org.hibernate.query.criteria.internal.BasicPathUsageException：无法加入基本类型的属性

Question

我对标准查询有疑问。

在MySQL数据库中，有专栏人员，并且与专栏护照相连。在标准api的帮助下，我试图通过对象/实体护照的“国家”属性获取对象/实体人。

我正在使用的代码示例：

public List<Guest> getByPassportCountry(String country) {
    try {
        CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
        CriteriaQuery<Guest> query = criteriaBuilder.createQuery(Guest.class);
        Root<Guest> root = query.from(Guest.class);
        Join<Guest, Passport> join = root.join(Guest_.PASSPORT).join(Passport_.COUNTRY);
        query.select(root).where(criteriaBuilder.equal(join.get(Passport_.COUNTRY), country));
        TypedQuery<Guest> result = entityManager.createQuery(query);
        return result.getResultList();
    } catch (NoResultException e) {
        return null;
    }
}

也存在此实体的连接：实体护照

@Entity
@Table(name = "passport_table")
public class Passport extends AEntity{

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

    @Column(name = "passport_number")
    private Integer passportNumber;

    @Column(name = "date_of_issue")
    @Convert(converter = StringToDataSQLConverter.class)
    private String dateOfIssue;

    @Column(name = "date_of_expiry")
    @Convert(converter = StringToDataSQLConverter.class)
    private String dateOfExpiry;

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

    //One guest to one passport
    @OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "guest_id", referencedColumnName = "id")
    private Guest guest;

＆实体访客

@Entity
@Table(name = "guest_table")
public class Guest extends AEntity {

    @Column(name = "guest_first_name", length = 45)
    private String guestFirstName;

    @Column(name = "guest_last_name", length = 45)
    private String guestLastName;

    @Column(name = "guest_birthdate")
    @Convert(converter = StringToDataSQLConverter.class)
    private String guestBirthDate;

    //Many guests to one room
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "room_id", referencedColumnName = "id")
    private Room room;

    //One passport to one guest
    @OneToOne(mappedBy = "guest")
    private Passport passport;

我做错了什么？？？请帮忙!!!

Answer 1

我做到了

public List<Guest> getByPassportCountry(String country) {
        try {
            CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
            CriteriaQuery<Guest> query = criteriaBuilder.createQuery(Guest.class);
            Root<Guest> root = query.from(Guest.class);
            Join<Guest, Passport> passport = root.join(Guest_.passport);
            query.select(root).where(criteriaBuilder.equal(passport.get(Passport_.COUNTRY),country));
            TypedQuery<Guest> result = entityManager.createQuery(query);
            return result.getResultList();
        } catch (NoResultException e) {
            return null;
        }
    }