In Mongo 2.8 RC2 there is a new data aggregation operator: $dateToString which can be used to group by a day and simply have a “YYYY-MM-DD” in the result: Example from the documentation: db.sales.aggregate( [ { $project: { yearMonthDay: { $dateToString: { format: “%Y-%m-%d”, date: “$date” } }, time: { $dateToString: { format: “%H:%M:%S:%L”, date: … Read more
The main purpose of the aggregation framework is to ease the query of a big number of entries and generate a low number of results that hold value to you. As you have said, you can also use multiple find queries, but remember that you can not create new fields with find queries. On the … Read more
I solved the same problem just last night, this way: > db.test.aggregate({$group:{_id:{$gt:[“$field”, null]}, count:{$sum:1}}}) { “_id” : true, “count” : 2 } { “_id” : false, “count” : 2 } See http://docs.mongodb.org/manual/reference/bson-types/#bson-types-comparison-order for a full explanation of how this works. Added From comment section: To check if the value doesn’t exist or is null use … Read more
The parameter in your Javascript is an array of 4 objects with one element each, in your PHP it’s an associative array (object) with 4 elements. This would represent your Javascript: $result = $c->aggregate(array( array( ‘$project’ => array( ‘day’ => array(‘$dayOfYear’ => ‘$executed’) ), ), array( ‘$group’ => array( ‘_id’ => array(‘day’ => ‘$day’), ‘n’ … Read more
The best way to do this is using the aggregation framework. You need to $group your documents by “user” and return the last document for each user using the $last accumulator operator but for this to work, you need a preliminary sorting stage using the $sort aggregation pipeline operator. To sort your documents, you need … Read more
Your current schema has the marks field data type as string and you need an integer data type for your aggregation framework to work out the sum. On the other hand, you can use MapReduce to calculate the sum since it allows the use of native JavaScript methods like parseInt() on your object properties in … Read more
General Problem of Dealing with “local dates” So there is a short answer to this and a long answer as well. The basic case is that instead of using any of the “date aggregation operators” you instead rather want to and “need to” actually “do the math” on the date objects instead. The primary thing … Read more
MongoDB 4.0 adds the $convert aggregation operator and the $toString alias which allows you to do exactly that: db.getCollection(‘example’).aggregate([ { “$match”: { “example”:1 } }, { “$project”: { “_id”: { “$toString”: “$_id” } } } ]) A main usage would most likely be though to use the _id value as a “key” in a document. … Read more
For any particular person document, you can use the populate() function like var query = mongoose.model(“person”).find({ “name”: “foo” }).populate(“projects.tags”); And if you want to search for any persons that have any tag with ‘MongoDB’ or ‘Node JS’ for example, you can include the query option in the populate() function overload as: var query = mongoose.model(“person”).find({ … Read more
Starting with Mongo 2.6.0 you can do this natively without any additional manipulation. db.<collection>.aggregate( [ { <operation> }, { <operation> }, …, { $out : “<output-collection>” } ] ) Check the new aggregation operator $out for more detailed example. P.S. using this way you are not limited to 16Mb size.