I have an entity with a composite key
#Entity
data class Page(
#EmbeddedId
val pageId : PageId,
...
)
#Embeddable
data class PageId (
#Column(name = "id")
val id: UUID,
#Column(name = "is_published")
val isPublished: Boolean
)
But I need to respect the existing column names in the db table, which are 'id' and 'is_published'
But querying the db with a JDBCRepository I get the error:
SQL Error executing Query: ERROR: column page_.page_id_published does
not exist
Is there any way that I can map the columns correctly?
Try and error led me to the answer, somehow Micronaut does not like a Boolean to be named 'isPublished', when I rename it to 'published' it works fine:
data class PageId (
#MappedProperty(value = "id")
val id: UUID,
#MappedProperty(value = "is_published")
val published: Boolean)
Related
we're using Postgres and JPA/Hibernate to import a lot of data on a biweekly basis (~50-100M rows per import). We're trying to partition our tables per import, which has us running into some Hibernate PK/FK column mapping problems. The setup is essentially this on the SQL side
CREATE TABLE row (
import_timestamp timestamp,
id uuid,
PRIMARY KEY (import_timestamp, id)
) PARTITION BY LIST (import_timestamp);
CREATE TABLE row_detail (
import_timestamp timestamp,
id uuid,
row_id uuid,
PRIMARY KEY(import_timestamp, id),
CONSTRAINT row_detail_row_fk FOREIGN KEY (row_id, import_timestamp) REFERENCES row (id, import_timestamp)
) PARTITION BY LIST (import_timestamp);
and this on the Java side:
#Entity(name = "row")
public class RowEntity {
#EmbeddedId
private PartitionedId id;
#OneToMany(cascade = ALL, mappedBy = "row")
private List<RowDetailEntity> details;
}
#Entity(name = "row_detail")
public class RowDetailEntity {
#EmbeddedId
private PartitionedId id;
#ManyToOne
#JoinColumns({
#JoinColumn(name = "row_id", referencedColumnName = "id"),
#JoinColumn(name = "importTimestamp", referencedColumnName = "importTimestamp")
})
private RowEntity row;
}
#Embeddable
public class PartitionedId implements Serializable {
private Instant importTimestamp;
private UUID id;
}
Hibernate then complains on boot that:
column: import_timestamp (should be mapped with insert="false" update="false")
I can silence that error by doing as it says, but that makes little sense, because I am forced to set insertable=false and updatable=false for both #JoinColumn()s, which would mean row_id isn't populated on insert.
I could go the #MapsId route, but only if I give the row_detail table a PK that includes all 3 properties (import_timestamp, id, row_id), and I don't really want or need that.
So the question is, how do I get Hibernate to understand my overlapping, but not entirely nested PK/FK?
I am trying to save an entity that has a many-to-many association to another entity and cascade the persistence to the associated entity and create the association using spring data jpa repository.
I can insert the parent entity_a which contains a set of entity_b using entityARepository.save(entityA). Spring jpa is taking care of all the inserts needed in the transaction. All the entity_b's get inserted, entity_a's get inserted and the join table in the middle has the association inserted as well. If I update the same entity_a with a new value in, say timestamp column, the same entityARepository.save(entityA) handles this and does a corresponding update.
The problem happens when there already exists entity_b (which has an association between some entity_a) and I try to insert a new entity_a with the same entity_b. It is many to many so this is how the data model is supposed to be. But instead of updating the existing entity_b during this entityA save() transaction, it tries to do inserts on entity_b and a constraint violation exception on the primary key is thrown.
org.springframework.transaction.TransactionSystemException: Could not commit JPA transaction; nested exception is javax.persistence.RollbackException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.0.v20150309-bf26070): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLIntegrityConstraintViolationException: ORA-00001: unique constraint (USER1.SYS_C0013494) violated
Error Code: 1
Call: INSERT INTO ENTITY_B (ID, NAME, VALUE, TIME_STAMP) VALUES (?, ?, ?, ?)
bind => [4 parameters bound]
Query: InsertObjectQuery(EntityB [name=shape, value=circle])
The problem is that spring doesn't have update(). It only has save which should handle update if it receives the same primary key. It's not doing that when a new entity_a is saved and has a collection of entity_b, if any entity_b exists, the whole transaction is failing sure to primary key constraint violation of entity_b.
