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Create a new JPA Entity to produce a new object from two tables

Hello programming council, this is my first use of JPA in anger.
I have 2 Tables:
Entity
#Table(name="category")
public class Category {
#Id
#Column(name="id")
#GeneratedValue(strategy=GenerationType.IDENTITY)
private int id;
#Column(name="category")
private String category;
#Column(name="budget")
private double budget;
#Column(name="savings")
private String savings;
#Column(name="archive")
private String archive;
Entity
#Table(name="Transaction")
public class Transaction {
#Id
#Column(name="transaction_no")
#GeneratedValue(strategy=GenerationType.IDENTITY)
private long transactionNo;
#Column(name="transaction_date")
private String transactionDate;
#Column(name="transaction_category")
private String transactionCategory;
#Column(name="transaction_description")
private String transactionDescription;
#Column(name="transaction_amount")
private double transcationAmount;
#Column(name="transaction_auto")
private String transactionAuto;
I want to create a new object called Tile which will contain String category and String balance, the SQL for which would be:
select t.transaction_category as category, sum(t.transaction_amount) as balance
from budgeteer.category c
join budgeteer.transaction t
on c.category = t.transaction_category
group by t.transaction_category;
What is the easiest/best way for me to accomplish this?
Thanks in advance.

Ok, so after a little more research, I discovered that I could actually just do this with the same Entity, repository and service without generating a table. You just need to leave out the #Table annotation when you create your entity.

How to create hibernate composite key and get values from table

I am trying to use hibernate annotations for getting data from a MySQL database table which doesn't have a primary key defined.
However the fact is 2 columns of that table together are unique in the table. How can I achieve the same using hibernate annotation?
This is my code
#Entity
#Table(name = "coc_order_view")
public class CoCOrderDetailsTest {
#EmbeddedId
private MyJoinClassKey key;
#Column(name = "coupon_code")
private String couponCode;
some other columns and their getters and setters .....
#Embeddable
public class MyJoinClassKey implements Serializable {
private static final long serialVersionUID = -5L;
#Column(name = "product_id")
private int productId;
#Column(name = "order_id")
private int orderId;
gettes and setters....
And here is my criteria query
Criteria criteria = getHibernatetemplate().getSession().createCriteria(CoCOrderDetailsTest.class);
criteria.add(Restrictions.eq("status", "New"));
ArrayList<CoCOrderDetailsTest> orderDet = (ArrayList<CoCOrderDetailsTest>) getHibernatetemplate().get(criteria);
I am unable to get all the values from db. Kindly suggest some solutions.

After reading through your question again not sure this will help. You can't have a table without primary key(s). Read the first couple of paragraphs in this article
That said, if you can alter the table and add primary keys on those fields you need to add #IdClass annotation to your class signature for CoCOrderDetailsTest and then get rid of the #embeddable and #embeddedId notation in your classes.
Another alternative, if you can add a field to the table, would be to use an #GeneratedValue on that added primary key field and of course annotate it with #Id.
If you can't alter the table then you can't use JPA and you'll have to use JDBC.
See http://docs.oracle.com/javaee/5/api/javax/persistence/IdClass.html
A working example:
#Entity
#Table(name = "player_game_log")
#IdClass(PlayerGameLogId.class)
public class PlayerGameLog {
#Id
#Column(name = "PLAYER_ID")
private Integer playerId;
#Id
#Column(name = "GAME_ID")
private String gameId;
....
and the id class (note there are no annotations on the id class)....
public class PlayerGameLogId implements Serializable {
private static final long serialVersionUID = 1L;
private Integer playerId;
private String gameId;
Try:
String hql = "FROM CoCOrderDetailsTest WHERE status = :status";
Query query = session.createQuery(hql);
query.setParameter("status","New");
List results = query.list();
I usually use EntityManager rather than session so I'm not familiar with this syntax - and I have typically added a type to the list to be returned - like:
List<CoCOrderDetailsTest> results = query.list();

