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Solucionado
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respostas

Erro ao criar tabela

Ao criar algumas classes para inserir as mesmas no banco é apresentado uma mensagem de erro no console, e acaba não criando uma tabela específica, a tabela livro.

set 24, 2017 1:15:21 AM org.hibernate.tool.schema.internal.ExceptionHandlerLoggedImpl handleException
WARN: GenerationTarget encountered exception accepting command : Error executing DDL via JDBC Statement
org.hibernate.tool.schema.spi.CommandAcceptanceException: Error executing DDL via JDBC Statement
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:67)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.applySqlString(AbstractSchemaMigrator.java:559)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.applySqlStrings(AbstractSchemaMigrator.java:504)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.createTable(AbstractSchemaMigrator.java:277)
    at org.hibernate.tool.schema.internal.GroupedSchemaMigratorImpl.performTablesMigration(GroupedSchemaMigratorImpl.java:71)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.performMigration(AbstractSchemaMigrator.java:207)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.doMigration(AbstractSchemaMigrator.java:114)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:183)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:72)
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:313)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:452)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:889)
    at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:58)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:55)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:39)
    at br.com.fema.biblioteca.util.JPAUtil.<clinit>(JPAUtil.java:10)
    at br.com.fema.biblioteca.util.PopulaBanco.main(PopulaBanco.java:14)
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Incorrect column specifier for column 'id'
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:423)
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:411)
    at com.mysql.jdbc.Util.getInstance(Util.java:386)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4120)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4052)
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2503)
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2664)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2809)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2758)
    at com.mysql.jdbc.StatementImpl.execute(StatementImpl.java:894)
    at com.mysql.jdbc.StatementImpl.execute(StatementImpl.java:732)
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:54)
    ... 16 more
2 respostas
solução!

Bom dia João, você colocou @Id nas suas pk?

Olá Guilherme, descobri o que erroneamente passou despercebido e coloquei String na pk ao invés da Integer rsrs, mas obrigado.

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