public class EntityA {
#Id
#SequenceGenerator( name = "EntityASeq", sequenceName = "SQ_ENTITY_A", allocationSize = 1, initialValue = 1 )
#GeneratedValue(strategy = GenerationType.IDENTITY, generator = "EntityASeq")
#Column(name = "ID")
private Integer id;
#ManyToMany(cascade = {CascadeType.PERSIST, CascadeType.MERGE}, fetch = FetchType.LAZY)
#JoinTable(name = "MY_JOINED_TABLE",
joinColumns = {
#JoinColumn(name = "a_id", referencedColumnName = "ID")},
inverseJoinColumns = {
#JoinColumn(name = "b_id", referencedColumnName = "ID")})
private Set<EntityB> attributes;
// These three columns below have a unique constraint together.
#Column(name = "name")
private String name;
#Column(name = "tenant")
private String tenant;
#Column(name = "type")
private String type;
#Column(name = "timestamp")
private Timestamp timestamp;
}
public class EntityB {
#Id
#SequenceGenerator( name = "EntityBSeq", sequenceName = "SQ_ENTITY_B", allocationSize = 1, initialValue = 1 )
#GeneratedValue(strategy = GenerationType.IDENTITY, generator = "EntityBSeq")
#Column(name = "ID")
private Integer id;
#ManyToMany(mappedBy = "attributes")
private Set<EntityA> aSet;
// These two columns below have a unique constraint together.
#Column(name = "name")
private String name;
#Column(name = "value")
private String value;
#Column(name = "timestamp")
private Timestamp timestamp;
}
The id for each is generated by default. I also have a unique constraint on a few columns, which means if an EntityB has the same name/value as an existing one in the database, I want to just update the timestamp. That works if entity_a is already in the table and it has the same entity_b's. A and B's timestamp are updated and no error when I persist with entityARepository.save(entityA). (I do some checking on the db with findOne because the id is auto generated an not known. So if a name/value exist, I don't try to insert with a new id, I use the same one in the db and it works (similarly with entity_atenant/name/type.
It also works when I persist an existing entity_a with updated entity_b's. So if a new entity_b is associated with entity_a (that exists as an association with a different entity_a), etc, that works and the persistence is working.
The issue again, is just on INSERT of entityA via repo.save() when some entity_b
s already exist for other associations. It should be doing:
INSERT INTO entity_a ...
UPDATE entity_b ...
INSERT INTO MY_JOINED_TABLE ...
But it seems like it's doing
INSERT INTO entity_a ...
INSERT INTO entity_b ... -- fails because primary key constraint fails
INSERT INTO MY_JOINED_TABLE ...
EDIT: I tried removing CascadeType.PERSIST but I get an error saying
During synchronization a new object was found through a relationship that was not marked cascade PERSIST: EntityB [name=color, value=blue].
I wanted to try to manually insert/update but I couldn't do that. It wants me to have the EntityA specified with PERSIST because it has associations to the entityB
I tried inserting in the reverse and now I'm having issues inserting from entityB.save() when there already exists some entityA and I'm adding a new entityA to entityB
I have the following structure to store data for tracking devices (reduced complexity to simplify) in a MySQL database.
I use Hibernate to handle this data in a web application.
Tables structure
tag
id
name (string)
last_tag_detail_id (foreign key on tag_detail table)
tag_detail
id
tag_id (foreign key on tag table)
date (UTC date)
lat (gps latitude, may be null if no GPS position)
lng (gps longitude, may be null is no GPS position)
Problem
I have a requirement that when I retrieve a tag, I should get the last tag_detail (which I have thanks to the tag_detail_id foregin key).
Now I also have to get the most recent tag_detail with a valid GPS position (latitude and longitude != null).
Is it possible to do so with Hibernate annotations without having to add a last_tag_detail_gps_id foreign key in the tag table ?
For now I have the following mapping :
Tag class
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id", unique = true, updatable = false, nullable = false)
private Long id;
#Column(name = "mac")
private String name;
#OneToOne(fetch = FetchType.LAZY, cascade = { CascadeType.MERGE, CascadeType.REFRESH })
#JoinColumn(name = "last_tag_detail")
private TagDetail lastTagDetail;
TagDetail class
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id", unique = true, updatable = false, nullable = false)
private Long id;
#ManyToOne(fetch = FetchType.EAGER, cascade = { CascadeType.REFRESH, CascadeType.MERGE })
#JoinColumn(name = "tag_id", nullable = false)
private Tag tag;
#Column(name = "lat")
private BigDecimal latitude;
#Column(name = "lng")
private BigDecimal longitude;
#Column(name = "data", nullable = false)
#Type(type = "org.jadira.usertype.dateandtime.joda.PersistentDateTime")
private DateTime date;
Is it possible without adding a foreign key in the tag table to add a lastTagDetailWithGPS field in the Tag class, which would retrieve the last TagDetail by date which have a non null latitude and longitude ? If so, what would be the proper Jpa or Hibernate annotation to do so ?
