In my spring mvc app, i have the following object. I am trying to make a visual of data using devtool in my app.
#Entity
#Data
public class ConsultationRequest {
#Id
#GeneratedValue
private Long id;
private String name;
private String email;
private String purpose;
private String programme;
private int year;
private String language;
private String comments;
#Enumerated(EnumType.STRING)
private ConsultationStatus status;
}
Then i used the jpa to make the entity:
#Repository
public interface ConsultationRequestRepository extends JpaRepository<ConsultationRequest, Long> {
}
The problem is when i load my application, i face with 2 errors:
Unsuccessful: drop sequence hibernate_sequence
[36morg.hibernate.tool.hbm2ddl.SchemaExport Sequence "HIBERNATE_SEQUENCE" not found; SQL statement:
Then when i open the
http://localhost:8080/h2-console/
I cannot see the table.
It seems that the in the boot process, table is not made.
Update your code as below:
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
As you have not specified a sequence table name, hibernate will look for a sequence table named as hibernate_sequence and use it as default.
For Oracle/Postgres, increment fields used are sequence tables.
In MySql, there are increment fields that automatically increment.
If someone is getting this error with Spring Boot testing(with H2) , we have to use following in the application.properties(or whatever profile we are using) :
spring.jpa.hibernate.ddl-auto=create-drop
Setting the following property in the properties file helped me solve the hibernate_sequence problem for hibernate 5.4v
spring:
jpa:
hibernate:
use-new-id-generator-mappings: false
Check persistence.xml
property name="hibernate.hbm2ddl.auto" value="create"
not hdm2ddl
This worked in my case.
SQL operation ERROR when start app spring-boot.
I added the setting in spring properties and solved: in the spring:jpa.hibernate.ddl-auto= create-drop to hibernate be able create or drop table automatically.
If you use a 2nd cache with liquidbase, you have to add the sequence in the changelog like this:
<changeSet author="liquibase-docs"
id="createSequence-example">
<createSequence catalogName="cat" cycle="false"
incrementBy="1" ordered="true" schemaName="public"
sequenceName="hibernate_sequence" startValue="0" />
</changeSet>
For spring-boot 2.7.x and h2 2.x you need to add MODE=LEGACY; in the database connection:
example application.yml:
spring:
datasource:
url: jdbc:h2:mem:test;MODE=LEGACY;
exemple application.properties:
spring.datasource.url=jdbc:h2:mem:test;MODE=LEGACY;
For Mysql:
the auto-increment is not added, modify ur table:
ALTER TABLE table_name MODIFY COLUMN id BIGINT AUTO_INCREMENT=1
Related
I use springboot 2.6 and postgres 15 and hibernate 5.7
I use database Generate persistence Mapping to make entity from database schema;
When I run application with this config:
spring.jpa.hibernate.ddl-auto=validate
table properties name not work.
for example this is my field:
#Basic
#Column(name = "isEnabled", nullable = false)
private boolean isEnabled;
When I run applicataion get this error:
Caused by: org.hibernate.tool.schema.spi.SchemaManagementException: Schema-validation: missing column [is_enabled] in table [analytics]
Can you try to add the following annotation to your Analytics entity definition?
#Table(name = "analytics", schema = "public")
there are 2 problems. first is field name. seccond is table name;
first problem fix with this config in application.properties:
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
this fix first problem. but seccond problem not fix.
you should change table name with this trick:
#Table(name = ""analytics"")
Table 'liquibase_db.hibernate_sequence' doesn't exist.So I am getting this type of error while I post data from the postman. I tried to solve that problem I tried to add hibernate sequence It's not working.
With the generation GenerationType.AUTO hibernate will look for the default hibernate_sequence table , so change generation to IDENTITY as below :
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
Reference:
200+ accepted as answer to: Table 'DBNAME.hibernate_sequence' doesn't exist
I'm trying to launch a SQL file at my database initialization.
Here is my configuration:
spring:
profiles: local
jpa:
properties:
hibernate.temp.use_jdbc_metadata_defaults: false
generate-ddl: true
hibernate:
ddl-auto: update
database: h2
show-sql: true
autoCommit: false
datasource:
platform: h2
url: jdbc:h2:mem:db;DB_CLOSE_DELAY=-1;DATABASE_TO_UPPER=false;
driver-class-name: org.h2.Driver
initialization-mode: always
data: classpath:/sql/CreateGeographicZones.sql
My script is just this line (atm):
INSERT INTO GEOGRAPHIC_ZONE (name) VALUES ('EUROPE');
And the related entity:
#NoArgsConstructor
#Data
#Entity
#Table(name = "GEOGRAPHIC_ZONE")
public class GeographicZone {
#Id
#GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "geo_zone_sequence")
#SequenceGenerator(name = "geo_zone_sequence", sequenceName = "geo_zone_id_seq", allocationSize = 1)
#Column(nullable = false)
private Long id;
...
}
The table is created as I can see in the logs:
Hibernate: create table geographic_zone (id bigint not null, name varchar(100) not null, primary key (id))
But I have an SQL error when the script is executed:
Table "GEOGRAPHIC_ZONE" not found; SQL statement:
INSERT INTO GEOGRAPHIC_ZONE (name) VALUES ('EUROPE')
In the logs I can see that my table is created before the script execution, so why it's not working ?
According with your entity's metadata Hibernate is querying geo_zone_id_seq sequence's next value and using it for the ID on each insert.
If you would like to use the same approach when inserting directly in your database then you will need to implement a H2 Trigger
Also you may use either the EntityManager bean or your Spring JPA Repository to insert your data after application startup via CommandLineRunner interface.
Using EntityManager:
#Bean
CommandLineRunner registerZonesDataRunner(EntityManager entityManager, TransactionTemplate transactionTemplate) {
return args -> transactionTemplate.execute(new TransactionCallbackWithoutResult() {
#Override
protected void doInTransactionWithoutResult(TransactionStatus status) {
// presuming that GeographicZone has a constructor expecting NAME
Stream.of("AFRICA", "EUROPE")
.map(GeographicZone::new)
.forEach(entityManager::persist);
}
});
Using Spring JPA Repository:
#Bean
CommandLineRunner registerZonesDataRunner(GeographicZoneRepository repository) {
// presuming that GeographicZone has a constructor expecting NAME
return args -> repository.saveAll(Stream.of("AFRICA", "EUROPE")
.map(GeographicZone::new)
.collector(Collectors.toList()));
}
minimal, reproducible example
You don't show how you've defined the id column but the schema indicates there is no auto-generation scheme. So, try:
INSERT INTO GEOGRAPHIC_ZONE (id, name) VALUES (1, 'EUROPE');
in your data file. If that works, you'll need to either manually set the id in your inserts or add something like #GeneratedValue(strategy = AUTO) to your #Id property.
Looking for next debugging steps. I'm using Hibernate with Spring Boot, with my ddl setting set to update (I realize the hibernate setting is redundant but I just wanted to be sure):
application.properties
spring.jpa.generate-ddl=true
spring.jpa.hibernate.ddl-auto=update
hibernate.hbm2ddl.auto=update
I've just added a new Entity, which succesfully created. But when adding an additional field, profileImageURI2, nothing is ever run to attempt to add it:
#Entity
#Data
public class User {
#Id
#GeneratedValue(strategy = GenerationType.SEQUENCE)
private Long id;
private String profileImageURI;
private String profileImageURI2;
}
Is there any additional config that will control the modification of existing tables? The user also has all permissions for the database.
I have the following entity defined:
#Entity
#Table(name = "EmailTemplate")
public class EmailTemplate {
Despite the table annotation, I receive java.sql.SQLException: Invalid object name 'email_template'. How can I prevent an entity class such as EmailTemplate being transformed into email_template table name?
Edit:
I'm using Spring Boot: start JPA. From my build.gradle file,
compile("org.springframework.boot:spring-boot-starter-data-jpa")
Spring by default uses org.springframework.boot.orm.jpa.SpringNamingStrategy which splits camel case names with underscore. Try setting spring.jpa.hibernate.naming-strategy=org.hibernate.cfg.EJB3NamingStrategy in application.properties. Check out this and this for more info.
For hibernate v5:
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
For Spring Boot 2 (checked with 2.2.6.RELEASE) it should be configuration yml file:
spring:
jpa:
hibernate:
naming:
physical-strategy: org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
So you could have model like:
#Table(name = "tblDepartments")
public class Department {
#Id
#Column(name = "dpID")
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#NotEmpty
#Size(min = 1, max = 25)
#Column(name = "dpName", length = 25)
private String name;
and populate tables at startup with data.sql:
INSERT INTO tblDepartments (dpName) VALUES ('Gryffindor');
INSERT INTO tblDepartments (dpName) VALUES ('Hufflepuff');
Use this in your appplication.properties.
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
spring.jpa.hibernate.naming.implicit-strategy=org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl
Use
spring.jpa.hibernate.naming-strategy=org.hibernate.cfg.DefaultNamingStrategy
In application.properties set
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
There are two most common org.hibernate.boot.model.naming.PhysicalNamingStrategys:
org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
org.hibernate.boot.model.naming.CamelCaseToUnderscoresNamingStrategy
# also deprecated in 2.6 in favor of CamelCaseToUnderscoresNamingStrategy
# for removal in 2.8
org.springframework.boot.orm.jpa.hibernate.SpringPhysicalNamingStrategy
org.springframework.boot.autoconfigure.orm.jpa.HibernateProperties holds:
private void applyNamingStrategies(Map<String, Object> properties) {
applyNamingStrategy(properties, AvailableSettings.IMPLICIT_NAMING_STRATEGY, this.implicitStrategy,
() -> SpringImplicitNamingStrategy.class.getName());
applyNamingStrategy(properties, AvailableSettings.PHYSICAL_NAMING_STRATEGY, this.physicalStrategy,
() -> CamelCaseToUnderscoresNamingStrategy.class.getName());
}
so by default CamelCaseToUnderscoresNamingStrategy is in use and you have underscores...
org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute statement; SQL [n/a];
nested exception is org.hibernate.exception.SQLGrammarException: could not execute statement
Both are required :
implicit-strategy
physical-strategy
Solved.
Invalid Object Name: Springboot with JPA(SQL server)
In application.yaml/properties specify the
spring.jpa.hibernate.naming.implicit-strategy
spring.jpa.hibernate.naming.physical-strategy
jpa:
show-sql: false
hibernate:
ddl-auto: none # Defaults to "none" when NOT in embedded mode
naming:
implicit-strategy: org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl
physical-strategy: org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl