我对Hibernate有问题。从昨天开始我一直在为此苦苦挣扎,这似乎很容易,但是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;
}
}
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;
}
}
这些实体通过@OneToOne关系连接。
我想做的是使用登录类(用户名)的字段查找我的办公室的名称(officeName)。
这是我在TourOfficeDAO.java中的功能:
public TourOffice findOfficeNameByLogin(String username) {
Criteria name = createCriteria();
name.add(Restrictions.eq("login.username", username));
return (TourOffice) name.uniqueResult();
}
它通过TourOfficeService到达我的rest控制器,在该控制器上调用此方法。但这没关系,因为在DAO中引发了感知:
无法解析以下属性:login.username:
offermanager.model.entity.TourOffice;嵌套异常为
org.hibernate.QueryException:无法解析属性:
login.username的用户:Offersmanager.model.entity.TourOffice
它找不到“ login.username”,也不知道为什么...一切似乎都很好。
我寻找了类似的主题,但仍然没有设法使它起作用。任何帮助,将不胜感激。
编辑1:
这是我的抽象类DAO.java,其中的功能createCriteria()
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;
}
}
最佳答案
我认为您首先需要创建一个别名:
public TourOffice findOfficeNameByLogin(String username) {
Criteria name = createCriteria();
name.createAlias("login", "login");
name.add(Restrictions.eq("login.username", username));
return (TourOffice) name.uniqueResult();
}