Test Hibernate persitence.xml

Version imprimable

Bonjour,

Je souhaite tester ma classe DAO :

Code:

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@ContextConfiguration(locations = {"classpath:META-INF/persistence.xml"})
public class UserControllerTest extends AbstractJUnit4SpringContextTests {
 
    @Autowired
    private UserDao userDao;
 
    public UserControllerTest() {
    }
 
    @Test
    public void addUser() {
        ..........
    }
}

Code:

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<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence
             http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
    <persistence-unit name="persistence" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.ejb.HibernatePersistence</provider>
        <class>entity.Authorities</class>
        <class>entity.Users</class>
        <exclude-unlisted-classes>false</exclude-unlisted-classes>
        <validation-mode>NONE</validation-mode>
        <properties>
          .......................
        </properties>
    </persistence-unit>
</persistence>

Caused by: org.springframework.beans.factory.xml.XmlBeanDefinitionStoreException: Line 5 in XML document from class path resource [META-INF/persistence.xml] is invalid; nested exception is org.xml.sax.SAXParseException; lineNumber: 5; columnNumber: 75; cvc-elt.1 : Déclaration de l'élément 'persistence' introuvable.

Caused by: org.xml.sax.SAXParseException: cvc-elt.1 : Déclaration de l'élément 'persistence' introuvable.

Pouvez-vous m'aider ?

21/03/2014, 09h28
jeffray03

salut,
cela veut dire tout simplement que la version de JPA n´est pas bonne ou n´est pas supporté.
essaies ceci:
- supprimes
- et cree la de nouveau en utilisant click droit sur le projet --> preferences --> projet facets --> selectionnes JPA et version --> apply.

et voir si cela continue.

Eric
21/03/2014, 09h59
chacalpuant1987

J'i supprimé l'ancien persistence.xml

new --> create a persistence Unit -->

name : persistence
persistance librairy : EclipseLink (JPA 2.0)
Database connection : MYSQL
Table generation Strategy : None

The*target*server*for*the*project*is*not*set*properly.*Java*Persistence*features are*server-dependent*and*some*wizards*in*the*IDE*might*not*behave*as*expected *if*the*target*server*is*not*specified

J'ai résolu le problème précédant mais j'ai toujours l'erreur :

cvc-elt.1 : Déclaration de l'élément 'persistence' introuvable.

Code:

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<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
  <persistence-unit name="persistence" transaction-type="RESOURCE_LOCAL">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <class>entity.Authorities</class>
    <class>entity.Users</class>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties>
.....
    </properties>
  </persistence-unit>
</persistence>

21/03/2014, 12h21
jeffray03

tu l´as mis dans quel repertoire?
ou peut-etre qu´il te manque quelques Jar.
21/03/2014, 12h36
chacalpuant1987

1 pièce(s) jointe(s)

Pièce jointe 141093
21/03/2014, 12h52
jeffray03

as tu essayé :

Code:

@ContextConfiguration(locations = {"classpath*:persistence.xml"})

mais une question, c´est du Spring que tu fais las?
si tel est le cas, alors pourquoi ne pas mettre le bon fichier de configuration:

Code:

@ContextConfiguration(locations={"classpath*:applicationContext.xml"})

et verifies aussi si ton fichier est present dans :

Code:

target/test-classes/

Eric

J'ai bien pris en compte ton analyse et j'ai décidé de refondre mon environnement comme ça :

http://www.mkyong.com/spring/maven-s...mysql-example/

Code:

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public class UserControllerTest {
 
    ApplicationContext appContext = 
    	  new ClassPathXmlApplicationContext("/config/BeanLocations.xml");
 
    UserBo userBo = (UserBo)appContext.getBean("UserBo");
 
    @Test
    public void testFindData() {
        Users user = userBo.findUserByUsername("romain");
        Assert.assertNotNull(user);  
}

beanLocations

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<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
                           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
 
    <!-- Database Configuration -->
    <import resource="../database/dataSource.xml"/>
    <import resource="../database/hibernate.xml"/>
 
    <!-- Beans Declaration -->
    <import resource="../spring/users.xml"/>
    <import resource="../spring/authorities.xml"/>
 
</beans>

database/hibernate.xml

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<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
                           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
 
    <!-- Hibernate session factory -->
    <bean id="sessionFactory" 
      class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
 
        <property name="dataSource">
            <ref bean="dataSource"/>
        </property>
 
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
            </props>
        </property>
 
        <property name="mappingResources">
            <list>
                <value>/hibernate/Authorities.hbm.xml</value>
                <value>/hibernate/Users.hbm.xml</value>
            </list>
        </property>	
 
    </bean>
</beans>

et j'obtiens cette erreur :

