org.hibernate.AnnotationException：使用@OneToMany 或@ManyToMany 定位未映射的 class：core.user.User.userRole[core.user.UserRole]

Question

这些实体类出现此错误。我查看了 this question，但我没有 persistence.xml，所以我不确定如何解决这个问题。如果有帮助，我正在使用 Dropwizard 构建这些实体。

问题似乎在于我如何将 User 映射到 UserRole。两者类在同一个包中。

User.java

@Entity
@Table(name = "USER")
public class User implements TimestampedItem {

    //region Fields
    protected String email;

    protected String password;

    protected String displayName;

    protected Long ID;

    protected Set<UserRole> userRole;

    protected Boolean isActive = true;

    protected Boolean isEmailConfirmed = false;

    protected DateTime added = new DateTime();

    protected DateTime modified = new DateTime();
    //endregion

    //region Constructors
    public User() {

        this("", "", "");
    }

    public User(String email, String password, String displayName) {

        this(email, password, displayName, true);
    }

    public User(String email, String password, String displayName, Boolean isActive) {

        this(email, password, displayName, new HashSet<>(), isActive, false);
    }

    public User(String email, String password, String displayName, Set<UserRole> userRole, Boolean isActive, Boolean isEmailConfirmed) {

        this(email, password, displayName, userRole, isActive, isEmailConfirmed, new DateTime(), new DateTime());
    }

    public User(String email, String password, String displayName, Set<UserRole> userRole, Boolean isActive, Boolean isEmailConfirmed, DateTime added, DateTime modified) {

        this.email = email;
        this.password = password;
        this.displayName = displayName;
        this.userRole = userRole;
        this.isActive = isActive;
        this.isEmailConfirmed = isEmailConfirmed;
        this.added = added;
        this.modified = modified;
    }

    //endregion

    //region Getters and Setters

    /**
     * Returns this Users email address
     *
     * @return this Users email address
     */
    @Email
    @Column(name = "EMAIL", unique = true, nullable = false, length = 255)
    public String getEmail() {

        return this.email;
    }

    /**
     * Sets this Users email address
     *
     * @param email The email address to set
     */
    public void setEmail(String email) {

        this.email = email;
    }

    /**
     * Returns this User's password
     *
     * @return this User's password
     */
    @Column(name = "PASSWORD", nullable = false, length = Passwords.DESIRED_KEY_LENGTH)
    public String getPassword() {

        return this.password;
    }

    /**
     * Sets this User's password
     *
     * @param password The password to set
     */
    public void setPassword(String password) {

        this.password = password;
    }

    /**
     * Returns this User's unique User ID
     *
     * @return this User's unique User ID
     */
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "USER_ID", unique = true, nullable = false)
    public Long getID() {

        return this.ID;
    }

    /**
     * Sets this User's unique User ID
     *
     * @param id The unique ID to set to this user
     */
    public void setID(Long id) {

        this.ID = id;
    }

    /**
     * Whether this User's profile is isActive
     *
     * @return true if this User's profile is isActive, false otherwise
     */
    @Column(name = "IS_ACTIVE", nullable = false)
    public Boolean isActive() {

        return this.isActive;
    }

    /**
     * Sets whether this User's profile is isActive
     *
     * @param active true if this User's profile is isActive, false otherwise
     */
    public void setActive(Boolean active) {

        this.isActive = active;
    }

    /**
     * Whether this User's email address has been confirmed.
     * To confirm an email, they must click the activation link that is sent to their email address
     *
     * @return true if this User's email has been confirmed, false otherwise
     */
    @Column(name = "IS_EMAIL_CONFIRMED", nullable = false)
    public Boolean isEmailConfirmed() {

        return this.isEmailConfirmed;
    }

    /**
     * Sets whether this User's email address has been confirmed.
     * To confirm an email, they must click the activation link that is sent to their email address
     *
     * @param isEmailConfirmed true if this User's email has been confirmed, false otherwise
     */
    public void setEmailConfirmed(Boolean isEmailConfirmed) {

        this.isEmailConfirmed = isEmailConfirmed;
    }

    /**
     * Returns this User's display name
     *
     * @return this User's display name
     */
    @Column(name = "DISPLAY_NAME", nullable = false)
    public String getDisplayName() {

        return this.displayName;
    }

    /**
     * Sets this User's display name
     *
     * @param displayName The display name to associate with this User
     */
    public void setDisplayName(String displayName) {

        this.displayName = displayName;
    }

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "user")
    public Set<UserRole> getUserRole() {
        return this.userRole;
    }

    public void setUserRole(Set<UserRole> userRole) {
        this.userRole = userRole;
    }

