I have 2 tables in MySql and I have mapped them using hibernate in Spring: users and roles .
I have created one more table: user_roles but I don't know how to map it in hibernate.
You can see the table structure below:
CREATE TABLE users (
username varchar(30) NOT NULL,
email varchar(50) NOT NULL,
password varchar(255) NOT NULL,
first_name varchar(40) NOT NULL,
last_name varchar(40) NOT NULL,
date_of_birth Date,
phone_number varchar(20),
PRIMARY KEY (username)
);
CREATE TABLE roles (
role_id INTEGER NOT NULL AUTO_INCREMENT,
name VARCHAR(255),
PRIMARY KEY (role_id)
);
CREATE TABLE user_roles (
username VARCHAR(30) NOT NULL,
role_id INTEGER NOT NULL,
PRIMARY KEY (username, role_id)
);
Here is the mapping for the roles tables:
#Entity
#Table(name = "roles")
public class Role implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name="role_id")
private Integer id;
private String name;
}
Here is the mapping for the users table:
#Entity
#Table(name = "users")
#JsonInclude(JsonInclude.Include.NON_NULL)
public class User implements Serializable {
#Id
#NotEmpty
#Size(min = 5, max = 15)
#Column(name = "username")
private String username;
#Email
#NotEmpty
#Column(name = "email")
private String email;
#NotEmpty
#Size(min = 5)
#Column(name = "password")
private String password;
#NotEmpty
#Size(max = 40)
#Column(name = "first_name")
private String firstName;
#NotEmpty
#Size(max = 40)
#Column(name = "last_name")
private String lastName;
...
}
I have created the POJO for the user_role table, but I don't know how to use hibernate on it, I am using #EmbeddedId annotation but it is not working. I don't know how to show the 2 classes above that they are embeddable
#Entity
#Table(name = "user_roles")
#JsonInclude(JsonInclude.Include.NON_NULL)
public class UserRole implements Serializable {
public UserRole() { }
public UserRole(User username, Role role_id) {
this.username = username;
this.role_id = role_id;
}
private static final long serialVersionUID = -2947211066236048069L;
#EmbeddedId
private User username;
#EmbeddedId
private Role role_id;
}
How can I map the "UserRole" class to 'user_role' in hibernate? Thank you!
There are two different ways which you can map user_roles table, which I suggest the first one :
1.
#Entity
#Table(name = "users")
public class User
{
#ManyToMany(targetEntity = Role.class)
#JoinTable(name = "users_roles", joinColumns = #JoinColumn(name = "user_id"), inverseJoinColumns = #JoinColumn(name = "role_id"))
#NotAudited
private Set<Role> roles = new HashSet<>();
}
2.
#Embeddable
public class UserRoleId implements java.io.Serializable
{
#Column(name = "user_id", nullable = false)
private long userId;
#Column(name = "role_id", nullable = false)
private long roleId;
public UserRoleId()
{
}
public UserRoleId(long userId, long roleId)
{
this.userId = userId;
this.roleId = roleId;
}
//hashcode equal
}
Then create the entity.
#Entity
#Table(name = "users_roles")
public class Userroles implements java.io.Serializable
{
#EmbeddedId
private UserRoleId id;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "user_id", nullable = false, insertable = false, updatable = false)
#NotNull
private Users users;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "role_id", nullable = false, insertable = false, updatable = false)
#NotNull
private Role role;
}
Related
I am creating entity relationships in Spring Boot data JPA. Since those tables being legacy I am not able to modify or add columns. Issue is I am getting error if point part of embedded Id.
My entity classes looks like below:
Class Customer {
#EmbededId
private CustomerPk id;
#Column("NAME")
private String name;
#OneToMany(fetch=FetchType.LAZY, cascade=CascadeType.ALL, mappedBy="customerDetails")
private List<Purchase> purchaseDetails;
...