Tests in error:
testFindData(UserControllerTest): Error creating bean with name 'sessionFactory' defined in class path resource [database/hibernate.xml]: Initialization of bean failed; nested exception is java.lang.NoClassDefFoundError: org/hibernate/cache/CacheProvider

salut c´est quelle version d´hibernate que tu utilises:
sinon essaies ceci:

Code:

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<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
                           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
 
    <!-- Hibernate session factory -->
    <bean id="sessionFactory" 
      class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
 
        <property name="dataSource">
            <ref bean="dataSource"/>
        </property>
 
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
            </props>
        </property>
 
        <property name="mappingResources">
            <list>
                <value>/hibernate/Authorities.hbm.xml</value>
                <value>/hibernate/Users.hbm.xml</value>
            </list>
        </property>	
 
    </bean>
</beans>

24/03/2014, 11h17
chacalpuant1987

1 pièce(s) jointe(s)

Même erreur :(

Pièce jointe 141279
24/03/2014, 11h24
jeffray03

ok,
as-tu stopper le serveur, fait un clean sur le projet et un clean sur le server?

-- quelle version d´hibernate utilises tu?

Eric
24/03/2014, 11h28
chacalpuant1987

Oui j'ai stoppé le serveur, clean and build.

<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>4.2.6.Final</version>
</dependency>

alors utilises:

Code:

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<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
                           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
 
    <!-- Hibernate session factory -->
    <bean id="sessionFactory" 
      class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
 
        <property name="dataSource">
            <ref bean="dataSource"/>
        </property>
 
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
            </props>
        </property>
 
        <property name="mappingResources">
            <list>
                <value>/hibernate/Authorities.hbm.xml</value>
                <value>/hibernate/Users.hbm.xml</value>
            </list>
        </property>	
 
    </bean>
</beans>

24/03/2014, 14h01
chacalpuant1987

1 pièce(s) jointe(s)

Je vais plus loin mais toujours une erreur :(

Pièce jointe 141286

testFindData(UserControllerTest): Error creating bean with name 'sessionFactory' defined in class path resource [database/hibernate.xml]: Invocation of init method failed; nested exception is org.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.component.PojoComponentTuplizer]

essaies d´ajouter ceci:

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<dependency>
    <groupId>javassist</groupId>
    <artifactId>javassist</artifactId>
    <version>3.12.1.GA</version>
</dependency>

24/03/2014, 14h36
chacalpuant1987

Même erreur :(
24/03/2014, 14h47
jeffray03

il ya sans doute une erreur dissimulé dans tes getter() et setter().
comme je n´ai pas de vision generale, je ne fais qe supposer. et aussi on n´arrive pas a ouvrir la derniere image envoyé.
24/03/2014, 14h50
chacalpuant1987

@Entity
@Table(name = "users")
public class Users implements Serializable {

private static final long serialVersionUID = 1L;

private static final Logger LOGGER = LoggerFactory.getLogger(Users.class);

@Id
@Column(name = "username")
private String username;
@Column(name = "password")
private String password;
@Column(name = "enabled")
private Boolean enabled;
@Column(name = "email")
private String email;

@OneToMany(mappedBy = "users", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private List<Authorities> authoritiesList;

/**
* Constructor
*/
public Users() {
}

/**
*
* @param pUsername username
*/
public Users(String pUsername) {
this.username = pUsername;
}

/**
*
* @param pUsername username
* @param pPassword password
*/
public Users(String pUsername, String pPassword) {
this.username = pUsername;
this.password = pPassword;
}

public String getUsername() {
return username;
}

public void setUsername(String pUsername) {
this.username = pUsername;
}

public String getPassword() {
return password;
}

public void setPassword(String pPassword) {
this.password = pPassword;
}

public Boolean getEnabled() {
return enabled;
}

public void setEnabled(Boolean pEnabled) {
this.enabled = pEnabled;
}

public String getEmail() {
return email;
}

public void setEmail(String pEmail) {
this.email = pEmail;
}

public List<Authorities> getAuthoritiesList() {
return authoritiesList;
}

public void setAuthoritiesList(List<Authorities> pAuthoritiesList) {
this.authoritiesList = pAuthoritiesList;
}

@Override
public int hashCode() {
int hash = 0;
hash += username != null ? username.hashCode() : 0;
return hash;
}

@Override
public boolean equals(Object object) {
if (!(object instanceof Users)) {
return false;
}
Users other = (Users) object;
if ((this.username == null && other.username != null) || (this.username != null && !this.username.equals(other.username))) {
return false;
}
return true;
}

@Override
public String toString() {
return "entity.Users[ username=" + username + " ]";
}
}

@Embeddable
public class AuthoritiesPK implements Serializable {

private static final long serialVersionUID = 1L;

@Column(name = "username")
private String username;
@Column(name = "authority")
private String authority;

