Mongodb Aggregation : How to return only matching elements of an array [duplicate]

Answer 1

Building on the response from cirrus, I recommend doing the $unwind first to avoid the redundant $match. Something like:

db.books.aggregate(
    {$unwind:"$indexTokens"},
    {$match:{indexTokens:/^Par/}},
    {$group:{_id:null,indexTokens:{$push:"$indexTokens"}}
})

How do you do this in Java? You can use the DBCollection.aggregate(...) method of the MongoDB v2.9.0 driver. Each pipeline operator, eg. $unwind or $match, corresponds to a DBObject object.

Answer 2

You could do this with the 2.2 aggregation framework. Something like this;

db.books.runCommand("aggregate", {
    pipeline: [
        {   // find docs that contain Par*
            $match: { "indexTokens" : { "$regex" : "^Par" , "$options" : "i"}},
        },
        {   // create a doc with a single array elemm for each indexToken entry
            $unwind: "$indexTokens" 
        },
        {   // now produce a list of index tokens
            $group: {
                _id: "$indexTokens",
            },
        },
    ],
})

Or this might be even closer to what you're after if you really want the array without the doc;

db.books.runCommand("aggregate", {
    pipeline: [
        {   // find docs that contain Par*
            $match: { "indexTokens" : { "$regex" : "^Par" , "$options" : "i"}},
        },
        {   // create a doc with a single array elemm for each indexToken entry
            $unwind: "$indexTokens" 
        },
        {   // now throw out any unwind's that DON'T contain Par*
            $match: { "indexTokens": { "$regex": "^Par", "$options": "i" } },
        },
        {   // now produce the list of index tokens
            $group: {
                _id: null,
                indexTokens: { $push: "$indexTokens" },
            },
        },
    ],
})