I want to clone a MongoDB collection and save it on the same server with a different name. So for example right now I have the following collections: demo1.categories, demo1
The most simple & efficient way is by using copyTo(), so you can use:
db.source.copyTo("target");
& if "target"
doesn't exist, it will be created
-- Update --
According to CopyTo Documentation, Because copyTo()
uses eval internally, the copy operations will block all other operations on the mongod instance. So it shouldn't be used on production environment.
-- Update --
Because CopyTo()
uses eval() internally & eval()
is deprecated since version 3.0, so CopyTo()
is also deprecated since version 3.0.
The fastest option is
db.myoriginal.aggregate([ { $out: "mycopy" } ])
In the mongo console, you can do the following as well, where db_host is the machine where db_host has the db with the collection that you want to clone.
use db.cloneCollection(, )
Yet again the MongoDB documentation comes to the rescue
assuming that the collection actually is named "demo1.categories":
db.demo1.categories.find().forEach( function(x){db.demo2.categories.insert(x)} );
This is the fastest way to clone your collection:
mongoexport -d db_name -c src_collection | mongoimport -d db_name -c dst_collection --drop
it will clone src_collection in db_name to dst_collection. Or you can do it in two steps on bson level:
mongodump -d db_name -c src_collection
mongorestore --drop -d db_name -c dst_collection ./dump/db_name/src_collection.bson
If you're concerned about speed then I found that by using aggregate
with $project
and $out
to be a 100 times faster, not sure if there are restrictions though, but you would have to create a set of fields that you'd want to copy
For example:
// Set of fields in the categories collection
var setOfFields = {field1:1, field2:1.......}
db.demo1.categories.aggregate([{ "$project": setOfFields},{ $out: "demo2.categories"}]);
This copies (projects) the selected set of fields for all documents from demo1.categories
to demo2.categories