我有一個User
實體,UserToApplication
實體和Application
實體。使用CriteriaBuilder的彈簧數據JPA規範w /一對多關係
單個User
可以訪問多個Application
。並且一個Application
可以被多個User
使用。
這裏是User
實體。
@Entity
@Table(name = "USER", schema = "UDB")
public class User {
private Long userId;
private Collection<Application> applications;
private String firstNm;
private String lastNm;
private String email;
@SequenceGenerator(name = "generator", sequenceName = "UDB.USER_SEQ", initialValue = 1, allocationSize = 1)
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "generator")
@Column(name = "USER_ID", unique = true, nullable = false)
public Long getUserId() {
return userId;
}
public void setUserId(Long userId) {
this.userId = userId;
}
@OneToMany(mappedBy = "user", fetch = FetchType.LAZY)
public Collection<Application> getApplications() {
return applications;
}
public void setApplications(Collection<Application> applications) {
this.applications = applications;
}
/* Other getters and setters omitted for brevity */
}
這裏是UserToApplication
實體。
@Entity
@Table(name = "USER_TO_APPLICATION", schema = "UDB")
public class Application {
private Long userToApplicationId;
private User user;
private Application application;
@SequenceGenerator(name = "generator", sequenceName = "UDB.USER_TO_APP_SEQ", initialValue = 0, allocationSize = 1)
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "generator")
@Column(name = "USER_TO_APPLICATION_ID", unique = true, nullable = false)
public Long getUserToApplicationId() {
return userToApplicationId;
}
public void setUserToApplicationId(Long userToApplicationId) {
this.userToApplicationId = userToApplicationId;
}
@ManyToOne
@JoinColumn(name = "USER_ID", referencedColumnName = "USER_ID", nullable = false)
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
@ManyToOne
@JoinColumn(name = "APPLICATION_ID", nullable = false)
public Application getApplication() {
return application;
}
}
這裏是Application
實體。
@Entity
@Table(name = "APPLICATION", schema = "UDB")
public class Application {
private Long applicationId;
private String name;
private String code;
/* Getters and setters omitted for brevity */
}
我有以下Specification
,我使用由firstNm
,lastNm
和email
來搜索User
。
public class UserSpecification {
public static Specification<User> findByFirstNmLastNmEmail(String firstNm, String lastNm, String email) {
return new Specification<User>() {
@Override
public Predicate toPredicate(Root<User> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
final Predicate firstNmPredicate = null;
final Predicate lastNmPredicate = null;
final Predicate emailPredicate = null;
if (!StringUtils.isEmpty(firstNm)) {
firstNmPredicate = cb.like(cb.lower(root.get(User_.firstNm), firstNm));
}
if (!StringUtils.isEmpty(lastNm)) {
lastNmPredicate = cb.like(cb.lower(root.get(User_.lastNm), lastNm));
}
if (!StringUtils.isEmpty(email)) {
emailPredicate = cb.like(cb.lower(root.get(User_.email), email));
}
return cb.and(firstNmPredicate, lastNmPredicate, emailPredicate);
}
};
}
}
這裏是我到目前爲止的User_
元模型。現在
@StaticMetamodel(User.class)
public class User_ {
public static volatile SingularAttribute<User, String> firstNm;
public static volatile SingularAttribute<User, String> lastNm;
public static volatile SingularAttribute<User, String> email;
}
,我想也通過在應用程序標識的Specification
列表,使得它的方法簽名是:
public static Specification<User> findByFirstNmLastNmEmailApp(String firstNm, String lastNm, String email, Collection<Long> appIds)
所以,我的問題是,我可以添加@OneToMany
映射到我的User
實體的Collection<Application> applications
字段的User_
元模型,然後我將如何參考Specification
?
我現在Specification
將類似於下面的SQL查詢:
select * from user u
where lower(first_nm) like '%firstNm%'
and lower(last_nm) like '%lastNm%'
and lower(email) like '%email%';
而我想在新Specification
實現將是這樣的:
select * from user u
join user_to_application uta on uta.user_id = u.user_id
where lower(u.first_nm) like '%firstNm%'
and lower(u.last_nm) like '%lastNm%'
and lower(u.email) like '%email%'
and uta.application_id in (appIds);
是否有可能在元模型中做這種映射,我怎麼能在我的Specification
中實現這個結果?
在Application_類變量聲明應該如下,公共靜態揮發性SingularAttribute <應用程序,龍>的applicationID;而不是公共靜態volatile SingularAttribute applicationId;整體答案是完美的,雖然 –
Avi
@Avi謝謝指出!我在我的回答中更正了它。 –