Creating a table dynamicaly with hibernate - java

I am learning Hibernate ORM(v. 3 ) now and I've a got a question.
I have a table called USERS, created with annotations :
package com.hibernatedb.entities;
import javax.persistence.*;
#Entity
#Table(name = "USERS",uniqueConstraints = {#UniqueConstraint(columnNames={"USER_LOGIN", "USER_EMAIL"})})
public class User {
#Column(name = "USER_LOGIN", length=80, nullable=false)
private String login;
#Column(name = "USER_PASS", length=80, nullable=false)
private String password;
#Column(name = "USER_EMAIL", length=80, nullable=false)
private String email;
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
#Column(name = "USER_ID", nullable=false)
private Long id;
...
// some getters and setters, toString() and other stuff
...
}
And a Product entity :
#Entity
#Table(name = "PRODUCTS",uniqueConstraints = {#UniqueConstraint(columnNames={"PRODUCT_ID", "PRODUCT_NAME"})})
public class Product {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
#Column(name="PRODUCT_ID")
private long id;
#Column(name="PRODUCT_NAME", length=85, nullable=false)
private String name;
#Column(name="PRODUCT_DESCRIPTION", columnDefinition="mediumtext", length=1000)
private String description;
#Column(name="PRODUCT_COST", nullable=false)
private double cost;
So my question is : How can a create a table called like "USER +
User.getId()
BUYS", which contains a 2 foreign keys (USER_ID and PRODUCT_ID) for user in entity (table record) "User" without raw SQL table creation, but using Hibernate annotations or XML mapping.So i want to have something like
public class TransactionEntityBulider() {
public TransactionEntityBulder(User user)
// something that build me "USER + User.getId() BUYS" table and
}
public TransactionEntity getEntity() {
// something that return a "USER + User.getId() BUYS" table entity
}
Also i would like to see some another ways to solve my problem.


I think hibernate is not done for that kind of usage, because you would have to use dynamic mappings. Hibernate provide ways to specify mapping statically (xml and annotations).
I suggest you modify your approach. It normally should not be harmfull to have all the "USER_BUY" in the same table. Example :
#Entity
public class User {
...
#OneToMany(cascade=CascadeType.ALL, mappedBy="user")
List<UserBuys> buys = new ArrayList<UserBuys>();
...
}
#Entity
public class Product { ... }
#Entity
public class UserBuys {
...
#ManyToOne
Product product;
#ManyToOne
User user;
}

Related

JPA StackOverflow -- Many to Many and one to one mapping

I am trying to save a JPA entity which has ManytoMany Relationship (Consumer and Product table) and OnetoOne relation with ConsumerDetailstable.Below are my entities
#JsonInclude(JsonInclude.Include.NON_NULL)
#JsonIgnoreProperties(ignoreUnknown = true)
#JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class)
#Entity
public class Consumer {
#Id
#GeneratedValue
private Long id;
private String firstName;
private String lastName;
#JsonManagedReference
#OnToMany(mappedBy = "consumer")
private Set<ConsumerProduct> consumerProducts;
#OneToOne
private CustomerDetails consumerDetails;
}
#Entity
public class Product {
#Id
#GeneratedValue
private Long productId;
private String productCode;
#OneToMany(mappedBy = "product")
private Set<ConsumerProduct> consumerProducts;
}
#JsonInclude(JsonInclude.Include.NON_NULL)
#JsonIgnoreProperties(ignoreUnknown = true)
#JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class)
#Entity(the join table)
public class ConsumerProduct {
#EmbeddedId
ConsumerProductKey id;
#JsonBackReference
#ManyToOne
#MapsId("id")
#JoinColumn(name = "id")
private Consumer consumer;
#ManyToOne
#MapsId("productId")
#JoinColumn(name = "product_id")
private Product product;
}
#Embeddable (forgein keys combined as embeded id)
public class ConsumerProductKey implements Serializable {
#Column(name="id")
private Long id;
#Column(name = "product_id")
private Long productId;
}
#Enitity (one to one relation table)
public class CustomerDetails {
#Id
#GeneratedValue
private Long consumerDtlId;
#OneToOne
private Consumer consumer;
private String city;
private String state;
private String country;
}
To save the entity am have just extended JPARepository and called save method
public class ConsumerRepository<Consumer> Implements JPARepository<Consumer, Long> {
#Override
public Consumer S save(Consumer entity) {
return save(entity);
};
}
I get java.lang.StackOverFlowError at save method.
Anything wrong with my Mappings ?
Question: Since this will be save operation and since Consumer Id is yet to be generated how do I assign to below Entities
ConsumerProduct.ConsumerProductKey (how do i assign Id of consumer table once it is inserted to join table ? will JPA take care of it)
CustomerDetails (how do i assign Id of consumer table once it is inserted to join table ? will JPA take care of it)
EDIT: I have updated the entity with JsonManagedReference and JsonBackedReference but still i have am facing stackoverflow error

