Spring-Boot 2.0.0 seems to have modified the way Hibernate is auto configured.
Let's suppose two simple and independent JPA entities:
#Entity
class Car {
#Id
#GeneratedValue
private long id;
//....
}
#Entity
class Airplane {
#Id
#GeneratedValue
private long id;
//....
}
Prior, using Spring-Boot 1.5.10, I was able to generate separate sequences of auto-increments, meaning that I can get a Car with 1 as primary key and an Airplane with 1 as primary key too.
No correlation between them, e.g no shared sequence.
Now, with 2.0.0, when I sequentially create a very first Car then a very first Airplane, the car gets 1 as id and airplane gets 2.
It seems that he has to deal with the GeneratedType.AUTO, that is the "used by default" specified within the #GeneratedValue annotation source.
However, my reasoning seems to stop here since GeneratedType.AUTO was also set as default with the 1.5.10.
A simple workaround to fulfil my expectation is to specify the IDENTITY strategy type of generation like so:
#Entity
class Car {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
//....
}
#Entity
class Airplane {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
//....
}
I can't figure out an explanation of this behavior.
What has Spring-boot 2.0.0 changed, explaining this scenario?
Spring Boot 2.0 uses Hibernate 5.2 (https://github.com/spring-projects/spring-boot/wiki/Spring-Boot-2.0-Release-Notes).
Hibernate changes its GeneratedType.AUTO strategy since 5.2. Any database that does not support sequences natively (e.g. MySQL), they use the TABLE generator instead of IDENTITY. (https://hibernate.atlassian.net/browse/HHH-11014)
That's why GeneratedType.AUTO does not work as you expected.
You can use
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
to use MySQL autoincrement.
If you are in need for a quick, not future-proof solution to prevent this issue from happening:
spring.jpa.hibernate.use-new-id-generator-mappings=false, as from the Spring Boot 2 docs:
spring.jpa.hibernate.use-new-id-generator-mappings= # Whether to use Hibernate's newer IdentifierGenerator for AUTO, TABLE and SEQUENCE.
This will prevent from using the new generators and keep the old functionality included in Spring boot 1.x.x.
Please note that this is probably not the best solution, but it is very helpful on short term
As Andrew has pointed out in the comment, if you don't want the id to be incremented while values are created in other tables, you can specify your ID like this:
#Id
#GeneratedValue(
strategy= GenerationType.AUTO,
generator="native"
)
#GenericGenerator(
name = "native",
strategy = "native"
)
private Long id;
Doing this will make each table has its unique id beginning with 1,2,3 ... and so on.
By default spring-boot uses the auto and increment the value based on the order the objects are saved.
To provide unique id based on each object, use the following
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
Related
I have the next issue -> I have a table on my db with a composite id...
Supose (Id1,Id2,Id3,Id4), the Id4 is generated by a sequence on the db...
My question is, in spring boot, I generate the entity 'Table1' and the corresponding 'Table1Id',
but when i want to add the corresponding GeneratedValue from the sequence, is not generating anything.
I was looking for in the internet and i found that the GeneratedValue is not working without the #Id anotation, but maybe there are some way to fix this issue.
Thank's and sorry for my english.
SOLVED:
When a composite id is required in your project, it is impossible with an embeddedId. It is necesary to use #IdClass on my compossiteId instead #EmbeddedId, because the second does not work with #GeneratedValues for example my solution was:
#Data
#Entity(name = "table_name")
#IdClass(CompositeIdTest.class)
public class TestClass implements Serializable {
#Id
#Column(name = "column", nullable = false)
private String column;
#Id
#SequenceGenerator(name = "sequence", sequenceName = "sequence", allocationSize = 1)
#GeneratedValue(generator = "sequence")
private int idGenerated;
Anyway, thank's
consider following model:
#Entity
#Getter
#Setter
public class Person {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#ElementCollection
#Column(name = "phone")
private List<String> phoneNumber = new LinkedList<>();
}
In front-end user should be able define order of phone numbers in this sample with drag and drop or something like that. Can you tell me what is the most effective way to handle this use case? I found solution with jpa annotation #OrderColumn which generates additional column. But I guess if I need re-order items, solution is delete all from collection and save it again with new order right? I afraid that is not very elegant solution. Can you give me your advice? Thank you.