Thank you
You say you wanted to:
retrieve the last TagDetail by date which have a non null latitude and
longitude
I don't think you need to add an additional foreign key to your tag table, since in your Tag entity you already have a reference to a TagDetail entity mapped as one-to-one. In your JPA query, you can just simply traverse to your TagDetail entity and check on the values of its properties: latitude, longitude and date.
Here's your JPA query:
SELECT td FROM Tag t INNER JOIN TagDetail td
WHERE td.date = :date AND
td.latitude IS NOT NULL AND
td.longitude IS NOT NULL
I have been trying to figure out how to do this for sometime without any luck and have not managed to find anything useful while search on Google either.
I have THREE tables:
HOTEL
- id
- name
- local_id (foreign key)
DESCRIPTION
- id
- description
- hotel_id (foreign key)
- locale_id (foreign key)
LOCALE
- id
- local
I also have the following HOTEL DAO model:
#Entity
#Table(name = "HOTEL")
public class Hotel implements Serializable {
#Column(name = "id")
private long id;
#Column(name = "description")
private HotelDescription description;
}
Using JPA, how can I retrieve the data from table DESCRIPTION based on hotel_id and locale_id to populate description in DAO model hotel?
Well, you also have HotelDescription JPA entity, right? So you can define bidirectional mapping for entities.
instead of
#Column(name = "description")
private HotelDescription description;
you should have something like
#OneToOne(mappedBy = "hotel", cascade = CascadeType.ALL)
private HotelDescription desc;
and on the other side, in HotelDescription you should have back mapping
#OneToOne
#JoinColumn(name = "hotel_id")
private Hotel hotel;
When you will extract Hotel entity, JPA will also fetch child entity (HotelDescription) for you.
if you want to use #OneToMany mapping it will be (many descriptions for one hotel)
#OneToMany(mappedBy = "hotel", cascade = CascadeType.ALL)
private HotelDescription desc;
and on the other side
#ManyToOne
#JoinColumn(name = "hotel_id")
private Hotel hotel;
In JPA you can use several types of mapping like OneToMany, ManyToMany... That's only basics. Find a tutorial. You may start here: http://docs.oracle.com/javaee/6/tutorial/doc/bnbqa.html (not the best one probably)
Oh. And make sure you annotate id with #Id
I would consider ditching the Locale table and working with java.util.Locale directly. Hibernate (not sure about other JPA implementations) has auto type conversion from char column to java.util.Locale. This would then look something like:
DESCRIPTION
- id
- description
- hotel_id (foreign key)
- locale
And Entity:
import java.util.Locale;
#Entity
#Table(name = "HOTEL")
public class Hotel implements Serializable {
#Column(name = "id")
private long id;
#OneToMany
#JoinColumn(name = "holiday_id", nullable = false)
#MapKeyColumn(name = "locale_id")
private Map<Locale, HotelDescription> descriptions;
public String getDescriptionForLocale(Locale locale){
//try an exact match e.g. en_us
if(descriptions.containsKey(locale){
return descriptions.get(locale).getDescription();
}
//try language only e.g. en
else if (decsriptions.containsKey(locale.getLanguage){
return descriptions.get(locale.getlanguage()).getDescription();
}
//return a default or null
return ??
}
}
How do I represent similar multiple row data in one hibernate pojo?
For example lets have a table say
PKEY | REFS | UNIQUEID
Now I envision a scenario in which I expect multiple REFS for same set of {PKEY, UNIQUEID} values. This will cause multiple rows in the database with same PKEY and UNIQUEID but differing REFS.
In such scenario how do I map that table to my hibernate pojo. Will the following work?
#Entity
#Table(name = "MYTABLE")
public class TablePojo
{
#Id
#Column(name = "PKEY")
int pkey;
#Column(name = "REFS")
List<String> refsList;
#Column(name = "USERID")
Long userId;
..........
..........
..........
};