I'm using Hibernate Envers for Auditing Change Data, I have a Class that store information about companies like this :
#Getter
#Setter
#Entity
#Table(name = "COMPNAY")
#Audited
public class Compnay {
private String name;
private String code;
}
and it's using Envers for keeping the changes of companies.
also, I have a class for Keep the data of items that manufacture in any of this company, the class will be like this :
#Getter
#Setter
#Entity
#Table(name = "COMPNAY")
#Audited
public class Item {
#Column(name = "NAME", nullable = false)
private String name ;
#Column(name = "CODE", nullable = false)
private String code;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "COMPANY_ID", nullable = false)
private Compnay compnay;
}
Consider that there is a company in company table like this :
ID
NAME
CODE
1
Apple
100
2
IBM
200
and the data in the item's table will be like this :
ID
NAME
CODE
COMPANY_ID
3
iPhone
300
1
4
iPad
400
1
if I edit the information of Apple company and change the code from 100 to 300 how can I fetch the information of Items that were saved before this change with the previous code? Is there is any way to reference to audit table?
Yes, you can write a HQL query that refers to the audited entities. Usually, the audited entities are named like the original ones, with the suffix _AUD i.e. you could write a query similar to the following:
select c, i
from Company_AUD c
left join Item_AUD i on i.id.revision < c.id.revision
where c.originalId = :companyId
Related
I have been playing around with JPA and came across this scenario and wanted to know how to solve this.
I have 2 tables namely Company and Employee.
So here an employee can work for only 1 company, therefore #OneToOne uni-directional mapping is done in Employee class. And the company details in the Company table would already exist.
So when I try to insert a record into Employee table, I even create a reference to the company, add it to the Employee class and save it using my EmployeeRepository's save method. Since the company_id would already exist in Company table, it should just refer it instead of creating a new company. But this is not happening.
I get an exception saying company object is still in the transient state. I can solve this by adding CascadeType.All but I don't need to insert the company rather I have to just link the company to the employee.
Here are the code snippets.
Employee Class
#Entity
#Table(name = "employee")
#Setter
#Getter
public class Employee
{
#Id
#GeneratedValue
private Long id;
#Column(name = "employee_id")
private Integer employeeId;
#Column(name = "employee_name")
private String employee_name;
#OneToOne
#JoinColumn(name = "company_id")
private Company company;
}
Company class
#Entity
#Table(name = "company")
#Setter
#Getter
public class Company
{
#Id
#GeneratedValue
#Column(name = "company_id")
private Long id;
#Column(name = "company_name")
private String companyName;
}
Company Table
company_id company_name
1 Dell
2 Apple
3 Microsoft
Here is how I am creating the Employee object
Employee emp = new Employee();
emp.setEmployeeId(10);
emp.setEmployeeName("Michael");
Company company = new Company();
company.setId(1);
emp.setCompany(company);
employeeRepository.save(emp);
Please, someone, let me know how to link the company class to an employee class rather than saving one more company.
The best solution for me is to lazy load your company with a proxy. To do it with Spring Data JPA, you need to make your company repository extends JpaRepository. That give you access to the method getReferenceById :
Employee emp = new Employee();
emp.setEmployeeId(10);
emp.setEmployeeName("Michael");
emp.setCompany(companyRepository.getReferenceById(1))
employeeRepository.save(emp);
If there is no company for the given id, an EntityNotFoundException is throw.
With a proxy, you avoid the request to the database in most case because Hibernate use its cache for check the existence of the company. But if my memory serves me correctly, Hibernate gonna make a request to the database at each call to a getter of the proxy, except for getId(). So, in your case it's a good solution, but don't use it all the time.
Assuming the Company may have more than one Employee the relation is a ManyToOne, not a OneToOne.
If you want to reference an existing entity, load it instead of creating a new one:
Employee emp = new Employee();
// ...
emp.setCompany(companyRepository.findById(1));
employeeRepository.save(emp);
I have such a case where I need to have internally many-to-one using hibernate proxies and only id externally, here using MapsId. The issue appears when I try to save something, because the target entity is not fetched, when I set the value only on the id.
Let's take an example: I have an Account table and DeviceConfig table. Inside the DeviceConfig's class definition, I add account in a many-to-one relation and accountId in relation with #MapsId.
Now when creating, I always set a value to accountId, but never the value is picked up, and the backend throws an SQL error, because the field cannot be null.
#Table(name = "djl_device_config")
#Entity
#Getter
#Setter
#ToString
#RequiredArgsConstructor
public class DeviceConfig extends CoreEntity {
...
#JsonIgnore
#ManyToOne
#MapsId("accountId")
#JoinColumn(name = "account_id")
private Account account;
#Column(name = "account_id", insertable = false, updatable = true, nullable = true)
private UUID accountId;
}
So I suppose this is a config error on my side, but I've been reading the JPA for these three days and I still don't know what's wrong or what I should do to achieve the behaviour I expect.
