Java 无法解析属性-休眠

我对Hibernate有问题。我从昨天开始就在努力解决这个问题，看起来很简单，但我不知道为什么它不起作用

我有entityLogin.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;
    }
}

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

这些实体通过@OneToOne关系连接。 我要做的是用Login类（username）字段查找我办公室的名称（officeName）

这是我在TourOfficeDAO.java中的函数：

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;
    }
}

   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rest控制器。但这并不重要，因为在DAO中抛出了异常：

无法解析属性：login.username的： offersmanager.model.entity.TourOffice；嵌套异常是 org.hibernate.QueryException:无法解析属性： login.username of:offersmanager.model.entity.TourOffice

它找不到“login.username”，也不知道为什么。。。一切似乎都很好。 我找了一些类似的话题，但我还是没能做到这一点。任何帮助都将不胜感激

编辑1:

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;
    }
}

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

这是我的抽象类DAO.java，其中是createCriteria（）函数

公共抽象类DAO实现了可序列化{
公共抽象类getEntityClass（）；
@自动连线
受保护的SessionFactory SessionFactory；
受保护会话getSession（）{
返回sessionFactory.getCurrentSession（）；
}
受保护的查询createQuery（字符串查询）{
返回getSession（）.createQuery（查询）；
}
受保护的SQLQuery createSQLQuery（字符串查询）{
返回getSession（）.createSQLQuery（查询）；
}
受保护的条件createCriteria（）{
返回getSession（）.createCriteria（getEntityClass（））；
}
@抑制警告（“未选中”）
公共模型findById（整数id）{
返回（模型）getSession（）.get（getEntityClass（），id）；
}
公共作废保存（模型实体）{
getSession（）.save（实体）；
getSession（）.flush（）；
}
公共无效更新（模型实体）{
getSession（）.update（实体）；
getSession（）.flush（）；
}
公共作废保存或更新（模型实体）{
getSession（）.saveOrUpdate（实体）；
getSession（）.flush（）；
}
公共作废删除（模型实体）{
getSession（）.delete（实体）；
getSession（）.flush（）；
}
公开名单（）{
Criteria=createCriteria（）；
@抑制警告（“未选中”）
List=criteria.List（）；
退货清单；
}
}

我认为您首先需要创建这样一个别名：

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

---

从问题中删除样板getter/setters代码，并提供createCriteria（）的代码，您认为“样板getter/setters”是什么？我添加了createCriteria（）代码。这意味着它们与您的问题无关，所以您应该删除它们。我粘贴它们是因为在某个主题中有人说getter和setter对Hibernate QueriesOfftopic有影响问题：为什么您的属性不是私有的？是的，现在它可以工作了。非常感谢。但我不知道为什么我必须创建别名。我的朋友有非常相似的代码，但与一个omany（或许多-不记得很清楚）有关，他不必创建别名。当您创建别名时，Hibernate发出“加入”请求（如果您想查看Hibernate生成的sql，可以显示sql）。对于您的好友连接信息，可以在OneToMany或ManyToMany属性上设置（检查@JoinTable信息）