[FIXED] Cannot define analyzer in hibernate search 6

Issue

I define an analyzer on hibernate search 6

public class MyLuceneAnalysisConfigurer implements LuceneAnalysisConfigurer {
    @Override
    public void configure(LuceneAnalysisConfigurationContext context) {
        context.analyzer( "english" ).custom()
                .tokenizer( StandardTokenizerFactory.class )
                .charFilter( HTMLStripCharFilterFactory.class )
                .tokenFilter( LowerCaseFilterFactory.class )
                .tokenFilter( SnowballPorterFilterFactory.class )
                .param( "language", "English" )
                .tokenFilter( ASCIIFoldingFilterFactory.class );
        ;


        context.analyzer( "french" ).custom()
                .tokenizer( StandardTokenizerFactory.class )
                .charFilter( HTMLStripCharFilterFactory.class )
                .tokenFilter( LowerCaseFilterFactory.class )
                .tokenFilter( SnowballPorterFilterFactory.class )
                .param( "language", "French" )
                .tokenFilter( ASCIIFoldingFilterFactory.class );
    }
}

After I add this to application.properties:

hibernate.search.backend.indexes.<Product>.analysis.configurer = com.example.springmvcrest.services.MyLuceneAnalysisConfigurer 

but its not resolved by IntelliJ,

and after running:

 Analyzer not found org.hibernate.search.util.common.SearchException: HSEARCH000353: Unknown analyzer: 'english'. Make sure you defined this analyzer.

Any solution? I use spring boot 2.4.0

Solution

The following should work:

In your application.properties, change this

hibernate.search.backend.indexes.<Product>.analysis.configurer

to:

hibernate.search.backend.analysis.configurer

In summary:

spring.jpa.properties.hibernate.search.backend.analysis.configurer=com.example.springmvcrest.services.MyLuceneAnalysisConfigurer

Source: Hibernate Search 6.0.1.Final: Reference Documentation

Answered By - manu
Answer Checked By - Timothy Miller (JavaFixing Admin)