【发布时间】：2014-06-15 22:49:54
【问题描述】：
我正在使用 Hibernate Search 4.5.1 和 Spring 4.0.5.RELEASE 构建一个应用程序。我正在尝试索引以下类：
@Entity
@Indexed
@Analyzer(impl= org.apache.lucene.analysis.standard.StandardAnalyzer.class)
@Table(name="SONG")
@XmlRootElement(name="song")
public class Song
{
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "ID", updatable = false, nullable = false)
private Long id;
@Field(store = Store.YES)
@Column(name="NAME", length=255)
private String name;
@Field(store = Store.YES)
@Column(name="ALBUM", length=255)
private String album;
@Field(store = Store.YES)
@Column(name="ARTIST", length=255)
private String artist;
@NotNull
@Column(name="PATH", length=255)
private String path;
@NotNull
@Column(name="PATH_COVER", length=255)
private String cover;
@NotNull
@Column(name="LAST_VOTE")
private Date date;
@Field(store = Store.YES)
@NotNull
@Column(name="N_VOTES")
private int nvotes;
@NotNull
@Column(name="ACTIVE", nullable=false, columnDefinition="TINYINT(1) default 0")
private boolean active;
@OneToOne(fetch=FetchType.LAZY)
@JoinColumn(name="IMAGE_ID",insertable=true,updatable=true,nullable=false,unique=false)
private Image image;
@IndexedEmbedded
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "PLAYLIST_ID", nullable = false)
private PlayList playList;
@OneToMany(mappedBy = "song")
private Set<UserVotes> userServices = new HashSet<UserVotes>();
我正在构建一个如下所示的 junit 测试用例：
@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = {"classpath:jukebox-servlet-test.xml"})
@Transactional
public class SongDaoTest {
@Autowired
public I_PlaceDao placeDao;
@Autowired
public I_PlayListDao playListDao;
@Autowired
public I_SongDao songDao;
@Before
public void prepare() throws Exception
{
Operation operation = sequenceOf(CommonOperations.DISABLE_CONTRAINTS, CommonOperations.DELETE_ALL,CommonOperations.INSERT_SONG_DATA, CommonOperations.ENABLE_CONTRAINTS);
DbSetup dbSetup = new DbSetup(new DriverManagerDestination("jdbc:mysql://localhost:3306/jukebox", "root", "mpsbart"), operation);
dbSetup.launch();
FullTextSession fullTextSession = Search.getFullTextSession(placeDao.getSession());
fullTextSession.createIndexer().startAndWait();
}
@Test
@Rollback(false)
public void searchTest()
{
PlayList playList = playListDao.read(1l);
List<Song> songs = songDao.search(playList, "offspring", 1, 10);
assertEquals(10, songs.size());
}
搜索方法实现为：
@SuppressWarnings("unchecked")
public List<Song> search(PlayList playlist, String searchTerm,int page,int limit)
{
FullTextSession fullTextSession = Search.getFullTextSession(getSession());
QueryBuilder queryBuilder = fullTextSession.getSearchFactory().buildQueryBuilder().forEntity(Song.class).get();
BooleanQuery luceneQuery = new BooleanQuery();
luceneQuery.add(queryBuilder.keyword().onFields("name","album","artist").matching("*"+searchTerm+"*").createQuery(), BooleanClause.Occur.MUST);
luceneQuery.add(queryBuilder.phrase().onField("playList.place.id").sentence("\""+playlist.getPlace().getId()+"\"").createQuery(), BooleanClause.Occur.MUST);
luceneQuery.add(queryBuilder.phrase().onField("playList.id").sentence("\""+playlist.getId()+"\"").createQuery(), BooleanClause.Occur.MUST);
// wrap Lucene query in a javax.persistence.Query
FullTextQuery query = fullTextSession.createFullTextQuery(luceneQuery, Song.class);
org.apache.lucene.search.Sort sort = new Sort(new SortField("n_votes",SortField.INT));
query.setSort(sort);
List<Song> songs = query.setFirstResult(page*limit).setMaxResults(limit).list();
return songs;
}
测试结果失败，没有找到任何匹配的对象。使用 luke lucene 时，我可以看到有结果，如果我尝试在 luke 上由 hibernate 生成的查询，它确实会返回元素。 hibernate生成的查询是：+(name:metallica album:metallica artist:metallica) +playList.place.id:"1" +playList.id:"1"
我还在 luke lucene 上注意到一些索引词的长度最多为六个字符，例如，一首歌的艺术家是“The Offspring”，而存储在索引中的词是“the”和“offspr”。第一个没关系，但第二个词不应该是“后代”。为什么要截断名称？
【问题讨论】：
标签： spring lucene hibernate-search