hibernate - could not execute statement; SQL [n/a] - saving nested object

I'm trying to save a nested object using hibernate and I receive could not execute statement; SQL [n/a] Exception
CODE
#Entity
#Table(name = "listing")
#Inheritance(strategy = InheritanceType.JOINED)
public class Listing implements Serializable {
#Id
#Column(name = "listing_id")
private String listingId;
#Column(name = "property_type")
private PropertyType propertyType;
#Column(name = "category")
private Category category;
#Column(name = "price_currency")
private String priceCurrency;
#Column(name = "price_value")
private Double priceValue;
#Column(name = "map_point")
private MapPoint mapPoint;
#Column(name = "commission_fee_info")
private CommissionFeeInfo commissionFeeInfo;
}
public class MapPoint implements Serializable {
private final float latitude;
private final float longitude;
}
public class CommissionFeeInfo implements Serializable {
private String agentFeeInfo;
private CommissionFeeType commissionFeeType;
private Double value;
private Double commissionFee;
}
public enum CommissionFeeType implements Serializable { }
Using RazorSQL I saw that hibernate defines MapPoint and CommissionFee as VARBINARY
What I can't understand, is the fact that hibernate manages to save it when commissionFeeInfo is not present. It has no problem with saving MapPoint
Does anyone have an idea about what I do wrong?
UPDATE
I found out that if all attributes of CommissionFeeInfo excepting agentFeeInfoare null, the object will be saved without problems. If one of the other attributes is != null, the errors occur.
UPDATE 2
I changed the type of all attributes of CommissionFeeInfo into String and the object will be saved without problem, but I can't let the attributes as String.

I solved the problem by adding setting
#Column(name = "commission_fee_info", columnDefinition = "LONGVARBINARY")
as annotation for the field commisionFeeInfo in the class Listing

For me,
#Column(columnDefinition="text")
solves my problem.

That solution could help for a different reason. One other reason could be Column length. Check your column length. I had the same error the reason was my data exceed the size of the column.
setSignInProvider("String length > 15 ")
Before
#Column(name = "sing_in_provider", length = 15)
and then
#Column(name = "sing_in_provider", length = 100)

I was also facing the same issue . and then I solved the problem
spring.jpa.hibernate.ddl-auto=update

For me I'm using current_date for a field in my sql table. but this is a keyword in SQL so I can't use this name. I changed the field name to current_date_and_time it works for me. also I added the column name on my entity class.
#Column(name = "current_date_and_time")

Joining two table in spring data JPA and Querying two table data from repository

I am using Sprind JPA, Spring 3.1.2(in future 3.2.3), Hibernate 4.1 final.
I am new to Sprind Data JPA. I have tow Table Release_date_type and Cache_media which entities are as follows :
ReleaseAirDate.java
#Entity
#Table(name = "Release_date_type")
public class ReleaseDateType {
#Id
#GeneratedValue(strategy=GenerationType.TABLE)
private Integer release_date_type_id;
#Column
private Integer sort_order;
#Column
private String description;
#Column
private String data_source_type;
#Column(nullable = true)
private Integer media_Id;
#Column
private String source_system; with getters and setters..
and CacheMedia as
#Entity
#Table(name = "Cache_Media")
public class CacheMedia {
#Id
#GeneratedValue(strategy=GenerationType.TABLE)
private Integer id;
#Column(name="code")
private String code;
#Column(name="POSITION")
private Integer position;
#Column(name="DESCRIPTION")
private String media_Description; with setter and getters.
Now my repository interface is as follows :
public interface ReleaseDateTypeRepository extends CrudRepository<ReleaseDateType, Long>{ }
Now i want to write a method(Query) in ReleaseDateTypeRepository interface which can get all the data from Release_Date_Type table including appropriate media_description from Table 'Cache_Media' based on media_id of Release_date_type table.
So my select (SQL)query looks like
SELECT * from Release_Date_Type r left join Cache_Media c on r.media_id=c.id
I don't know how to map entities.
I tried so many thing but not luck.
Any help is appreciated.

Its not the answer for joining via Hibernate, but alternatively you can create a view with your join and map the view to your objects

Hibernate Criteria select using embedded object (tuple)

In my case I have a SQL query which looks like:
select * from event_instance where (object_id, object_type) in
(<LIST OF TUPLES RETRIEVED FROM SUBQUERY>);
I want to map this on Hibernate Entities and I have a problem with this query. My mapping looks like that:
#Entity
#Table(name="event_instance")
public class AuditEvent {
<OTHER_FIELDS>
#Column( name = "object_type", nullable = false)
private String objectType;
#Column( name ="object_id" , nullable = false)
private Integer objectId;
}
and second entity:
#Entity
#Table(schema = "els" ,name = "acg_objects")
public class AcgObject implements Serializable{
#Id
#Column(name = "acg_id")
private String acgId;
#Id
#Column(name="object_type")
private String objectType;
#Id
#Column(name="object_id")
private Integer objectId;
<OTHER FIELDS>
}
I already run query for getting AcgObjects and for my DAO I'm getting List only thing I want to do is query a touple using criteria like:
crit.add(Restrictions.in("objectType,objectId",<List of tuples>);
Is it possible? I was trying to use #Embedded object but don't know how exactly construct a query for it. Please help

You can do that not in standard SQL nor using criteria; you have to split in two distinct restrictions or using a Session.SQLQuery() if you want to use specific RDBMS (look at SQL WHERE.. IN clause multiple columns for an explanation)