Most search engines can't directly index relational content, as documents in the index logically behave like a single flat database table. Yet, relational content is everywhere! A job listing site has each company joined to the specific listings for that company. Each resume might have separate list of skills, education and past work experience. A music search engine has an artist/band joined to albums and then joined to songs. A source code search engine would have projects joined to modules and then files.
Perhaps the PDF documents you need to search are immense, so you break them up and index each section as a separate Lucene document; in this case you'll have common fields (title, abstract, author, date published, etc.) for the overall document, joined to the sub-document (section) with its own fields (text, page number, etc.). XML documents typically contain nested tags, representing joined sub-documents; emails have attachments; office documents can embed other documents. Nearly all search domains have some form of relational content, often requiring more than one join.
If such content is so common then how do search applications handle it today?
One obvious "solution" is to simply use a relational database instead of a search engine! If relevance scores are less important and you need to do substantial joining, grouping, sorting, etc., then using a database could be best overall. Most databases include some form a text search, some even using Lucene.
If you still want to use a search engine, then one common approach is to denormalize the content up front, at index-time, by joining all tables and indexing the resulting rows, duplicating content in the process. For example, you'd index each song as a Lucene document, copying over all fields from the song's joined album and artist/band. This works correctly, but can be horribly wasteful as you are indexing identical fields, possibly including large text fields, over and over.
Another approach is to do the join yourself, outside of Lucene, by indexing songs, albums and artist/band as separate Lucene documents, perhaps even in separate indices. At search-time, you first run a query against one collection, for example the songs. Then you iterate through all hits, gathering up (joining) the full set of corresponding albums and then run a second query against the albums, with a large OR'd list of the albums from the first query, repeating this process if you need to join to artist/band as well. This approach will also work, but doesn't scale well as you may have to create possibly immense follow-on queries.
Yet another approach is to use a software package that has already implemented one of these approaches for you! elasticsearch, Apache Solr, Apache Jackrabbit, Hibernate Search and many others all handle relational content in some way.
BlockJoinQueryyou can now directly search relational content yourself!
stock keeping units or SKUs, which have their own fields like size, color, inventory count, etc. The SKUs are what you actually sell, and what you must stock, because when someone buys a shirt they buy a specific SKU (size and color).
Maybe you are lucky enough to sell the incredible Mountain Three-wolf Moon Short Sleeve Tee, with these SKUs (size, color):
- small, blue
- small, black
- medium, black
- large, gray
name:wolf AND size=small AND color=bluewhich should match this shirt.
nameis a shirt field while the
colorare SKU fields.
But if the user drills down instead on a small gray shirt:
name:wolf AND size=small AND color=graythen this shirt should not match because the small size only comes in blue and black.
How can you run these queries using
BlockJoinQuery? Start by indexing each shirt (parent) and all of its SKUs (children) as separate documents, using the new
IndexWriter.addDocumentsAPI to add one shirt and all of its SKUs as a single document block. This method atomically adds a block of documents into a single segment as adjacent document IDs, which
BlockJoinQueryrelies on. You should also add a marker field to each shirt document (e.g.
type = shirt), as
Filteridentifying the parent documents.
To run a
BlockJoinQueryat search-time, you'll first need to create the parent filter, matching only shirts. Note that the filter must use
FixedBitSetunder the hood, like
Filter shirts = new CachingWrapperFilter( new QueryWrapperFilter( new TermQuery( new Term("type", "shirt"))));Create this filter once, up front and re-use it any time you need to perform this join.
Then, for each query that requires a join, because it involves both SKU and shirt fields, start with the child query matching only SKU fields:
BooleanQuery skuQuery = new BooleanQuery(); skuQuery.add(new TermQuery(new Term("size", "small")), Occur.MUST); skuQuery.add(new TermQuery(new Term("color", "blue")), Occur.MUST);Next, use
BlockJoinQueryto translate hits from the SKU document space up to the shirt document space:
BlockJoinQuery skuJoinQuery = new BlockJoinQuery( skuQuery, shirts, ScoreMode.None);The
ScoreModeenum decides how scores for multiple SKU hits should be aggregated to the score for the corresponding shirt hit. In this query you don't need scores from the SKU matches, but if you did you can aggregate with
Finally you are now free to build up an arbitrary shirt query using
skuJoinQueryas a clause:
BooleanQuery query = new BooleanQuery(); query.add(new TermQuery(new Term("name", "wolf")), Occur.MUST); query.add(skuJoinQuery, Occur.MUST);You could also just run
skuJoinQueryas-is if the query doesn't have any shirt fields.
Finally, just run this
querylike normal! The returned hits will be only shirt documents; if you'd also like to see which SKUs matched for each shirt, use
BlockJoinCollector c = new BlockJoinCollector( Sort.RELEVANCE, // sort 10, // numHits true, // trackScores false // trackMaxScore ); searcher.search(query, c);The provided
Sortmust use only shirt fields (you cannot sort by any SKU fields). When each hit (a shirt) is competitive, this collector will also record all SKUs that matched for that shirt, which you can retrieve like this:
hits = c.getTopGroups( skuJoinQuery, skuSort, 0, // offset 10, // maxDocsPerGroup 0, // withinGroupOffset true // fillSortFields );
skuSortto the sort order for the SKUs within each shirt. The first
offsethits are skipped (use this for paging through shirt hits). Under each shirt, at most
maxDocsPerGroupSKUs will be returned. Use
withinGroupOffsetif you want to page within the SKUs. If
fillSortFieldsis true then each SKU hit will have values for the fields from
The hits returned by
BlockJoinCollector.getTopGroupsare SKU hits, grouped by shirt. You'd get the exact same results if you had denormalized up-front and then used grouping to group results by shirt.
You can also do more than one join in a single query; the joins can be nested (parent to child to grandchild) or parallel (parent to child1 and parent to child2).
However, there are some important limitations of index-time joins:
- The join must be computed at index-time and "compiled" into the
index, in that all joined child documents must be indexed along
with the parent document, as a single document block.
- Different document types (for example, shirts and SKUs) must
share a single index, which is wasteful as it means non-sparse
data structures like
FieldCacheentries consume more memory than they would if you had separate indices.
- If you need to re-index a parent document or any of its child
documents, or delete or add a child, then the entire block must be
re-indexed. This is a big problem in some cases, for example if
you index "user reviews" as child documents then whenever a user
adds a review you'll have to re-index that shirt as well as all
its SKUs and user reviews.
- There is no
QueryParsersupport, so you need to programmatically create the parent and child queries, separating according to parent and child fields.
- The join can currently only go in one direction (mapping child
docIDs to parent docIDs), but in some cases you need to map parent
docIDs to child docIDs. For example, when searching songs,
perhaps you want all matching songs sorted by their title. You
can't easily do this today because the only way to get song hits
is to group by album or band/artist.
- The join is a one (parent) to many (children), inner join.
There is work underway to create a more flexible, but likely less performant, query-time join capability, which should address a number of the above limitations.