}
#Embeddable
Class CustomerPk {
#Column("CUSTOMER_ID")
private String customerId
#Column("PURCHASE_ID")
private String productId;
#Column("PURCHASE_DATE")
private String date;
}
Purchase Entity looks like below:
Class Purchase {
#EmbededId
private PurchasePK id;
#Column("TRANSACTION_NAME")
private String transactionName;
#ManyToOne(fetch=FetchType.LAZY, cascade=CascadeType.ALL)
#JoinColumns({
#JoinColumn(name="CUSTOMER_ID" referencedColumnName="CUSTOMER_ID")
#JoinColumn(name="PURCHASE_ID" referencedColumnName="PURCHASE_ID")
)}
private Customer customerDetails;
...
}
#Embeddable
Class PurchasePK {
#Column("CUSTOMER_ID")
private String customerId
#Column("PURCHASE_ID")
private String productId;
#Column("TRANSACTION_DATE")
private String date;
}
With above structure I am getting org.hibernate.AnnotationException: referencedColumnNames(CUSTOMER_ID, PURCHASE_ID) of Purchase.customerDetails referencing Customer not mapped to a single property.
If I remove date property from CustomerPK, I am able to make the server up. But with current requirement I need date to be part of the CustomerPK class.
I think if I use part of the composite key as Join Columns I am getting this error.
Working version:
#Entity
public class Customer {
#EmbeddedId
private CustomerPk id;
#Column(name = "NAME")
private String name;
#OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "customerDetails")
private List<Purchase> purchaseDetails;
}
#Embeddable
public class CustomerPk implements Serializable {
#Column(name = "CUSTOMER_ID")
private String customerId;
#Column(name = "PURCHASE_ID")
private String productId;
#Column(name = "PURCHASE_DATE")
private String date;
}
#Entity
public class Purchase {
#EmbeddedId
private PurchasePK id;
#Column(name = "TRANSACTION_NAME")
private String transactionName;
#ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
#JoinColumns({
#JoinColumn(name = "CUSTOMER_ID", referencedColumnName = "CUSTOMER_ID", insertable = false, updatable = false),
#JoinColumn(name = "PURCHASE_ID", referencedColumnName = "PURCHASE_ID", insertable = false, updatable = false),
#JoinColumn(name = "PURCHASE_DATE", referencedColumnName = "PURCHASE_DATE", insertable = false, updatable = false)
})
private Customer customerDetails;
}
#Embeddable
public class PurchasePK implements Serializable {
#Column(name = "CUSTOMER_ID")
private String customerId;
#Column(name = "PURCHASE_ID")
private String productId;
#Column(name = "TRANSACTION_DATE")
private String date;
}
Conclusion:
The provided information from #Ray was valid, you missed adding the required join columns to represent the full entity relation, regarding your note for the same #Ray point, yes you are right both columns usage is different but also both columns have their own name which it will not override any row value on runtime.
The result of the above tables and representation is as follows:
create table customer
(
customer_id varchar(255) not null,
purchase_date varchar(255) not null,
purchase_id varchar(255) not null,
name varchar(255),
primary key (customer_id, purchase_date, purchase_id)
);
create table purchase
(
customer_id varchar(255) not null,
transaction_date varchar(255) not null,
purchase_id varchar(255) not null,
transaction_name varchar(255),
purchase_date varchar(255),
primary key (customer_id, transaction_date, purchase_id)
);
alter table purchase
add constraint FK6rkrb8rq8x56kai7g5gm32d1y foreign key (customer_id, purchase_date, purchase_id) references customer;
I have 2 entities User and Venue. One user can have multiple venues. The table looks like this:
user_id is the foreign key in the Venue table.
I am trying to create a query for retrieving all Venues by user_id. This is what I have:
#Override
public List<Venue> findVenueByOwnerId(Long userId) {
return em.createQuery("select v from Venue v where v.user.id = :ownerId", Venue.class)
.setParameter("ownerId", userId)
.getResultList();
}
What am I doing wrong?
EDIT:
These are my mappings:
For User class:
#Entity
public class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", updatable = false, nullable = false)
private Long id;
private String email;
private String username;
private String password;
private String firstName;
private String lastName;
#OneToMany
#JoinColumn(name = "user_id")
private List<Venue> venues;
For Venue class:
#Entity
public class Venue {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", updatable = false, nullable = false)
private Long id;
private String name;
private String location;
private int capacity;
#ManyToOne
#Column(name = "user_id", updatable = false, nullable = false)
private User user;
And if I am trying to add getters and setters for the User in the Venue class, it throws me this error:
Caused by: org.hibernate.MappingException: Could not determine type for: razvan.sem3.project.model.User, at table: Venue, for columns: [org.hibernate.mapping.Column(user)]
I have a users table, roles table and a notifications table. The user_id is the foreign key for linking users to notifications.
