I read that specifying optional = false in the #ManyToOne association annotation could help Spring improve the performance of the queries.
In a Kotlin data class entity, do I actually need to specify the parameter in the annotation, or can Spring figure this out by itself using the nullability of the item field?
For instance, if I have the following declaration:
#Entity
#Table(name = ACCESS_LOGS_ARCHIVES_TABLE, indexes = [
Index(name = "access_logs_archives_item_idx", columnList = "access_item_id")
])
data class AccessLogArchive(
val date: LocalDate,
#ManyToOne(optional = false)
#JoinColumn(name = "access_item_id", nullable = false)
val item: AccessLogItem,
val occurrences: Int
) {
#Id
#GeneratedValue
var id: Long? = null
}
#Entity
#Table(name = ACCESS_ITEMS_TABLE)
data class AccessLogItem(
#Column(length = 3) val code: String,
#Column(columnDefinition = "text") val path: String,
#Column(length = 10) val verb: String
) {
#Id
#GeneratedValue
var id: Long? = null
}
In this case, I would for instance expect Spring to know that the item field is not nullable, and thus the relationship should be understood as optional=false even without specifying it as I did. Is this the case?
Same question goes for the #JoinColumn's nullable = false, by the way.
Consider a simple entity like a Room which has a #ManyToOne relationship to House.
#Entity
class Room(
#ManyToOne(optional = true)
val house: House
) {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
val id: Long = 0
}
JPA will create a room table with a column
`house_id` bigint(20) DEFAULT NULL
If you specify #ManyToOne(optional = false)
the column will look like this:
`house_id` bigint(20) NOT NULL
By specifiying optional you tell JPA how the schema should be generated, whether the column can be NULL or not.
At runtime trying to load a Room without a House will cause an Exception if the house property is not nullable (House instead of House?) even when value of optional is true.
The same applies to #JoinColumn.
Is #ManyToOne's “optional” param automatically set using Kotlin's
nullability?
No it is not. It is independent from that and by default set to true.
Conclusion: In order for you schema to reflect your entities it is a good idea to use optional = true if the house property would be nullable and optional = false if the house property would be non-nullable.
Related
I have the following entities in a fairly simple and straightforward Spring Boot application in Kotlin:
#Entity
class Target(
#Id #GeneratedValue var id: Long? = null,
// ... other stuff
)
#Entity
class Ruleset(
#OneToOne(fetch = FetchType.LAZY) #MapsId
var target: Target,
#Id #GeneratedValue var id: Long? = null,
// ... other stuff
)
And I have the following code to create them upon startup of a #Component:
#PostConstruct
#Transactional
fun init() {
val target = Target()
targetRepository.save(target)
val rule = Ruleset(target)
rulesetRepository.save(rule)
}
And when this runs I get the "detached entity passed to persist: com.mystuff.Target" error. I've used this approach in the past (see here: https://vladmihalcea.com/the-best-way-to-map-a-onetoone-relationship-with-jpa-and-hibernate/) without issue, although never in trying to create them at the same time in the same method. I've also tried using the entity passed back by the .save() call on the Target repository in the persist of the Ruleset object with no success.
I am able to fix this if I go back to the "normal" way of doing a OneToOne relationship:
#Entity
class Target(
#OneToOne(mappedBy = "target", cascade = [CascadeType.ALL],
fetch = FetchType.LAZY, optional = false)
var ruleset: Ruleset?
#Id #GeneratedValue var id: Long? = null
)
#Entity
class Ruleset(
#OneToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "target_id")
var target: Target,
#Id #GeneratedValue var id: Long? = null,
)
But this is annoying as it forces me to pass a null into the Target constructor and then update it immediately after creating the Ruleset. I can't figure out why the other, simpler approach doesn't work.
I am facing a wierd issue in my implementation where I am persisitng data to a PostgresSQL DB using Spring data JpaRepository
In my Entity class I have the below columns:
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", unique = true, nullable = false)
private int id;
#Column(name = "field1", nullable = false, length = 16)
private String field1;
#Column(name = "field2", nullable = false, length = 16)
private String field2;
#Column(name = "field3", nullable = false, length = 16)
private String field3;
I initially avoided declaring the fields above as composite since there were many fields to be dealt with as composite keys. I thought the java code check would do the trick in all scenarios
So basically, I have to maintain the uniqueness of each row based on field1,field2 and field3. That was the basic requirement for which I had checks in my java code that if any entry exists in the DB for the combination of field1,field2 and field3 then I used to throw java exceptions
No two rows can have these values repeating. All was good until the application was tested under some errorneous business scenarios which would never happen in production but got run by mistake
Whats happening now is that if 2 requests are triggered at the exact same instance with the exact same 3 fields above (through a script) then they both enter into the Database since both get the entry check as false
Would declaring all of them as one composite key resolve the situation?
You should define the unique constraint in your database in addition of JPA constraint.
#Entity
#Table(uniqueConstraints={
#UniqueConstraint(columnNames = {"field1", "field2", "field3"})
})
public class MyEntity {
...
}
I'm working on oracle database to manage a JPA entity with a String Primary key.
I cannot modify the type on the PK to a Long or int in the database, so i want to know how to configure the pk sequence in my JPA entity,
i've tried this :
#Id
#SequenceGenerator(name="SEQ_ID", sequenceName = "SEQ_ID" )
#GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SEQ_ID")
#Column(name="SEQ_ID",unique=true, nullable = false,updatable = false)
private String id;
but when persisting a new entity i got the error : Unknown integral data type for ids : java.lang.String
someone can help me please ?
Try removing #GeneratedValue and #SequenceGenerator
Also, a remark, #Id will automatically set unique=true, nullable = false,updatable = false so you can remove them from #Column.
Otherwise, you can check this article for more details about creating a custom string generator https://vladmihalcea.com/how-to-implement-a-custom-string-based-sequence-identifier-generator-with-hibernate/
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 !
I have written a spring boot app with oracle db.
Below is my entiry class.
#Entity
public class SystemTypeLookup{
#Id
#GeneratedValue(generator = "UUID")
#GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
#Type(type = "uuid-char")
#Column(name = "ID", updatable = false, nullable = false)
protected UUID id;
#Column(name = "CODE")
private String code;
}
And in passing my own UUID as primary key value.
In oracle db ID is considered as RAW and the UUID stored in oracle is differently.
There is no - separation in oracle and all the UUID chars are in upper case.
When i try to find the entity using primary key it is not fetching the row with id. I'm always getting null.
#Resource(name = "coreRepository")
private ErpEntityRepository coreRepositoryBase;
SystemTypeLookup systemTypeLookup = coreRepositoryBase.findOne("WHERE o.id='"+id+"'", SystemTypeLookup.class);
when is pass 76c03cd9-3d96-40c5-8df9-aad8f2369453 as id value then the oracle will insert the id without '-' and all chars will be in upper case.
So how to solve this issue?
First of all you should use parameters in your query and second make sure that the id you are passing is of type UUID and not String.