当通过hibernate在DB中发生任何更新时,lucene索引不会同步

时间:2014-01-03 07:15:33

标签: hibernate lucene hibernate-search

我正在使用以下env在基于Lucene的Hibernate Search 上处理一些POC:

  • 冬眠-搜索引擎-4.4.2.Final.jar
  • lucene的核 - 3.6.2.jar
  • MySQL 5.5
  • 在域类上使用@Indexed注释。
  • 在字段上使用@Field(index=Index.YES, analyze=Analyze.YES, store=Store.NO)
  • 使用@IndexedEmbedded而不是收集不同域类的实例。

我在应用程序启动时做了明确的索引 ONLY (因为这是在Hibernate Search API中编写的,Hibernate Search将透明地索引通过Hibernate Core持久化,更新或删除的每个实体,{{ 3}})通过以下代码:

private static void doIndex() throws InterruptedException {
    Session session = HibernateSearchUtil.getSession();

    FullTextSession fullTextSession = Search.getFullTextSession(session);
    fullTextSession.createIndexer().startAndWait();

    fullTextSession.close();
}

索引编制时工作正常搜索已在已播种的数据库以及创建&通过应用程序中的hibernate核心删除操作。

但不是通过Hibernate Core对现有数据进行任何更新操作,因为我在通过hibernate搜索进行搜索时没有获得更新的数据

不知道Hibernate Search或lucene是否有问题的原因,意味着索引没有得到更新或者由于没有通过Hibernate Search得到更新结果的其他原因。

用户类:

@Entity
@Table(name = "user")
@Indexed
public class User implements java.io.Serializable {

private static final long serialVersionUID = 5753658991436258019L;
private Integer idUser;

@Field(index = Index.YES, analyze = Analyze.YES, norms = Norms.NO, store = Store.NO)
private String Name;

private Set<UserInfo> userInfos = new HashSet<UserInfo>(0);

public User() {
}

@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "iduser", unique = true, nullable = false)
public Integer getIdUser() {
    return idUser;
}

public void setIdUser(Integer idUser) {
    this.idUser = idUser;
}

@Column(name = "name", nullable = false, length = 256)
public String getName() {
    return this.Name;
}

public void setName(String tenantName) {
    this.Name = tenantName;
}

@IndexedEmbedded
@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "user")
public Set<UserInfo> getUserInfos() {
    return userInfos;
}

public void setUserInfos(Set<UserInfo> userInfos) {
    this.userInfos = userInfos;
    }
}

UserInfo类:

@Entity
@Table(name = "userInfo")
public class UserInfo implements java.io.Serializable {

private static final long serialVersionUID = 5753658991436258019L;
private Integer iduserInfo;
private User user;
private String address; 

public UserInfo() {
}

@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "iduserInfo", unique = true, nullable = false)
public Integer getIduserInfo() {
    return iduserInfo;
}

public void setIduserInfo(Integer iduserInfo) {
    this.iduserInfo = iduserInfo;
}

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

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

@Field(index=Index.YES, analyze=Analyze.YES, store=Store.NO, norms=Norms.NO)
@Column(name = "address", nullable = false, length = 256)
public String getAddress() {
    return address;
}

public void setAddress(String address) {
    this.address = address;
}
}

1 个答案:

答案 0 :(得分:0)

尝试添加

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