JPA Criteria API where subclass - 出现错误：无法针对路径 [null] 解析属性 [lastName]

Posted 2023-04-14

【中文标题】JPA Criteria API where subclass - 出现错误：无法针对路径 [null] 解析属性 [lastName]

【英文标题】：JPA Criteria API where subclass - Getting error : Unable to resolve attribute [lastName] against path [null]

【发布时间】：2013-08-12 01:01:34

【问题描述】：

我在编写条件查询时遇到以下问题 我正在使用 Hibernate JPA 2.0，我正在尝试创建一个条件查询来获取债务

DebtHolder 是一个连接表 债务(1)-->(N)DebtHolders(N)-->(1)当事人

Party 有两个子类 Person 和 Org

Person 有属性 lastName

当我在 lastName 上尝试 where 子句时，我收到错误消息“java.lang.IllegalArgumentException: Unable to resolve attribute [lastName] against path [null]”

这是进行查询的方法：

  public List<Debt> searchByDebtNumber(String debtNumber, String taxId, String lastName, String phoneNumber, String address, String zip, String state) 

    CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<Debt> c = cb.createQuery(Debt.class);
    Root<Debt> debt = c.from(Debt.class);
    Join debtHolders = (Join)debt.fetch("debtHolders", JoinType.LEFT);
    Join debtor = (Join)debtHolders.fetch("debtor", JoinType.LEFT);
    c.select(debt);
    c.distinct(true);


    List<Predicate> criteria = new ArrayList<Predicate>();
    if (debtNumber != null) 
        ParameterExpression<String> p = cb.parameter(String.class, "debtNumber");
        criteria.add(cb.equal(debt.get("debtNumber"), p));
        

    if (taxId != null) 
        ParameterExpression<String> p = cb.parameter(String.class, "taxId");
        criteria.add(cb.equal(debtor.get("taxId"), p));
        

    if (lastName != null) 

        ParameterExpression<String> p = cb.parameter(String.class, "lastName");
        criteria.add(cb.equal(debtor.get("lastName"), p));
        


    if (criteria.size() == 0) 
           throw new RuntimeException("no criteria");
         else if (criteria.size() == 1) 
           c.where(criteria.get(0));
         else 
           c.where(cb.and(criteria.toArray(new Predicate[0])));
        


    TypedQuery<Debt> q = entityManager.createQuery(c);
    if (debtNumber != null)  
        q.setParameter("debtNumber", debtNumber);
        

    if (taxId != null)  
        q.setParameter("taxId", taxId);
        
    if (lastName != null)  
        q.setParameter("lastName", lastName);
        

    return  q.getResultList();

这是 JPA 实体的代码

            @Entity
            @Table(name = "debt")
            public class Debt 

                @NotNull
                @Column(name = "DEBT_NUMBER", nullable = false, unique = true)
                private String debtNumber;

                @OneToMany(mappedBy = "debt", fetch = FetchType.LAZY)
                private List<DebtHolder> debtHolders = new ArrayList<>();

            

            @Entity
            @Table(name = "debt_holder")
            public class DebtHolder 

                @ManyToOne
                @NotNull
                @Property
                @JoinColumn(name = "DEBT_ID")
                private Debt debt;

                @ManyToOne(fetch = FetchType.LAZY)
                @NotNull
                @Property
                @JoinColumn(name = "DEBTOR_ID")
                private Party debtor;

            

            @Entity
            @Inheritance(strategy = InheritanceType.JOINED)
            @Table(name = "PARTY")
            public class Party 

                @Basic
                @Column(name = "TAX_ID", length = 50)
                private String taxId;

            

            @Entity
            @Table(name = "PERSON")
            public class Person extends Party 

                @Basic
                @NotNull
                @Column(name = "LAST_NAME")
                private String lastName;

            

请让我知道我缺少什么...谢谢。

【问题讨论】：

您在这方面取得了进展吗？我相信我遇到了同样的问题。

【参考方案1】：

您在 Party、Person 和 Org 之间具有继承层次结构。并且只有 Person 具有属性名称“lastName”。 在您的 JPA 实体中，您指的是 私人债务人； 这意味着债务人是 Party 并且没有姓氏。 尝试使用 私人债务人；而不是私人债务人；