In my users class i am already accessing another table, roles via its foreignkey, role_id.
As shown
#Data
#Entity
#Table(name = "users")
public class User {`enter code here`
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column
private Long id;
#Column
#NotBlank
#Size(max = 40)
private String username;
// Valid From
#Column
private Date validFrom;
// Valid To
#Column
private Date validTo;
#Column
#NotBlank
#Size(max = 100)
private String password;
#OneToOne(fetch = FetchType.LAZY)
#JoinTable(name = "user_roles",
joinColumns = #JoinColumn(name = "user_id"),
inverseJoinColumns = #JoinColumn(name = "role_id"))
private Role role;
public User() {
}
public User(String username, String password) {
this.username = username;
this.password = password;
}
}
How can i use the #JoinTable annotation to connect to the notifications table?
Its not accepting duplicates.
You must not will use #JoinTable annotation. The #JoinTable annotation is used only to #ManyToMany relationship.
You need create a new Entity with three field, and each field must has the #ManyToOne and #JoinColumn annotation.
For Instance:
#Entity
#Table(name = "table_name")
class NewEntity {
//Id and anothers fields
#ManyToOne
#JoinColumn(name = "user_id")
private Users users;
#ManyToOne
#JoinColumn(name = "role_id")
private Roles roles;
#ManyToOne
#JoinColumn(name = "notification_id")
private Notifications notifications ;
//getters and setters
}
I have a SQL query like this:
"Select UIProfileID from UserTable where UPPER(UserID) = UPPER('?1')".
I want to convert it to Spring JPA.
I want to write getUIProfileId() and return Integer. But I don't know how to implement. Because User table doesn't have UIProfileId column that it was joined from UIProfileTable table. Please help me solve it.
Currently, I have tables:
User.java
#Entity
#Getter
#Setter
#NoArgsConstructor
#AllArgsConstructor
#Builder
#Table(name = "UserTable")
public class User {
#Column(name = "UserID", length = 32, nullable = false)
#Id
private String name;
#ManyToOne
#JoinColumn(name = "DomainID", nullable = false)
private Domain domain;
#Column(name = "Password", length = 32, nullable = false)
private String password;
#ManyToOne
#JoinColumn(name = "UIProfileID", nullable = false)
private UIProfile uiProfile;
#Column(name = "ResPerpage", nullable = false)
private Integer resperpage;
#Column(name = "DefaultTab")
private Integer defaulttab;
#ManyToOne
#JoinColumn(name = "AdminProfile")
private AdminProfiles adminProfile;
#Column(name = "LanguageId")
private Integer languageId;
}
UIProfile.java
#Entity
#Getter
#Setter
#Table(name = "UIProfileTable")
public class UIProfile implements Serializable {
#Id
#Column(name = "UIProfileID", length = 11, nullable = false)
private Integer id;
#Column(name = "UIProfileName", length = 32, nullable = false)
private String name;
#OneToMany(mappedBy = "id.uiProfile")
private List<UIProfileTopLevel> topLevels;
}
UserRepository.java
public interface UserRepository extends Repository<User, String> {
Optional<User> findOne(String name);
#Query("Select UIProfileID from User where UPPER(UserID) = UPPER('admin')")
Integer getUIProfileId();
}
You can try this:
#Query("SELECT u.uiProfile.id from User u where UPPER(u.name)=UPPER('admin')")
Integer getUIProfileId();
Here User is the domain class name and u is the reference of User. with u we will access User's field NOT the column name which are specified with #Column or #JoinColumn Ex : #JoinColumn(name = "UIProfileID", nullable = false).
I've read examples but have my personal question to you.
