Solucionado (ver solução)
Solucionado
(ver solução)
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resposta

Bom dia, nao consigo fazer login no sistema,

Hibernate: select usuario0_.email as email1_3_, usuario0_.nome as nome2_3_, usuario0_.senha as senha3_3_ from Usuario usuario0_ where usuario0_.email=?

Mon Mar 25 09:12:13 BRT 2019 WARN: Establishing SSL connection without server's identity verification is not recommended. According to MySQL 5.5.45+, 5.6.26+ and 5.7.6+ requirements SSL connection must be established by default if explicit option isn't set. For compliance with existing applications not using SSL the verifyServerCertificate property is set to 'false'. You need either to explicitly disable SSL by setting useSSL=false, or set useSSL=true and provide truststore for server certificate verification.

@EnableTransactionManagement
public class JPAConfiguration {

    @Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() {

        LocalContainerEntityManagerFactoryBean factoryBean = new LocalContainerEntityManagerFactoryBean();
        JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
        factoryBean.setJpaVendorAdapter(vendorAdapter);

        DriverManagerDataSource dataSource = new DriverManagerDataSource();
        dataSource.setUsername("matheus");
        dataSource.setPassword("password");
        dataSource.setUrl("jdbc:mysql://localhost/casadocodigo");
        dataSource.setDriverClassName("com.mysql.jdbc.Driver");
        factoryBean.setDataSource(dataSource);

        Properties properties = new Properties();
        properties.setProperty("hibernate.dialect", "org.hibernate.dialect.MySQL5Dialect");
        properties.setProperty("hibernate.show_sql", "true");
        properties.setProperty("hibernate.hbm2ddl.auto", "update");

        factoryBean.setJpaProperties(properties);
        factoryBean.setPackagesToScan("br.com.casadocodigo.loja.models");

        return factoryBean;
    }

    @Bean
    public JpaTransactionManager transactionManager(EntityManagerFactory emf) {
        return new JpaTransactionManager(emf);
    }


}
1 resposta
solução!

ja consegui descobriri o erro. obrigado

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