That for any help you'll provide.
I want to get rooms by HospitalId, there relation between classes like it looks below:
#Entity(name = "rooms")
#Table( name = "rooms",
uniqueConstraints = {
#UniqueConstraint(columnNames = "roomNumber")
})
public class Room {
#Id
#GeneratedValue(strategy = GenerationType.TABLE)
private Long idRoom;
private String roomNumber;
#ManyToOne
(fetch = FetchType.EAGER)
#JoinColumn(nullable = false, name = "idhospital")
private Hospital hospital;
this is the query i used :
#Query("select s from rooms s where s.hospital.idHospital =:hospital")
List<Room> findRoomsByHospital( #Param("hospital") Long hospital);
Error in the back :
java.lang.IllegalStateException: Cannot call sendError() after the response has been committed
at org.apache.catalina.connector.ResponseFacade.sendError(ResponseFacade.java:472) ~[tomcat-embed-core-9.0.24.jar:9.0.24]
Is send data but wrong form to the frontend :
error
I really need help can't understand how to deal with it.
I fount the problem :The repitation of data
Solution in to add this notation #JsonIgnore like this :
#ManyToOne
(fetch = FetchType.EAGER)
#JoinColumn(nullable = false, name = "idhospital")
#JsonIgnore
private Hospital hospital;
This is used for to restrict the data to repeat itself
I think you are query is not correct. You have used rooms there without using nativeQuery=true.
you can try below in RoomRepository
#Query("select s from Room s where s.hospital.idHospital =:hospital")
List<Room> findRoomsByHospital( #Param("hospital") Long hospital);
changed rooms to Room
Alternatively, you can try query by a method as shown below, you don't need to use the #Query in this case.
If you are passing hospitalId long use below
List<Room> findRoomsByHospitalIdHospital ( #Param("idHospital") Long hospital);
If you are passing hospital Object use below
List<Room> findRoomsByHospital ( #Param("Hospital ") Hospital hospital);
i'm using Spring Boot 2.4.2 and Data module for JPA implementation.
Now, i'm using an Oracle View, mapped by this JPA Entity:
#Entity
#Immutable
#Table(name = "ORDER_EXPORT_V")
#ToString
#Data
#NoArgsConstructor
#EqualsAndHashCode(onlyExplicitlyIncluded = true)
public class OrderExportView implements Serializable {
private static final long serialVersionUID = -4417678438840201704L;
#Id
#Column(name = "ID", nullable = false)
#EqualsAndHashCode.Include
private Long id;
....
The view uses an UNION which allows me to obtain two different attributes of the same parent entity, so for one same parent entity (A) with this UNION I get the attribute B in row 1 and attribute C in row 2: this means that the rows will be different from each other.
If I run the query with an Oracle client, I get the result set I expect: same parent entity with 2 different rows containing the different attributes.
Now the issue: when I run the query with Spring Data (JPA), I get the wrong result set: two lines but duplicate.
In debug, I check the query that perform Spring Data and it's correct; if I run the same query, the result set is correct, but from Java/Spring Data not. Why??
Thanks for your support!
I got it! I was wrong in the ID field.
The two rows have the same parent id, which is not good for JPA, which instead expects a unique value for each line.
So, now I introduced a UUID field into the view:
sys_guid() AS uuid
and in JPA Entity:
#Id
#Column(name = "UUID", nullable = false)
#EqualsAndHashCode.Include
private UUID uuid;
#Column(name = "ID")
private Long id;
and now everything works fine, as the new field has a unique value for each row.
How can I link an entity that already exists in another table into my important_table? I could insert the ID but then that would require a query. What I want is that the system automatically maps the element in the people_table to the important_table.
#Entity(name = "important_table")
data class ImportantEntity(
#Id #GeneratedValue #Column(name = "id")
val id: Short = 0,
#Embedded
val person: Person
)
Person Entity
#Entity(name = "person_table")
data class PersonEntity(
#Id #GeneratedValue #Column(name = "id")
val id: Long = 0,
...
)
I tried embedded but that creates a duplicate Person in the db. I want the link so that I can find "important" people easy and still get the same data.
Use a #OneToOne mapping to let Hibernate (or any ORM) knows you are linking the tables together.
#Entity(name = "important_table")
data class ImportantEntity(
#Id #GeneratedValue #Column(name = "id")
val id: Short = 0,
#OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
val person: Person
)
When you say: I could insert the ID but then that would require a query, indeed, it does... But with the FetchType.EAGER, it's your ORM that will execute the extra query for you... It is transparent.
Nonetheless, be careful with that, if you don't pay attention to your relationships, you can end up loading the entire db in memory...
As explained in this site, and this tuto, #Embedded is used to help having a nice and clean object definition, while storing the data into one table instead of 2... For instance, in your case, you would be storing the data from the class Person into the table ImportantTable.
Hope it helps !