I have 2 tables:
Role:
id, name
User:
id, login, name, role_id
Role entity
#Entity
#Table(name = "role")
public class Role {
#Id
#Column(name = "id")
private long id;
#Column(name = "name", length = 45)
private String name;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "role")
private Set<User> user = new HashSet<>();
//getters and setters
User entity
#Entity
#Table(name = "user")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id",insertable = false, updatable = false)
private long id;
#Column(name = "login")
private String login;
#Column(name = "user_name")
private String userName;
#ManyToOne(fetch = FetchType.LAZY)
private Role role;
//getters and setters
And repository:
public interface UserRepository extends JpaRepository<User, Long> {
String Q_GET_ALL_USERS = "from User u left join Role r on u.role_id=r.id";
#Query(Q_GET_ALL_USERS)
Collection<User> getAllUsers();
This code is showing: Caused by: java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [from com.example.jpa.model.User u left join Role r on u.role_id=r.id]
How I understand entity can't contains 'id' (in my case in Role) for references and I should remove this field. But entity should have '#Id'.
In this case I should create new column in 'Role'? or I can use more beautiful decision?
I put all project to bb
To use join in HQL (JPQL) you don't need on clause
String Q_GET_ALL_USERS = "select u from User u left join u.role";
This query doesn't have any sence because of you don't use role in the where clause.
If you want to get users with a fetched role you can use join fetch
String Q_GET_ALL_USERS = "select u from User u left join fetch u.role";
Update
Your schema for User and Role is not commonly used. I advice to you make #ManyToMany association from user to roles and remove any user association from the Role
#Entity
#Table(name = "user")
public class User {
#ManyToMany(fetch = FetchType.LAZY)
private Set<Role> roles;
}
#Entity
#Table(name = "role")
public class Role {
#Id
#Column(name = "id")
private long id;
#Column(name = "name", length = 45)
private String name;
}
No, you should create a new column in User.
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "role_id")
private Role role;
Thank you all for answers. Right entities and query below (plus tables schema).
Tables (queries)
CREATE TABLE role (
id INT NOT NULL PRIMARY KEY,
name VARCHAR(45) NOT NULL
);
CREATE TABLE user (
id INT NOT NULL PRIMARY KEY IDENTITY,
login VARCHAR(45) NOT NULL,
user_name VARCHAR(45) NOT NULL,
role_id INT NOT NULL,
FOREIGN KEY (role_id) REFERENCES role (id)
);
Entities:
#Entity
#Table(name = "user")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id",insertable = false, updatable = false)
private long id;
#Column(name = "login")
private String login;
#Column(name = "user_name")
private String userName;
#ManyToOne(fetch = FetchType.LAZY)
private Role role;
//getters and setters
}
and
#Entity
#Table(name = "role")
public class Role {
#Id
#Column(name = "id")
private long id;
#Column(name = "name", length = 45)
private String name;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "role")
private Set<User> user = new HashSet<>();
//getters and setters
}
Repository
public interface UserRepository extends JpaRepository<User, Long> {
String Q_GET_ALL_USERS = "select u from User u left join u.role";
#Query(Q_GET_ALL_USERS)
Collection<User> getAllUsers();
}
#v-ladynev proposed alternative decision(use only #ManyToMany in User). More details you can find in comments under this answer.
When I check this decision I will update this answer (I hope I don't forget it :-))
Models
#Entity
#Table(name = "sys_std_user")
public class StdUser {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "class_id")
public int classId;
#Column(name = "user_name")
public String userName;
}
#Entity
#Table(name = "sys_std_profile")
public class StdProfile {
#Id
#Column(name = "pro_id")
public int proId;
#Column(name = "full_name")
public String fullName;
}
Controllers
#PersistenceUnit
private EntityManagerFactory emf;
#GetMapping("/join")
public List actionJoinTable() {
EntityManager em = emf.createEntityManager();
List arr_cust = em
.createQuery("SELECT u.classId, u.userName, p.fullName FROM StdUser u, StdProfile p WHERE u.classId=p.proId")
.getResultList();
return arr_cust;
}
Result:
[
[
1,
"Ram",
"Ram Pukar Chaudhary"
],
[
2,
"Raja",
"Raja Kishor Shah"
]
]