演示问题的 GitHub 示例
https://github.com/jl431/Example.git
重现错误所需的所有代码都在这里。
还需要启动 mysql 服务器并更改 hibernate 文件中的连接详细信息。
错误
Initial SessionFactory creation failed.org.hibernate.MappingException: Could not determine type for: beans.User, at table: RATINGS, for columns: [org.hibernate.mapping.Column(USER_ID)]
Exception in thread "main" java.lang.ExceptionInInitializerError
at simple.HibernateUtil.buildSessionFactory(HibernateUtil.java:26)
at simple.HibernateUtil.<clinit>(HibernateUtil.java:10)
at simple.MovieManager.<init>(MovieManager.java:9)
at simple.MovieManager.main(MovieManager.java:13)
Caused by: org.hibernate.MappingException: Could not determine type for: beans.User, at table: RATINGS, for columns: [org.hibernate.mapping.Column(USER_ID)]
at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:336)
at org.hibernate.tuple.PropertyFactory.buildStandardProperty(PropertyFactory.java:353)
at org.hibernate.tuple.component.ComponentMetamodel.<init>(ComponentMetamodel.java:71)
at org.hibernate.mapping.Component.getType(Component.java:180)
at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:310)
at org.hibernate.mapping.RootClass.validate(RootClass.java:271)
at org.hibernate.cfg.Configuration.validate(Configuration.java:1360)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1851)
at simple.HibernateUtil.buildSessionFactory(HibernateUtil.java:19)
... 3 more
以下是我认为问题所在的几个关键文件。其余代码可从上面的 github 存储库 URL 获取。
hibernate.cfg.xml
<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<!-- Database connection settings -->
<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost:3306/testingbeta</property>
<property name="connection.username">root</property>
<property name="connection.password"></property>
<!-- JDBC connection pool (use the built-in) -->
<property name="connection.pool_size">1</property>
<!-- SQL dialect -->
<property name="dialect">org.hibernate.dialect.MySQLDialect</property>
<!-- Enable Hibernate's automatic session context management -->
<property name="current_session_context_class">thread</property>
<!-- Disable the second-level cache -->
<property name="cache.provider_class">org.hibernate.cache.internal.NoCacheProvider</property>
<!-- Echo all executed SQL to stdout -->
<property name="show_sql">true</property>
<!-- Drop and re-create the database schema on startup -->
<property name="hbm2ddl.auto">create</property>
<mapping resource="Movie.hbm.xml"/>
<mapping resource="User.hbm.xml"/>
<mapping resource="Rating.hbm.xml"/>
</session-factory>
</hibernate-configuration>
Rating.hbm.xml
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="beans">
<class name="Rating" table="RATINGS">
<composite-id name="RatingId" class="RatingId">
<key-property name="user" column="USER_ID" />
<key-property name="movie" column="MOVIE_ID" />
</composite-id>
<property name="rating" />
</class>
</hibernate-mapping>
评级.java
package beans;
import java.util.Date;
public class Rating {
private RatingId ratingId;
private int rating;
private Date timestamp;
public Rating() {} // No arg contructor for hibernate
public Rating(RatingId ratingId, int rating) {
super();
this.ratingId = ratingId;
this.rating = rating;
}
public RatingId getRatingId() {
return ratingId;
}
public void setRatingId(RatingId ratingId) {
this.ratingId = ratingId;
}
public int getRating() {
return rating;
}
public void setRating(int rating) {
this.rating = rating;
}
}
RatingId.java
package beans;
import java.io.Serializable;
public class RatingId implements Serializable {
private User user;
private Movie movie;
public RatingId(User user, Movie movie) {
this.user = user;
this.movie = movie;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Movie getMovie() {
return movie;
}
public void setMovie(Movie movie) {
this.movie = movie;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((movie == null) ? 0 : movie.hashCode());
result = prime * result + ((user == null) ? 0 : user.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
RatingId other = (RatingId) obj;
if (movie == null) {
if (other.movie != null)
return false;
} else if (!movie.equals(other.movie))
return false;
if (user == null) {
if (other.user != null)
return false;
} else if (!user.equals(other.user))
return false;
return true;
}
}
最佳答案
我怀疑复合键列是否可以位于不同的表中(在我们的例子中 - USERS 和 MOVIE 表)。据我了解,它们需要出现在评级表中。 我尝试将它们映射到同一个表中,并且能够成功地保留评级对象。我正在使用 JPA 注释。通过注释,配置更加简单。下面是代码。
注意 - 复合键是在 JPA 中使用 @Embeddeble
和 @EmbeddedId
注释定义的。
@Embeddable
public class User {
@Column(name = "USER_ID")
private long userId;
@Column(name = "EMAIL_ADDRESS")
private String email;
public User() {}
public User(long userId, String email, String password) {
this.userId = userId;
this.email = email;
}
public long getUserId() {
return userId;
}
public void setUserId(long userId) {
this.userId = userId;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
}
@Embeddable
public class Movie {
@Column(name = "MOVIE_ID")
private long movieId;
@Column(name = "MOVIE_NAME")
private String title;
public Movie() {}
public Movie(long movieId, String title) {
this.movieId = movieId;
this.title = title;
}
public long getMovieId() {
return movieId;
}
public void setMovieId(long movieId) {
this.movieId = movieId;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
}
@Embeddable
public class RatingId implements Serializable {
@Embedded
private User user; //Embeddable inside Embeddable.
@Embedded
private Movie movie;
public RatingId() {
}
public RatingId(User user, Movie movie) {
this.user = user;
this.movie = movie;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Movie getMovie() {
return movie;
}
public void setMovie(Movie movie) {
this.movie = movie;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((movie == null) ? 0 : movie.hashCode());
result = prime * result + ((user == null) ? 0 : user.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
RatingId other = (RatingId) obj;
if (movie == null) {
if (other.movie != null)
return false;
} else if (!movie.equals(other.movie))
return false;
if (user == null) {
if (other.user != null)
return false;
} else if (!user.equals(other.user))
return false;
return true;
}
}
@Entity
@Table(name = "RATING")
public class Rating {
@EmbeddedId
private RatingId ratingId;
private int rating;
private Date timestamp;
public Rating() {} // No arg contructor for hibernate
public Rating(RatingId ratingId, int rating) {
super();
this.ratingId = ratingId;
this.rating = rating;
}
public RatingId getRatingId() {
return ratingId;
}
public void setRatingId(RatingId ratingId) {
this.ratingId = ratingId;
}
public int getRating() {
return rating;
}
public void setRating(int rating) {
this.rating = rating;
}
}
调用entityManager.persist( rating)将评级对象保存在数据库中。总而言之,数据库中只有一个表 - RATINGS,它具有所有必需的属性 - 用户、电影和评级。 这也是你所想象的吗?
关于java - 如何修复 hibernate hbm.xml 文件以使复合主键正常工作? (映射异常),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27643483/