It is due to Consumer trying to access ConsumerProduct and ConsumerProduct trying to access consumer entity and end up with StackOverflow error.
You should use #JsonManagedReference and #JsonBackReference annotation in consumer and ConsumerProduct respectivly.

Spring data jdbc - Specific join, specific mapping

I've got following two tables:
Customer
id
name
Order
id
product_name
customer_id
with a 1 to 1 relation
and java entities:
#Data
public class Customer{
#Id
private Long id;
private String name;
}
#Data
public class Order{
#Id
private Long id;
#Column("id")
private Customer customer; //i want to somehow map this
private String productName;
}
and a controller
#Controller
public class MyController{
//...
#GetMapping("/")
public String getmap(Model m){
System.out.println(repository.findAll()) //prints "nullrows" due to wrong sql statement
return "mytemplate";
}
}
my current issue is, that spring is executing following sql statement:
SELECT Order.id, Order.product_name, Customer.id, Customer.name
FROM Order LEFT OUTER JOIN Customer ON Customer.id = Order.id
what i actually want is to join on Customer.id = Order.customer_id while leaving the classes as they are i.e. the customer reference needs to stay in order.
i've tried every annotation that i could find so far and have made no progress.
EDIT:
I am not allowed to use jpa/hibernate

One workaround is to do the following:
#Data
public class Customer{
#Id
private Long id;
private String name;
}
#Data
public class Order{
#Id
private Long customerId;
private Long id;
#Column("id")
private Customer customer; //i want to somehow map this
private String productName;
}
causing this to automatically join on Customer.id = Order.customer_id
This does not look like a good fix however.

You can use #OneToOne and #JoinColumn annotations for your One-to-One relationship:
#Data
public class Customer{
#Id
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = "order_id", referencedColumnName = "id")
private Order order;
}
#Data
public class Order{
#Id
#Column(name = "id")
private Long id;
#Column(name = "product_name")
private String productName;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = "customer_id", referencedColumnName = "id")
private Customer customer;
}

Java hibernate: #OneToMany association doesn't work

In my code i have a oneToMany relation between customer class and item class. This means that, a customer may have one or many items.
Here is the customer code:
#Entity
#Data
public class customer {
#Id
#GeneratedValue
int id;
String name;
String lastname;
#Embedded
Address address;
#OneToMany
#Column(name="ITEM_ID")
List<item> item;
}
and it's the item class:
#Entity
#Data
public class item {
#Id
#GeneratedValue
int id;
String name;
String Serialnumber;
int price;
#ManyToOne
customer customer;
}
Then i have made some tests to try my queries in the models.
insert into item(id,name,Serialnumber,price) values(1,'bike','123',200);
insert into item(id,name,Serialnumber,price) values(2,'car','123',200);
insert into customer(id,name,lastname,Country,City,Street,No,item_id)
values(1,'Salman','Lashkarara','Iran','Tehran','Shariati','12',1);
insert into customer(id,name,lastname,Country,City,Street,No,item_id)
values(2,'Saba','Lashkarara','Iran','Tehran','Shariati','12',2);
insert into customer(id,name,lastname,Country,City,Street,No,item_id)
values(3,'Saba','Lashkarara','Iran','Tehran','Shariati','12',1);
But when i run my code, i face with the following error:
Column "ITEM_ID" not found; SQL statement:
insert into customer(id,name,lastname,Country,City,Street,No,item_id) values(1,'Salman','Lashkarara','Iran','Tehran','Shariati','12',1)
Please pay especial attention, that it is a java mvc-spring application and i create my models using the code, so there is no database to check the field item_id.
As you can see i have already added the #Column(name="ITEM_ID") to define the column.

