persistence.xml - dinâmico

JavaC Membro Membro desde: 12/07/2015 19:18:33 Mensagens: 25 Offline

Boa tarde pessoal. Gostaria de deixar o ip do servidor do persistence.xml de forma dinâmica.

<property name="javax.persistence.jdbc.url" value="jdbc:postgresql://ip_do_servidor:5432/nomebd"/>

Sabe me dizer o que devo fazer?

Beleza? Isso depende de como vc controla sua sessão. Entity manager.

Se vc Cria o EntityManager na mão, vc pode colocar as propriedades em tempo real na montagem do Factory.
Ex:

Map<String, String> properties = new HashMap<String, String>();

properties.put("javax.persistence.jdbc.url", "jdbc:oracle:thin:@" + endereco + ":" + porta + ":" + banco);
        properties.put("javax.persistence.jdbc.user", usuario);
        properties.put("javax.persistence.jdbc.password", senha);
        properties.put("javax.persistence.jdbc.driver", "oracle.jdbc.OracleDriver");

properties.put("hibernate.dialect", "org.hibernate.dialect.OracleDialect");
        properties.put("hibernate.show_sql", "true");
        properties.put("hibernate.format_sql", "false");
        properties.put("hibernate.default_schema", schema);

properties.put("hibernate.connection.provider_class", "org.hibernate.connection.C3P0ConnectionProvider");
        properties.put("hibernate.c3p0.min_size", "1");
        properties.put("hibernate.c3p0.max_size", "1");
        properties.put("hibernate.c3p0.timeout", "0");
        properties.put("hibernate.c3p0.max_statements", "1");
        properties.put("hibernate.c3p0.validate", "true");

EntityManagerFactory factory = Persistence.createEntityManagerFactory("oracle", properties);
        
        //EntityManager
        factory.createEntityManager();

JavaC Membro Membro desde: 12/07/2015 19:18:33 Mensagens: 25 Offline

Bom dia, já tentei definir as propriedades na classe EntityManagerProducer, porém sem sucesso.
Segue também o meu persistence.xml.

Persistence.xml

JavaC Membro Membro desde: 14/08/2020 00:29:13 Mensagens: 2 Offline

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