我在为多对多关系(用户和消息表)建立新表(UserMessage)时遇到了问题。当我尝试从映射表中访问一行时,我得到一个例外:org.hibernate.PropertyAccessException: could not get a field value by reflection getter of com.package.model.User.userid
UserMessage实体是进一步向关系表添加属性所必需的。
通过以下方式访问数据:
Session session = (Session) entityManager.getDelegate();
String hql = "FROM Message M, UserMessage UM WHERE M.messageid = UM.pk.message and UM.pk.user = :id";
Query query = session.createQuery(hql);
query.setParameter("id", id);
query.list();
我的课程:
User.java
@Entity
@Table(name = "USER")
public class User {
@Id @GeneratedValue(generator="system-uuid")
@GenericGenerator(name="system-uuid", strategy = "uuid")
@Column(nullable = false)
private Long userid;
@Column(unique = true, nullable = false)
private String username;
@OneToMany(fetch = FetchType.EAGER, mappedBy = "pk.user", cascade = {CascadeType.PERSIST})
private Set<UserMessage> inbox = new HashSet<UserMessage>();
public User() {
}
public Long getUserid() {
return userid;
}
public void setUserid(Long id) {
this.userid = id;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public Set<UserMessage> getInbox() {
return inbox;
}
public void setInbox(Set<UserMessage> inbox) {
this.inbox = inbox;
}
public void addMessageToUser(Message message){
UserMessage userMessage = new UserMessage();
userMessage.setUser(this);
userMessage.setMessage(message);
inbox.add(userMessage);
}
}
Message.java
@Entity
@Table(name = "MESSAGE")
public class Message {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Long messageid;
private String subject;
private String body;
@OneToMany(fetch = FetchType.EAGER, mappedBy = "pk.message",cascade = {CascadeType.PERSIST})
private Set<UserMessage> toUser = new HashSet<UserMessage>();
public Set<UserMessage> getToUser() {
return toUser;
}
public void setToUser(Set<UserMessage> toUser) {
this.toUser = toUser;
}
public void addUserToMessage(User user){
UserMessage userMessage = new UserMessage();
userMessage.setUser(user);
userMessage.setMessage(this);
toUser.add(userMessage);
}
public Message() {
}
public Long getMessageid() {
return messageid;
}
public void setMessageid(Long id) {
this.messageid = id;
}
public String getSubject() {
return subject;
}
public void setSubject(String subject) {
this.subject = subject;
}
public String getBody() {
return body;
}
public void setBody(String body) {
this.body = body;
}
}
UserMessage.java
@Entity
@Table(name = "USER_MESSAGE")
@AssociationOverrides({
@AssociationOverride(name = "pk.user", joinColumns = @JoinColumn(name = "userid")),
@AssociationOverride(name = "pk.message", joinColumns = @JoinColumn(name = "messageid"))
})
public class UserMessage{
@EmbeddedId
private UserMessagePK pk = new UserMessagePK();
public UserMessagePK getPk(){
return pk;
}
public void setPk(UserMessagePK pk){
this.pk = pk;
}
public UserMessage(){
}
@Transient
public User getUser(){
return this.pk.getUser();
}
@Transient
public Message getMessage(){
return this.pk.getMessage();
}
public void setUser(User user){
this.pk.setUser(user);
}
public void setMessage(Message message){
this.pk.setMessage(message);
}
}
UserMessagePK.java
@Embeddable
public class UserMessagePK implements Serializable {
private static final long serialVersionUID = 1L;
@ManyToOne
private User user;
@ManyToOne(cascade = CascadeType.REFRESH)
private Message message;
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Message getMessage() {
return message;
}
public void setMessage(Message message) {
this.message = message;
}
public UserMessagePK(User user, Message message){
this.user = user;
this.message = message;
}
public UserMessagePK(){
}
}
有什么问题?
答案 0 :(得分:1)
问题出在查询中。
select distinct M from Message M join M.toUser t where t.pk.user = :user
效果很好。
答案 1 :(得分:0)
我建议采用不同的方法。 在User类中,不要持有一组UserMessage,而是写下:
@Entity
@Table(name = "Users")
public class User implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "User_Id")
private Long userid;
@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, orphanRemoval = true)
@JoinTable(name = "USER_MESSAGE", joinColumns = @JoinColumn(name = "User_Id"), inverseJoinColumns = @JoinColumn(name = "Message_Id"))
protected Set<Message> userMessages = new HashSet<>(0);
public void addUserMessage(Message toAdd) {
if (null != toAdd) {
this.userProviders.add(toAdd);
}
}
public void getUserMessages() {
return this.userMessages;
}
public void setUserMessages(Set<Message> userMessages) {
this.userMessages.clear();
this.userMessages.addAll(userMessages);
}
// add getter and setter to id
...
在Message类中,您应该:
@Entity
@Table(name = "User_Messages")
public class Message implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "Message_Id")
protected long id;
@ManyToOne(fetch = FetchType.LAZY)
@JoinTable(name = "Users_Messages", joinColumns = @JoinColumn(name = "Message_Id", insertable = false, updatable = false), inverseJoinColumns = @JoinColumn(name = "User_Id"))
protected User user;
// add getters and setters
...
您也可以摆脱UserMessagePK类和UserMessage(为什么需要这些?)。 现在,当您生成表格时(可以使用hbm2ddl tool),您应该有一个用户&lt; - &gt;消息关系的连接表。
现在您可以运行查询。