CODEWITHSUNDEEP
mongodbmongodb-query
xabi
xabi

Reputation: 193

How to get tag count in mongodb query results?

I'm trying to search in mongodb for articles based on tags:

> use test;
> db.articles.save( { name: "article1", tags: ['one', 'two', 'three'] } );
> db.articles.save( { name: "article2", tags: ['two', 'three'] } );
> db.articles.save( { name: "article3", tags: ['one', 'three'] } );
> db.articles.save( { name: "article4", tags: ['one', 'two', 'four'] } );
> db.articles.save( { name: "article5", tags: ['four', 'five'] } );
> db.articles.ensureIndex( { tags: 1 } );
> db.articles.find( { $or : [ { tags: 'four' }, { tags: 'five' } ] } )

{ "_id" : ObjectId("509d7b555bff77729a26a232"), "name" : "article4", "tags" : [ "one", "two", "four" ] }
{ "_id" : ObjectId("509d7b555bff77729a26a233"), "name" : "article5", "tags" : [ "four", "five" ] }

As you can see the query looks for "four" or "five", but I would like to get the results sorted by "best match" (two tags matched are better than one tag matched).

How can I get this done? map reduce?

Thanks in advance:

xabi

Upvotes: 3

Views: 379

Answers (1)

Salvador Dali
Salvador Dali

Reputation: 222751

I am really sorry to disappoint you but there is nothing mongo can do right now. The only option is to implement sorting on the application side. I know that this can be too inefficient but there is no other option.

The only thing I can help you with is to suggest to substitute 

$or : [ { tags: 'four' }, { tags: 'five' } ]

to 

tags:{$in:['four','five']}

this will not help you with your query (I mean to order results), but it will be easier to you to create it on the server if there will be a lot of different tags.

Upvotes: 1

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