Question

我的Hibernate Search分析器配置存在一些问题。我的一个索引实体（＆＃34; Hospital＆＃34;）有一个String字段（＆＃34; name＆＃34;），其中包含长度为1-40的值。我希望能够通过搜索一个字符找到一个实体（因为有可能，医院有单个字符名称）。

@Indexed(index = "HospitalIndex")
@AnalyzerDef(name = "ngram",
        tokenizer = @TokenizerDef(factory = StandardTokenizerFactory.class),
        filters = {
                @TokenFilterDef(factory = StandardFilterFactory.class),
                @TokenFilterDef(factory = LowerCaseFilterFactory.class),
                @TokenFilterDef(factory = NGramFilterFactory.class,
                        params = {
                                @Parameter(name = "minGramSize", value = "1"),
                                @Parameter(name = "maxGramSize", value = "40")})
        })
public class Hospital {

        @Field(index = Index.YES, analyze = Analyze.YES, store = Store.NO, analyzer = @Analyzer(definition = "ngram"))
        private String name = "";
}

如果我添加名为＆＃34;我的测试医院＆＃34; Lucene索引看起来像这样：

1   name    al
1   name    e
1   name    es
1   name    est
1   name    h
1   name    ho
1   name    hos
1   name    hosp
1   name    hospi
1   name    hospit
1   name    hospita
1   name    hospital
1   name    i
1   name    it
1   name    ita
1   name    ital
1   name    l
1   name    m
1   name    my
1   name    o
1   name    os
1   name    osp
1   name    ospi
1   name    ospit
1   name    ospita
1   name    ospital
1   name    p
1   name    pi
1   name    pit
1   name    pita
1   name    pital
1   name    s
1   name    sp
1   name    spi
1   name    spit
1   name    spita
1   name    spital
1   name    st
1   name    t
1   name    ta
1   name    tal
1   name    te
1   name    tes
1   name    test
1   name    y
1   name    a

这是我构建和执行搜索查询的方式：

QueryBuilder hospitalQb = fullTextEntityManager.getSearchFactory().buildQueryBuilder().forEntity(Hospital.class).get();
Query hospitalQuery = hospitalQb.keyword().onFields("name")().matching(searchString).createQuery();
javax.persistence.Query persistenceQuery = fullTextEntityManager.createFullTextQuery(hospitalQuery, Hospital.class);
List<Hospital> results = persistenceQuery.getResultList();

问题是我的搜索查询也使用相同的ngram分析器。因此，当我寻找＆＃34;医院＆＃34;我会在名称中找到所有包含＆＃34; a＆＃34; -character的医院。当我在其上调用toString方法时，这就是搜索查询的外观：

name:h name:ho name:hos name:hosp name:hospi name:hospit name:hospita name:hospital name:o name:os name:osp name:ospi name:ospit name:ospita name:ospital name:s name:sp name:spi name:spit name:spita name:spital name:p name:pi name:pit name:pita name:pital name:i name:it name:ita name:ital name:t name:ta name:tal name:a name:al name:l

所以问题是，是否有人知道更好的分析器配置或其他方式构建解决问题的搜索查询？

Answer 1

您可以设置第二个分析器，除了它没有ngram过滤器之外，然后覆盖用于查询的分析器：

QueryBuilder hospitalQb = fullTextEntityManager.getSearchFactory().buildQueryBuilder().forEntity(Hospital.class)
    .overridesForField( "name", "my_analyzer_without_ngrams" )
    .get();
// Then it's business as usual

此外，如果您要实施某种自动完成功能（foo*），而不是单词搜索（*foo*），则可能需要使用EdgeNGramFilterFactory代替{ {1}}：它只会生成作为索引标记前缀的ngram。

Hibernate搜索| ngram分析器与minGramSize 1

1 个答案: