Hibernate + JPA: Schema-validation: missing column

Question

I write an educational project. It's a simple chat. Stack of the backend technology is Java, Jetty (server, Web Socket), and Hibernate 5.3.7. Final, PostgreSQL.

I've got a runtime exception org.hibernate.tool.schema.spi.SchemaManagementException: Schema-validation: missing column [conversation_c_id] in table [conversation_reply]. The error had been making me have fun for a couple of days. I would be grateful for the help in solving this problem.

Here's the basic structure of the entities part:

@Entity
@Table(name = "conversation") //public class ConversationDataSet

@Id
@SequenceGenerator(name = "cIdSequence", sequenceName = "c_id_sequence", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "cIdSequence")
@Column(name = "c_id", nullable = false, updatable = false)
private long cId;

@OneToMany(mappedBy = "conversation", cascade = CascadeType.ALL, orphanRemoval = true)
private List<ConversationReplyDataSet> conversationReplies = new ArrayList<>();



@Entity
@Table(name = "conversation_reply") //public class ConversationReplyDataSet

@Id
@SequenceGenerator(name = "crIdSequence", sequenceName = "cr_id_sequence", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "crIdSequence")
@Column(name = "cr_id", nullable = false, updatable = false)
private long crId;

@Column(name = "c_id_fk")
private long cIdFk;

@ManyToOne(fetch = FetchType.LAZY)
private ConversationDataSet conversation;

The SQL schema is:

create table conversation
(
    c_id bigint not null primary key
);

create table conversation_reply
(
    cr_id bigint not null primary key,
    c_id_fk bigint not null,
    constraint fk_c_id_fk foreign key (c_id_fk) references conversation(c_id)
);

Stack trace:

org.hibernate.tool.schema.spi.SchemaManagementException: Schema-validation: missing column [conversation_c_id] in table [conversation_reply]
at org.hibernate.tool.schema.internal.AbstractSchemaValidator.validateTable(AbstractSchemaValidator.java:136)
at org.hibernate.tool.schema.internal.GroupedSchemaValidatorImpl.validateTables(GroupedSchemaValidatorImpl.java:42)
at org.hibernate.tool.schema.internal.AbstractSchemaValidator.performValidation(AbstractSchemaValidator.java:89)
at org.hibernate.tool.schema.internal.AbstractSchemaValidator.doValidation(AbstractSchemaValidator.java:68)
at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:191)
at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:72)
at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:310)
at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:467)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:708)
at dbService.DBServiceImpl.createSessionFactory(DBServiceImpl.java:155)
at dbService.DBServiceImpl.<init>(DBServiceImpl.java:42)
at main.Main.main(Main.java:23)

Just in case the Hibernate configuration is:

configuration.setProperty("hibernate.dialect", "org.hibernate.dialect.PostgreSQL94Dialect");
configuration.setProperty("hibernate.connection.driver_class", "org.postgresql.Driver");
configuration.setProperty("hibernate.connection.url", "jdbc:postgresql://localhost:5432/chatroomdb");
configuration.setProperty("hibernate.connection.username", "someusername");
configuration.setProperty("hibernate.connection.password", "somepassword");
configuration.setProperty("hibernate.show_sql", "true");
configuration.setProperty("hibernate.hbm2ddl.auto", "validate");
//   configuration.setProperty("hibernate.implicit_naming_strategy", "legacy-jpa");

Thank you.

Reza Nasiri · Accepted Answer · 2018-10-20 00:07:53Z

5

change your ManyToOne relation to this :

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "c_id_fk")
private ConversationDataSet conversation;

and remove this

@Column(name = "c_id_fk")
private long cIdFk;

you can read more here https://en.wikibooks.org/wiki/Java_Persistence/ManyToOne#Example_of_a_ManyToOne_relationship_annotations

edited Oct 20, 2018 at 0:07

answered Oct 19, 2018 at 0:50

Reza Nasiri

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4 Comments

Alexander Suvorov Over a year ago

Reza, thank you for your answer. I suggest it will not be work. As i think column cannot refer to its own entity. However I'll try it tonight.

Alexander Suvorov Over a year ago

Reza, it's working!! Could you please explain me what has happened? Why we must to refer to the column of the entity from the same entity? What kind of instruction is recivied by Hibernate implementing this decision? It's very interesting. Sorry for my premature skepticism. Many thanks for the help.

Reza Nasiri Over a year ago

you can read more here en.wikibooks.org/wiki/Java_Persistence/…

Alexander Suvorov Over a year ago

Oh yeah... I read it several times... Thanks for helping me get out of my stupor.

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