I have 2 diferent Entities:
USER:
#Entity
public class UserMW {
#Id
private Long id;
private String name;
private Long score;
...
}
USER_CHALLENGE
#Entity
public class UserChallengeMW {
#Id
private Long id;
#Index
private Ref<UserMW> user;
#Index
#Load
private Ref<ChallengeMW> challenge;
}
I want to be able to get one ChallengeMW object and query all users which have done this ChallengeMW. So far it's pretty simple. I just need to query USER_CHALLENGE filtering by "Ref challenge". Something like this:
ofy().load().type(UserChallengeMW.class).filter("challenge", challengeRef).list();
The problem is, I want to order it according to the property "score" into USER entity. Any idea about what is the correct way to go?
Unfortunately, there are no joins in the datastore. Either denormalize the 'score' into the UserChallengeMW object or query for all the data and sort it in memory.
Related
I'm working on saving schedule for shop into the db.
public class Shop {
#Id
private long id;
}
public class Schedule {
#Id
private long id;
#Column
#Enumerated(EnumType.STRING)
private DayOfWeek weekday;
#Column(name="time_from")
private Time from;
#Column(name="time_to")
private Time to;
#ManyToOne
#JoinColumn(name="shop_id")
private Shop shop;
}
What I want, is to have schedule available within Shop instance (one shop can have multiple to and from times, for example 09:00-13:00, 14:00-18:00), like this
public class Shop {
#Id
private long id;
private Map<DayOfWeek, List<Schedule>> scheduleList = new HashMap<>();
}
but I don't know how to map that. I was looking through the JPA 2.1 wiki page (https://en.wikibooks.org/wiki/Java_Persistence/Relationships#Nested_Collections.2C_Maps_and_Matrices), but the the criteria field is ENUM, and I think that there is no reason to create an extra table with 7 rows (for each weekday). Does anyone work with such features?
EDIT: the possible solution is to divide weekday to separate table:
Table shop
id
Table week_day
id
name
shop_id
Table hours
id
time_from
time_to
week_day_id
In this case the example from wiki would work, but is there any solution to use instead of intermediate table just enums?
If you used Java 8, then you could go with accessing the schedules on the shops with a simple inverse OneToMany relationship in your Shop object
#OneToMany
private List<Schedule> scheduleList;
and then creating a method like
public Map<DayOfWeek, List<Schedule>> getScheduleListByDaysOfWeek() {
return scheduleList.stream()
.collect(Collectors.groupingBy(Schedule::getWeekday);
}
I'm trying to build a Spring Boot data layer on top of another project's DB. I'm want to get to a point where I can consume their data via Restful endpoints rather than directly from the DB. Maximum abstraction is the goal. Here's my problem. Consider the following JPA entity:
#Entity
#Table(name = "PERSON", schema = "public")
public class Person {
#Id private long id;
private String name;
private long favoriteFood;
private Address address;
//Getters, Setter etc.....
}
Notice that favoriteFood is a long, not a String. This is because the DB uses a lookup table. Let's say Joe's favorite food is pizza. The person table stores a 1 in the favorite_food column which is the fk to the "pizza" value stored in the food_ref table. This pattern is repeated hundreds or times in the DB. What is the best way to model this in JPA/Hibernate? Change the variable to String and have the getter and setter do the lookup? I've not found any examples which seems strange. This is a common DB structure. Any advice on best practices would be appreciated. Thanks!
The best way in this scenario is to use one to one relationship in the JPA entity with the FoodRef class
#Entity
#Table(name = "PERSON", schema = "public")
public class Person {
#Id private long id;
private String name;
#OneToOne(fetch=FetchType.LAZY)
#JoinColumn(name="food_ref_id")
private FoodRef favoriteFood;
private Address address;
//Getters, Setter etc.....
}
i'm curious about how HQL would assert equality between an entity instances.
Let's say I have a Entity called Person
#Entity
public class Person{
#Id
private Long id;
private String name;
}
and Department
#Entity
public class Department {
#Id
private Long id;
#ManyToOne
private Person person;
}
then it's fine if I do the following statement:
Query query = getSession().createQuery("from Department d where d.person = ?");
query.setProperty(0,new Person(1L));
but, what if I have an Embedded entity and no pk defined? like
#Embeddable
public class Adress {
private String email;
private String street;
private Long identifier;
}
#Entity
public class Person{
#Id
private Long id;
private String name;
#Embedded
private Address address;
}
would have any way so I could tell JPA to make it work:
Query query = getSession().createQuery("from Person p where p.address = ?");
query.setProperty(0,new Address(1L));
even though it's not exactly a primary key?
For sure i know i'd work if I tried p.adress.identifier, and then passed just the Long value, but the point is, can I tell JPA provider how it's gonna kind of 'implement' equality my way?
Thank you all
No, it is not supported and it would be difficult in general or would not make sense in some situations, like when there are collections in the Embeddable.
If you find that you need this often though, consider converting such Embeddables to custom user types. Then you can perform comparisons the way you described.
Assuming theses Entities
#Entity
public class EntityNote implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#SequenceGenerator(name="SeqEntityNote", sequenceName="SeqEntityNote", allocationSize = 1)
#GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SeqEntityNote")
private long id;
private Date date;
private String subject;
private String content;
#ManyToMany
private List<EntityTopic> listEntityTopic;
//setters/getters
#Entity
public class EntityTopic implements Serializable {
#Id
#SequenceGenerator(name="SeqEntityTopic", sequenceName="SeqEntityTopic", allocationSize = 1)
#GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SeqEntityTopic")
private long id;
private String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
In my DB, a join table named "entity_note_list_entity_topic" records the ManyToMany relation.
This works correctly so far.
But I'd like to perform a count query like 'how many EntityNotes per EntitityTopic'
Unfortunatly I'm quite lost in this situation.
How this query can be written ?
Do I need other elements in my two entities ?
(In many examples I see a reverse relation using mappedBy attribute on ManyToMany.. Do I need this ?)
It will be the easiest if you make the many to many relation bidirectional. There are no serious extra costs involved, as it uses the same db structure, and the list are lazy loaded so if the relation is not being used the lists are not populated (you can hide the second direction by making accessors private).
Simply change:
#Entity
public class EntityTopic implements Serializable {
...
#ManyToMany(mappedBy="listEntityTopic")
private List<EntityNote> notes;
}
You can issue normal count jpql queries, for example:
SELECT count(n) from EntityTopic t INNER JOIN t.notes n where t.name =:name
so you don't neet to retrieve the notes and topics if don't need to.
But I also believe that your original mapping can also be queries with:
SELECT COUNT(n) FROM EntityNote n INNER JOIN n.listEntityTopic t WHERE t.name = :name
If you have the following code:
#Entity
public class EntityNote implements Serializable {
#ManyToMany(fetch = FetchType.LAZY)
private List<EntityTopic> topics;
}
#Entity
public class EntityTopic implements Serializable {
#ManyToMany(fetch = FetchType.LAZY)
private List<EntityNote> notes;
}
Then, topic.getNotes().size() will give you the number of notes associated with a topic. When using Hibernate as the JPA provider, a SELECT COUNT(...) query is issued for this instead of loading all the associated notes. If this does not work for you out-of-the-box, mark the collections as extra lazy using the instructions in this post.
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