So the query you have actually selects the “document” just like it should. But what you are looking for is to “filter the arrays” contained so that the elements returned only match the condition of the query. The real answer is of course that unless you are really saving a lot of bandwidth by filtering … Read more
As noted, the order of the arguments in the array of an $in clause does not reflect the order of how the documents are retrieved. That of course will be the natural order or by the selected index order as shown. If you need to preserve this order, then you basically have two options. So … Read more
$match: { $or: [{ author: ‘dave’ }, { author: ‘john’ }] } Like so, since the $match operator just takes what you would normally put into the find() function
Using aggregate is the right approach, but you need to $unwind the list array before applying the $match so that you can filter individual elements and then use $group to put it back together: db.test.aggregate([ { $match: {_id: ObjectId(“512e28984815cbfcb21646a7”)}}, { $unwind: ‘$list’}, { $match: {‘list.a’: {$gt: 3}}}, { $group: {_id: ‘$_id’, list: {$push: ‘$list.a’}}} ]) … Read more
Starting with MongoDB version 3.0, simply changing the order from collection.aggregate(…).explain() to collection.explain().aggregate(…) will give you the desired results (documentation here). For older versions >= 2.6, you will need to use the explain option for aggregation pipeline operations explain:true db.collection.aggregate([ { $project : { “Tags._id” : 1 }}, { $unwind : “$Tags” }, { $match: … Read more
2017 update $lookup can now directly use an array as the local field. $unwind is no longer needed. Old answer The $lookup aggregation pipeline stage will not work directly with an array. The main intent of the design is for a “left join” as a “one to many” type of join ( or really a … Read more
MongoDB has a distinct command which returns an array of distinct values for a field; you can check the length of the array for a count. There is a shell db.collection.distinct() helper as well: > db.countries.distinct(‘country’); [ “Spain”, “England”, “France”, “Australia” ] > db.countries.distinct(‘country’).length 4 As noted in the MongoDB documentation: Results must not be … Read more
In 4.2+, you can use the $set aggregation pipeline operator which is nothing other than an alias to $addFieldsadded in 3.4 The $addFields stage is equivalent to a $project stage that explicitly specifies all existing fields in the input documents and adds the new fields. db.collection.aggregate([ { “$addFields”: { “custom_field”: “$obj.obj_field1” } } ])
TLDR Summary In modern MongoDB releases you can brute force this with $slice just off the basic aggregation result. For “large” results, run parallel queries instead for each grouping ( a demonstration listing is at the end of the answer ), or wait for SERVER-9377 to resolve, which would allow a “limit” to the number … Read more
Use aggregation on name and get name with count > 1: db.collection.aggregate([ {“$group” : { “_id”: “$name”, “count”: { “$sum”: 1 } } }, {“$match”: {“_id” :{ “$ne” : null } , “count” : {“$gt”: 1} } }, {“$project”: {“name” : “$_id”, “_id” : 0} } ]); To sort the results by most to least … Read more