name="APPRVL")
@AssociationOverrides({
    @AssociationOverride(name="statement", joinColumns = @JoinColumn(name = "EXPENSE_STATEMENT_ID"))
})    
public class ExpenseApproval extends Approval<ExpenseStatement>{
    private ExpenseStatement statement;

}

Hibernate 和 Java 泛型 - 无法使用 MappedSuperClass 创建会话工厂

Hibernate and Java generics - Unable to create session factory using MappedSuperClass

java

persistence

hibernate

jpa