    @Override
    @Column(name = "ADDED", nullable = false)
    public DateTime getAdded() {

        return this.added;
    }

    @Override
    public void setAdded(DateTime added) {

        this.added = added;
    }

    @Override
    @Column(name = "MODIFIED", nullable = false)
    public DateTime getModified() {

        return this.modified;
    }

    @Override
    public void setModified(DateTime modified) {

        this.modified = modified;
    }
    //endregion


    @Override
    public boolean equals(Object o) {

        if (this == o) return true;
        if (!(o instanceof User)) return false;

        User user = (User) o;

        // These probably aren't needed, since a user is considered unique by ID alone
//        if (!displayName.equals(user.displayName)) return false;
//        if (!email.equals(user.email)) return false;
//        if (!password.equals(user.password)) return false;

        return getID().equals(user.getID());
    }

    @Override
    public int hashCode() {

        int result = email.hashCode();
        result = 31 * result + password.hashCode();
        result = 31 * result + displayName.hashCode();
        return result;
    }
}

UserRole.java

@Entity
@Table(
        name = "user_roles"
        , uniqueConstraints = @UniqueConstraint( columnNames = { "role", "USER_ID" } )
)
public class UserRole{

    private Integer userRoleId;
    private User user;
    private String role;

    public UserRole() {
    }

    public UserRole(User user, String role) {
        this.user = user;
        this.role = role;
    }

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "user_role_id", unique = true, nullable = false)
    public Integer getUserRoleId() {
        return this.userRoleId;
    }

    public void setUserRoleId(Integer userRoleId) {
        this.userRoleId = userRoleId;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "USER_ID", nullable = false)
    public User getUser() {
        return this.user;
    }

    public void setUser(User user) {
        this.user = user;
    }

    @Column(name = "role", nullable = false, length = 45)
    public String getRole() {
        return this.role;
    }

    public void setRole(String role) {
        this.role = role;
    }
}

Answer 1

您使用 spring 框架吗？如果你使用这个，你可以将这段代码添加到 appliactionContext.xml

<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="persistenceUnitName" value="dbke" />
    <property name="PersistenceXmlLocation" value="classpath:META-INF/jpa-persistence.xml" />
</bean>

您需要创建 jpa-persistence.xmlfirstly.For 示例：

<?xml version="1.0" encoding="UTF-8"?>

http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">

<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="dbke" transaction-type="RESOURCE_LOCAL">
        <class>core.user.Userrole</class>
        <class>core.user.User</class>       

        <provider>org.hibernate.ejb.HibernatePersistence</provider>     

    </persistence-unit>
</persistence>

Answer 2

您是否正确设置了 HibernateBundle？ Dropwizard Hibernate offers two bundles to set up the persistence unit with hibernate, one being the default HibernateBundle 需要在构造函数中传递的所有实体类 - 是否都传递了类？

或者，您可以使用 ScanningHibernateBundle 并将您的实体的包名称作为字符串传入，这样您就不会在将来遗漏任何内容。这在您的应用程序中看起来像这样：

private final HibernateBundle<MyConfiguration> hibernate =
        new ScanningHibernateBundle<MyConfiguration>("com.acme.entity") {

            @Override
            public DataSourceFactory getDataSourceFactory(MyConfiguration configuration) {
                return configuration.getDataSourceFactory();
            }
        };

org.hibernate.AnnotationException：使用@OneToMany 或@ManyToMany 定位未映射的 class：core.user.User.userRole[core.user.UserRole]

org.hibernate.AnnotationException: Use of @OneToMany or @ManyToMany targeting an unmapped class: core.user.User.userRole[core.user.UserRole]

java

hibernate

jpa

dropwizard

UserRole.java