java - 无法解析属性 - hibernate

Question

我有 Hibernate 的问题。从昨天开始我就在努力解决这个问题，这似乎很容易，但我不知道为什么它不起作用......

我有实体Login.java:

package offersmanager.model.entity;

import org.json.JSONObject;
import javax.persistence.*;

@Entity
public class Login {

    @Id
    @GeneratedValue
    private Integer id;
    @Column(nullable = false, unique = true)
    String username;
    @Column(nullable = false)
    String password;

    public Login(){
    }

    public Login(String username, String password){
        this.username = username;
        this.password = password;
    }

    public Login(JSONObject jsonObject) {
        this.id = (Integer) jsonObject.get("id");
        this.username = (String) jsonObject.get("username");
        this.password = (String) jsonObject.get("password");
    }

    public JSONObject toJsonObject() {
        JSONObject jsonObject = new JSONObject();
        jsonObject.put("id", this.id);
        jsonObject.put("username", this.username);
        jsonObject.put("password", this.password);
        return jsonObject;
    }

    public Integer getId() {
        return id;
    }

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

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }
}

和实体TourOffice.java:

package offersmanager.model.entity;

import org.json.JSONObject;

import javax.persistence.*;

@Entity
public class TourOffice {

    @Id
    @GeneratedValue
    private Integer id;
    @Column(nullable = false)
    String officeName;
    @Column(nullable = false)
    String eMail;
    @Column(nullable = false)
    String phoneNumber;
    @Column(nullable = false)
    String city;
    @Column(nullable = false)
    String zipCode;
    @Column(nullable = false)
    String address;

    @OneToOne(cascade = {CascadeType.ALL})
    @JoinColumn(name = "login_id")
    Login login;

    public TourOffice(){
    }

    public TourOffice(String officeName, String eMail, String phoneNumber, String city, String zipCode, String address) {
        this.officeName = officeName;
        this.eMail = eMail;
        this.phoneNumber = phoneNumber;
        this.city = city;
        this.zipCode = zipCode;
        this.address = address;
    }

    public TourOffice(JSONObject jsonObject) {
        this.id = (Integer) jsonObject.get("id");
        this.officeName = (String) jsonObject.get("officeName");
        this.eMail = (String) jsonObject.get("eMail");
        this.phoneNumber = (String) jsonObject.get("phoneNumber");
        this.city = (String) jsonObject.get("city");
        this.zipCode = (String) jsonObject.get("zipCode");
        this.address = (String) jsonObject.get("address");
        this.login = (new Login((JSONObject) jsonObject.get("login")));
    }

    public JSONObject toJsonObject() {
        JSONObject jsonObject = new JSONObject();
        jsonObject.put("id", this.id);
        jsonObject.put("officeName", this.officeName);
        jsonObject.put("eMail", this.eMail);
        jsonObject.put("phoneNumber", this.phoneNumber);
        jsonObject.put("city", this.city);
        jsonObject.put("zipCode", this.zipCode);
        jsonObject.put("address", this.address);
        jsonObject.put("login", this.login == null? null : login.toJsonObject());
        return jsonObject;
    }

    public Integer getId() {
        return id;
    }

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

    public String getOfficeName() {
        return officeName;
    }

    public void setOfficeName(String officeName) {
        this.officeName = officeName;
    }

    public String geteMail() {
        return eMail;
    }

    public void seteMail(String eMail) {
        this.eMail = eMail;
    }

    public String getPhoneNumber() {
        return phoneNumber;
    }

    public void setPhoneNumber(String phoneNumber) {
        this.phoneNumber = phoneNumber;
    }

    public String getCity() {
        return city;
    }

    public void setCity(String city) {
        this.city = city;
    }

    public String getZipCode() {
        return zipCode;
    }

    public void setZipCode(String zipCode) {
        this.zipCode = zipCode;
    }

    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }

    public Login getLogin() {
        return login;
    }

    public void setLogin(Login login) {
        this.login = login;
    }
}

这些实体与@OneToOne 关系相连。我想要做的是找到我的办公室名称(officeName)和登录类(用户名)字段。

这是我在 TourOfficeDAO.java 中的函数:

   public TourOffice findOfficeNameByLogin(String username) {
        Criteria name = createCriteria();
        name.add(Restrictions.eq("login.username", username));
        return (TourOffice) name.uniqueResult();
    }

它通过 TourOfficeService 到达调用此方法的其余 Controller 。但这并不重要，因为在 DAO 中抛出了异常:

could not resolve property: login.username of: offersmanager.model.entity.TourOffice; nested exception is org.hibernate.QueryException: could not resolve property: login.username of: offersmanager.model.entity.TourOffice

它找不到“login.username”并且不知道为什么......一切似乎都很好。我寻找了类似的主题，但我仍然无法做到这一点。任何帮助将不胜感激。

编辑 1:

这是我的抽象类 DAO.java 函数 createCriteria() 在哪里

public abstract class DAO<MODEL> implements Serializable {

    public abstract Class<MODEL> getEntityClass();

    @Autowired
    protected SessionFactory sessionFactory;

    protected Session getSession(){
        return sessionFactory.getCurrentSession();
    }

    protected Query createQuery(String query){
        return getSession().createQuery(query);
    }

    protected SQLQuery createSQLQuery(String query){
        return getSession().createSQLQuery(query);
    }

    protected Criteria createCriteria(){
        return getSession().createCriteria(getEntityClass());
    }

    @SuppressWarnings("unchecked")
    public MODEL findById(Integer id) {
        return (MODEL) getSession().get(getEntityClass(), id);
    }

    public void save(MODEL entity) {
        getSession().save(entity);
        getSession().flush();
    }

    public void update(MODEL entity) {
        getSession().update(entity);
        getSession().flush();
    }

    public void saveOrUpdate(MODEL entity) {
        getSession().saveOrUpdate(entity);
        getSession().flush();
    }

    public void delete(MODEL entity) {
        getSession().delete(entity);
        getSession().flush();
    }

    public List<MODEL> list(){
        Criteria criteria = createCriteria();
        @SuppressWarnings("unchecked")
        List<MODEL> list = criteria.list();
        return list;
    }
}

Answer 1

我认为您首先需要创建一个这样的别名:

  public TourOffice findOfficeNameByLogin(String username) {
    Criteria name = createCriteria();
    name.createAlias("login", "login");
    name.add(Restrictions.eq("login.username", username));
    return (TourOffice) name.uniqueResult();
}