What is the default sort order when none is specified? The default internal sort order (or natural order) is an undefined implementation detail. Maintaining order is extra overhead for storage engines and MongoDB’s API does not mandate predictability outside of an explicit sort() or the special case of fixed-sized capped collections which have associated usage … Read more
db.messages.find( { headers : { From: “[email protected]” } } ) This queries for documents where headers equals { From: … }, i.e. contains no other fields. db.messages.find( { ‘headers.From’: “[email protected]” } ) This only looks at the headers.From field, not affected by other fields contained in, or missing from, headers. Dot-notation docs
2017 update $lookup can now directly use an array as the local field. $unwind is no longer needed. Old answer The $lookup aggregation pipeline stage will not work directly with an array. The main intent of the design is for a “left join” as a “one to many” type of join ( or really a … Read more
There’s a more efficient way to do this in MongoDB 2.2+ now that you can use numeric array indexes (0 based) in query object keys. // Find all docs that have at least two name array elements. db.accommodations.find({‘name.1’: {$exists: true}}) You can support this query with an index that uses a partial filter expression (requires … 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
Starting with the 3.2 release of MongoDB, you can get N random docs from a collection using the $sample aggregation pipeline operator: // Get one random document from the mycoll collection. db.mycoll.aggregate([{ $sample: { size: 1 } }]) If you want to select the random document(s) from a filtered subset of the collection, prepend a … Read more
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
You could do this with MapReduce: mr = db.runCommand({ “mapreduce” : “my_collection”, “map” : function() { for (var key in this) { emit(key, null); } }, “reduce” : function(key, stuff) { return null; }, “out”: “my_collection” + “_keys” }) Then run distinct on the resulting collection so as to find all the keys: db[mr.result].distinct(“_id”) [“foo”, … Read more
There are a couple of ways to do this. The first is with Date Aggregation Operators, which allow you to dissect the “date” values in documents. Specifically for “grouping” as the primary intent: db.collection.aggregate([ { “$group”: { “_id”: { “year”: { “$year”: “$created_at” }, “dayOfYear”: { “$dayOfYear”: “$created_at” }, “hour”: { “$hour”: “$created_at” }, “interval”: … Read more
You can use a $where. Just be aware it will be fairly slow (has to execute Javascript code on every record) so combine with indexed queries if you can. db.T.find( { $where: function() { return this.Grade1 > this.Grade2 } } ); or more compact: db.T.find( { $where : “this.Grade1 > this.Grade2” } ); UPD for … Read more