Save Subset of MongoDB Collection to Another Collection
MongodbMongodb Problem Overview
I have a set like so
{date: 20120101}
{date: 20120103}
{date: 20120104}
{date: 20120005}
{date: 20120105}
How do I save a subset of those documents with the date '20120105' to another collection?
i.e db.subset.save(db.full_set.find({date: "20120105"}));
Mongodb Solutions
Solution 1 - Mongodb
I would advise using the aggregation framework:
db.full_set.aggregate([ { $match: { date: "20120105" } }, { $out: "subset" } ])
It works about 100 times faster than forEach
at least in my case. This is because the entire aggregation pipeline runs in the mongod process, whereas a solution based on find()
and insert()
has to send all of the documents from the server to the client and then back. This has a performance penalty, even if the server and client are on the same machine.
Solution 2 - Mongodb
Here's the shell version:
db.full_set.find({date:"20120105"}).forEach(function(doc){
db.subset.insert(doc);
});
Note: As of MongoDB 2.6, the aggregation framework makes it possible to do this faster; see melan's answer for details.
Solution 3 - Mongodb
Actually, there is an equivalent of SQL's insert into ... select from
in MongoDB. First, you convert multiple documents into an array of documents; then you insert the array into the target collection
db.subset.insert(db.full_set.find({date:"20120105"}).toArray())
Solution 4 - Mongodb
The most general solution is this:
Make use of the aggregation (answer given by @melan):
db.full_set.aggregate({$match:{your query here...}},{$out:"sample"})
db.sample.copyTo("subset")
This works even when there are documents in "subset" before the operation and you want to preserve those "old" documents and just insert a new subset into it.
Care must be taken, because the copyTo()
command replaces the documents with the same _id
.
Solution 5 - Mongodb
There's no direct equivalent of SQL's insert into ... select from ...
.
You have to take care of it yourself. Fetch documents of interest and save them to another collection.
You can do it in the shell, but I'd use a small external script in Ruby. Something like this:
require 'mongo'
db = Mongo::Connection.new.db('mydb')
source = db.collection('source_collection')
target = db.collection('target_collection')
source.find(date: "20120105").each do |doc|
target.insert doc
end
Solution 6 - Mongodb
Mongodb has aggregate along with $out operator which allow to save subset into new collection. Following are the details :
$out Takes the documents returned by the aggregation pipeline and writes them to a specified collection.
- The $out operation creates a new collection in the current database if one does not already exist.
- The collection is not visible until the aggregation completes.
- If the aggregation fails, MongoDB does not create the collection.
Syntax :
{ $out: "<output-collection>" }
Example A collection books contains the following documents:
{ "_id" : 8751, "title" : "The Banquet", "author" : "Dante", "copies" : 2 }
{ "_id" : 8752, "title" : "Divine Comedy", "author" : "Dante", "copies" : 1 }
{ "_id" : 8645, "title" : "Eclogues", "author" : "Dante", "copies" : 2 }
{ "_id" : 7000, "title" : "The Odyssey", "author" : "Homer", "copies" : 10 }
{ "_id" : 7020, "title" : "Iliad", "author" : "Homer", "copies" : 10 }
The following aggregation operation pivots the data in the books collection to have titles grouped by authors and then writes the results to the authors collection.
db.books.aggregate( [
{ $group : { _id : "$author", books: { $push: "$title" } } },
{ $out : "authors" }
] )
After the operation, the authors collection contains the following documents:
{ "_id" : "Homer", "books" : [ "The Odyssey", "Iliad" ] }
{ "_id" : "Dante", "books" : [ "The Banquet", "Divine Comedy", "Eclogues" ] }
In the asked question, use following query and you will get new collection named 'col_20120105' in your database
db.products.aggregate([
{ $match : { date : "20120105" } },
{ $out : "col_20120105" }
]);