/**
* Constructor
*/
public AuthoritiesPK() {
}

/**
*
* @param username username
* @param authority authority
*/
public AuthoritiesPK(String username, String authority) {
this.username = username;
this.authority = authority;
}

public String getUsername() {
return username;
}

public void setUsername(String username) {
this.username = username;
}

public String getAuthority() {
return authority;
}

public void setAuthority(String authority) {
this.authority = authority;
}

@Override
public int hashCode() {
int hash = 0;
hash += username != null ? username.hashCode() : 0;
hash += authority != null ? authority.hashCode() : 0;
return hash;
}

@Override
public boolean equals(Object obj) {

if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final AuthoritiesPK other = (AuthoritiesPK) obj;
if ((this.username == null) ? (other.username != null) : !this.username.equals(other.username)) {
return false;
}
return true;
}

@Override
public String toString() {
return "entity.AuthoritiesPK[ username=" + username + ", authority=" + authority + " ]";
}
}

@Entity
@Table(name = "authorities")
public class Authorities implements Serializable {

private static final long serialVersionUID = 1L;

@EmbeddedId
protected AuthoritiesPK authoritiesPK;

@ManyToOne
@JoinColumn(name = "username", insertable = false, updatable = false)
private Users users;

/**
* Constructor
*/
public Authorities() {
}

/**
*
* @param pAuthoritiesPK authoritiesPK
*/
public Authorities(AuthoritiesPK pAuthoritiesPK) {
this.authoritiesPK = pAuthoritiesPK;
}

/**
*
* @param pUsername username
* @param pAuthority authority
*/
public Authorities(String pUsername, String pAuthority) {
this.authoritiesPK = new AuthoritiesPK(pUsername, pAuthority);
}

public AuthoritiesPK getAuthoritiesPK() {
return authoritiesPK;
}

public void setAuthoritiesPK(AuthoritiesPK pAuthoritiesPK) {
this.authoritiesPK = pAuthoritiesPK;
}

@Override
public int hashCode() {
int hash = 0;
hash += authoritiesPK != null ? authoritiesPK.hashCode() : 0;
return hash;
}

public Users getUsers() {
return users;
}

public void setUsers(Users users) {
this.users = users;
}

@Override
public String toString() {
return authoritiesPK.getAuthority();
}

@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final Authorities other = (Authorities) obj;
if (this.authoritiesPK != other.authoritiesPK && (this.authoritiesPK == null || !this.authoritiesPK.equals(other.authoritiesPK))) {
return false;
}
return true;
}
}
24/03/2014, 15h07
jeffray03

Une question:
tu utilises les annotations ou les fichiers de mapping pour creer tes entités?
24/03/2014, 15h08
chacalpuant1987

annotations
25/03/2014, 09h50
chacalpuant1987

Pas d'autres idées ?
25/03/2014, 15h12
jeffray03

salut, la j´avoue etre un peu perdu.
1.votre fichier de configuration de la persistance est il persistence.xml ou hibernate.xml?
2. Dites nous quelles sont les versions de Springxxxx.jar que vous utiliser
3. vos classes DAOs, les entités sont deja bonnes, car je viens de creer un test avec et ca marche, certe avec HSQL mais le principe est le meme.

Eric

hibernate.xml est mon fichier de configuration de la persistance :

Code:

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<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
                           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
 
    <!-- Hibernate session factory -->
    <bean id="sessionFactory"
          class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
 
        <property name="dataSource">
            <ref bean="dataSource"/>
        </property>
 
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
            </props>
        </property>
 
        <property name="mappingResources">
            <list>
                <value>/hibernate/Authorities.hbm.xml</value>
                <value>/hibernate/Users.hbm.xml</value>
            </list>
        </property>
 
    </bean>
</beans>

Les versions de spring sont :

spring-beans-3.2.4.RELEASE.jar
spring-core-3.2.4.RELEASE.jar
spring-orm-3.2.4.RELEASE.jar
spring-web-3.2.4.RELEASE.jar

25/03/2014, 16h26
jeffray03

Ok, je crois qu´on ne se comprends pas:
voici la liste que doit contenir ton projet:
1. un fichier de configuration de context generalement appelé : applicationContext.xml
2. un fichier de configuration de la persistance des entités avec Hibernate c´est generalement hibernate-cfg.xml

Crees ses 2 fichiers et vois si cela produit des erreurs..

Eric.
25/03/2014, 16h46
chacalpuant1987

1 pièce(s) jointe(s)

Voici mon Projet :

Pièce jointe 141371

Je pense qu'il me manque ces deux fichiers ;)

Peux-tu me montrer un exemple ?
26/03/2014, 09h57
jeffray03

1 pièce(s) jointe(s)

je remarque 1 chose:

si tu utilises les annotations pourquoi avoir les fichier *.hbm.xml.
bon voici un exemple: