The standard driver takes java.util.date types and serializes as BSON dates. So with a collection object to “example” Date now = new Date(); BasicDBObject timeNow = new BasicDBObject(“date”, now); example.insert(timeNow); If you are looking for a way to use the “server” time in operations, there is the $currentDate operator, but this works with “updates”, so … Read more
MongoDB will not create an index on a collection if the index entry for an existing document exceeds the index key limit (1024 bytes). You can however create a hashed index or text index instead: db.logcollection.createIndex({“Module”:”hashed”}) or db.logcollection.createIndex({“Module”:”text”})
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 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 $sort aggregation pipeline operator. To sort your documents, you need … Read more
Your current schema has the marks field data type as string and you need an integer data type for your aggregation framework to work out the sum. On the other hand, you can use MapReduce to calculate the sum since it allows the use of native JavaScript methods like parseInt() on your object properties in … Read more
Use python regular expressions. import re db.col.find({‘music_description’: {‘$in’: [ re.compile(‘.*heavy.*’), re.compile(‘.*metal.*’)]}})
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
in this case there is required a nice play with $unwind and $project in aggregation framework please see below: db.alumni.aggregate([ {$match: {_id: ‘john’}}, {$unwind:”$items”}, {$unwind:”$items.items”}, {$lookup: { from: ‘schools’, localField: ‘items.items.school’, foreignField: ‘_id’, as: ‘schoolInfo’}}, {$unwind:”$schoolInfo”}, {$project:{ “_id”:1, “items”:[{ “name”:”$items.name”, “items”:[{ “school”:”$schoolInfo._id” , “grad”:”$items.items.grad” , “schoolInfo”:”$schoolInfo” }] }] }} ]).pretty() to see how it works … Read more
Good and bad news here. Bad news is that AFAIK you can’t do it with a single update() call – mongo doesn’t support referring to current object in update. Good news is that there are other ways to do it, e.g. you can run a forEach loop: db.item.find(conditions…).snapshot().forEach( function (doc) { doc.field1 = doc.field2; db.item.save(doc); … Read more