I'm using #version annotation in spring data so I have a parent entity, and it has list of child entity. when I delete an element from child list the parent version doesn't increase. can anyone clarify for me this #version alternative,
why the versing in this case doesn't increase, is it a good way to manage versioning or should I use trasaction "lock".
in the documentation i read that the version update only on updating a row
in the databse but in my case i put version on parent entity and i want
note: i searched a lot in the internet but i didnt find a clear solution, can any one help me.
I assume you are using Hibernate. Lets say that the "UnderlyingPerTradingAccount" table has a column called "trading_account_id", which is a foreign key to the TradingAccount table. In order to achieve the behavior you described, you need to change the mapping. Can you try this:
public class TradingAccount {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
#JoinColumn(name="trading_account_id", referencedColumnName = "trading_account_id", insertable = false, updatable = false)
private List<UnderlyingPerTradingAccount> underlyingPerTradingAccounts;
#Version
private Long version;
}
and
public class UnderlyingPerTradingAccount {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#ManyToOne
#JoinColumn(name="trading_account_id", nullable = false)
private TradingAccount tradingAccount;
private Boolean enableBuy;
private Boolean enableSell;
}
This should mark the parent entity as "dirty" when the child entity is updated and trigger the version increment.
However, I would think of some other method to track "version" changes of the parent entity as this would just cause an additional overhead and update statements to the parent.
I'm using Spring Boot JPA with Gradle. I'm struggling to find a guide that I can follow which focusses on creating a relational database with the correct Syntax for Spring Boot. I had a go but I get this error
No property idTestCase found for type TestRun!
I want TestRun and TestData entities with a OneToOne relationship with each other, and a TestCase entity that has a OneToMany relationship with TestRun. I reckon that TestRun should contain the foreign keys for TestData and TestCase.
Many times I make changes and it will not build, and when it does build the tables do not look correct, this is what I created:
#Entity
public class TestRun {
#Id #GeneratedValue(strategy = GenerationType.AUTO)
private long testRunId;
private Boolean result;
#OneToOne #JoinColumn(name="testData_id")
private TestData testData;
#ManyToOne #JoinColumn(name="testCase_id")
private TestCase testCase;
}
#Entity
public class TestCase {
#Id #GeneratedValue(strategy = GenerationType.AUTO)
private long testCaseId;
private String name;
private String description;
#OneToMany(cascade=CascadeType.ALL, mappedBy="testCase",targetEntity=TestRun.class)
private Collection<TestRun> testRun;
}
#Entity
public class TestData {
#Id #GeneratedValue(strategy = GenerationType.AUTO)
private long testDataId;
#OneToOne(cascade=CascadeType.ALL, mappedBy="testData",targetEntity=TestRun.class)
private TestRun testRun;
}
From the guides it isn't clear to me what goes in #JoinColumn(name= some say it needs to link to a field on your POJO and some say it doesn't. If I create the foreign key field in my POJO then I get two foreign key fields in the database table and if I don't it doesn't build at all.
For example from the error I can infer it wants me to add the following fields to TestRun:
private long idTestCase;
private long idTestData;
But then my database appears as:
SELECT * FROM TEST_RUN;
TEST_RUN_ID
ID_TEST_CASE
ID_TEST_DATA
RESULT
TEST_CASE_ID
TEST_DATA_ID
(no rows, 3 ms)
I tried setting #JoinColumn(name= to the name of the primary key field on the other side of the relationship but again it did not build.
Many thanks
I found the guide at JBoss to be the most helpful in describing the different mappings.
I have created many entities in jpa.
When i checked in the database, i don't see any foreign key.
#Entity
public class Lodger implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long lodgerId;
#OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "lodger")
private List<AccountOperation> accountOperationList;
...
}
#Entity
public class AccountOperation {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private long accountOperationId;
#ManyToOne
#JoinColumn(name = "lodger_id")
private Lodger lodger;
...
}
In this example i was thinking to get a foreign key in the account operation class.
table automaticaly created
http://www.wepaste.com/table_example/
Why?
Hibernate does not automatically generate foregein keys when generating dlls. I would recommend to turn of the generate-dll option as it may create inconsistent databases as the complexity increases. also check out either spring boot default database administration options:
Flyway
Liquibase
The main difference between the two relies in the fact that while both may be administrared with SQL, Liquibase offers a more database agnostic formats such as XML, and YML for the creation of your database