You have to use #JoinColumn for association columns:
#OneToMany
#JoinColumn(name="ITEM_ID")
List<item> item;

some other options
#OneToMany(cascade=CascadeType.All, fetch=FetchType.EAGER)
#JoinColumn(name="ITEM_ID")
List<item> item;
in Item class
#ManyToOne(mappedBy="item")
customer customer;

you could do this i also have user class and bcr class, one user have many bcr so below code will help you
bcr.java
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "user_who_enter_demand", nullable = false)
public User getUserByUserWhoEnterDemand() {
return this.userByUserWhoEnterDemand;
}
public void setUserByUserWhoEnterDemand(User userByUserWhoEnterDemand) {
this.userByUserWhoEnterDemand = userByUserWhoEnterDemand;
}
user.java
#OneToMany(fetch = FetchType.LAZY, mappedBy = "userByUserWhoEnterDemand")
public Set<BudgetControlRegister> getBudgetControlRegistersForUserWhoEnterDemand() {
return this.budgetControlRegistersForUserWhoEnterDemand;
}
public void setBudgetControlRegistersForUserWhoEnterDemand(Set<BudgetControlRegister> budgetControlRegistersForUserWhoEnterDemand) {
this.budgetControlRegistersForUserWhoEnterDemand = budgetControlRegistersForUserWhoEnterDemand;
}

You can't map a table column without table:
#Entity
#Table(name = "ITEM_TABLE")
public class item {
...
#ManyToOne(fetch=FetchType.LAZY)
#JoinColumn(name = "CUSTOMER_ID_ITEM_TABLE")
private customer customer;
...
}
#Entity
#Table(name = "CUSTOMER_TABLE")
public class customer {
...
#OneToMany
#Column(name="ITEM_ID")
List<item> item;
}

Unknown Entity Error With ManyToOne Relationship

OK, so I've designed a basic CRUD an an exercise. It has 2 tables Jobs and Employees. I'm trying to create a many to one relationship, but when I click the link to go to the Employee Entry page it throws an error that kicks off with the #ManyToOne referencing an Unknown Entity.
Here is what I've got in my Employees.java
String jobName;
#ManyToOne(fetch=FetchType.EAGER)
#Fetch(value = FetchMode.JOIN)
#JoinColumn(name = "Job_Name")
#Column (name='jobName')
public String getJobName() {
return jobName;
}
public void setJobName(String jobName) {
this.jobName = jobName;
}
Any idea what i"m doing wrong and how to resolve this?

As per your comment,i think you can define relationship between these two entities like below.
#Entity
#Table(name="employee")
class Employee{
#Id
#GeneratedValue
private Integer id;
#ManyToOne
#JoinColumn(name = "job_name")
private Job job;
// other column and getter and setter
}
#Entity
#Table(name="job")
class Job{
#Id
#GeneratedValue
private Integer id;
#Column(name="job_name")
private String jobName;
//provide other column and getter setter
}

OneToOne with the same PrimaryKey and ForeignKey

Database
*user_account*
id (PK)
email
password
*user_detail*
id(PK)(FK)
name
city
Entities
#Table(name="user_detail")
public class UserDetail implementsSerializable{
#Id private Integer id;
...
#OneToOne
#JoinColumn(name="id")
private UserAccount userAccount;
}
#Table(name="user_account")
public class UserAccount implementsSerializable{
#Id private Integer id;
#OneToOne(mappedBy="userAccount")
private UserDetail userDetails;
}
Error
Exception Description: Multiple writable mappings exist for the field [user_detail.ID]. Only one may be defined as writable, all others must be specified read-only.

If the ID in UserAccount is both a primary key and a foreign key, then you should declare it as a single field and map it appropriately. Like this:
#Entity
public class UserAccount implements Serializable {
#Id
#OneToOne(mappedBy="userAccount")
private UserDetail userDetails;
}
Or else using #MapsId.
However, i suspect that what you really want is a single class spread over two tables:
#Entity
#Table(name = "user_account")
#SecondaryTable(name = "user_detail")
public class User implements Serializable {
#Id
private int id;
private String email;
private String password;
#Column(table = "user_detail")
private String name;
#Column(table = "user_detail")
private String city;
}

You cannot have both #Id private Integer id; and #JoinColumn(name="id"), you must remove one of them: I doubt that you really need a primary key in the details, so just remove the #Id line from there.

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