Question

我是单元测试的初学者。我在Netbeans 8.1和Maven中构建了一个Java Web Application。我创建了这个类来尝试在PostgreSQl数据库中测试查询：

public class DatesUtil {

    private Collection<Menusistema> listamenus = new LinkedList<>();

    private Menusistema menu= new Menusistema();

    private String nombreBuscar = null;

    private static final EntityManager entityManager;
    static {
        entityManager = Persistence.createEntityManagerFactory("com.controlpersonal.pu").
        createEntityManager();
    }

   public static EntityManager getEntityManager() { 
        return entityManager;
    };

    public DatesUtil() {

    }

    @BeforeClass
    public static void setUpClass() {


    }

    @AfterClass
    public static void tearDownClass() {
    }

    @Before
    public void setUp() {
    }

    @After
    public void tearDown() {
    }

    // TODO add test methods here.
    // The methods must be annotated with annotation @Test. For example:
    //
     @Test
     public void hello() {
        buscarTodo();

     }


     public void buscarTodo(){
        findAll(Menusistema.class);
     }



    public <T> List<T> findAll(Class<T> clazz) {
        return (List<T>) getEntityManager().createQuery("SELECT p FROM " + clazz.getSimpleName() + " p", clazz).getResultList();
    }  


}

当我右键单击netbeans并选择：Test File：

时，我收到了下一个错误

[EL Info]：2018-04-30 20：30：57.78 - ServerSession（252553541） - EclipseLink，版本：Eclipse 持久性服务 - 2.5.2.v20140319-9ad6abd [EL Severe]：ejb： 2018-04-30 20：30：57.836 - ServerSession（252553541） - 例外 [EclipseLink-7060]（Eclipse持久性服务 - 2.5.2.v20140319-9ad6abd）：org.eclipse.persistence.exceptions.ValidationException异常说明：无法获取数据源[jdbc / controlpersonal]。内部异常：javax.naming.NoInitialContextException：需要在环境或系统属性中指定类名，或作为applet指定参数，或在应用程序资源文件中： java.naming.factory.initial的

我的persistence.xml是：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
  <persistence-unit name="com.controlpersonal.pu" transaction-type="JTA">
    <jta-data-source>jdbc/controlpersonal</jta-data-source>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties/>
  </persistence-unit>
</persistence>

我真的想测试这是否可行。提前谢谢！

Answer 1

我刚创建了替代persistente .xml：

<?xml version="1.0" encoding="UTF-8"?> <persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
    <persistence-unit name="com.controlpersonal.pu.test" transaction-type="RESOURCE_LOCAL">
        <!--<jta-data-source>jdbc/controlpersonal</jta-data-source>-->
        <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
        <class>com.your.class1</class>
        <class>com.your.class2</class>  
        <properties>
            <property name="eclipselink.target-database" value="PostgreSQL" />
            <property name="eclipselink.logging.level" value="INFO" />
            <property name="javax.persistence.jdbc.driver"     value="org.postgresql.Driver" />
            <property name="javax.persistence.jdbc.url"     value="jdbc:postgresql://localhost:5432/controlpersonal" />
            <property name="javax.persistence.jdbc.user" value="postgres" />
            <property name="javax.persistence.jdbc.password" value="xxx" />
        </properties>
    </persistence-unit> </persistence>

在：src / test / resources / META-INF / persistence.xml

Junit测试与实体mananger

1 个答案: