What is a proper way to duplicate a collection in Mongodb on the same server using C#? MongoVUE has an option 'Duplicate collection', is there something similar for C#?
问题:
回答1:
There isn't a built-in way to copy collections with the C# driver, but you can still do it pretty simply as:
var source = db.GetCollection("test");
var dest = db.GetCollection("testcopy");
dest.InsertBatch(source.FindAll());
Note, however, that this won't copy any indexes from the source collection. The shell's copyTo
method has the same limitation so it's likely implemented similarly.
回答2:
I had the exact same problem, but while the accepted answer works, I also needed to make it as fast as possible.
The fastest way to copy a collection is apparently using an aggregate with an $out
pipeline stage. This way, you won't have to download all the documents and then re-upload them, they are just copied inside the database.
This is trivial to execute inside the mongo shell:
db.originalColl.aggregate([ { $match: {} }, { $out: "resultColl"} ]);
However, I had a lot of trouble running this from C#. Since eval
has now been deprecated, you can't just stuff the above in a string to be executed on the server. Instead you need to construct a Bson document that represents the above code.
Here's how I made it work:
var aggDoc = new Dictionary<string,object>
{
{"aggregate" , "originalCollection"},
{"pipeline", new []
{
new Dictionary<string, object> { { "$match" , new BsonDocument() }},
new Dictionary<string, object> { { "$out" , "resultCollection"}}
}
}
};
var doc = new BsonDocument(aggDoc);
var command = new BsonDocumentCommand<BsonDocument>(doc);
db.RunCommand(command);
This turns out to be very fast (about 3 minutes to copy 5M documents), and no data is transferred between the db and the application running the above code. One drawback is that it creates a temporary collection, so the resultCollection
will be empty (or not existing) until the operation completes. So if you have a progress bar that is based on the size of the resultCollection
it will no longer work.