如何使用mongo db解决唯一索引中的缺失值？

Question

The mongo documentation states that "When a document is saved to a collection with unique indexes, any missing indexed keys will be inserted with null values. Thus, it won't be possible to insert multiple documents missing the same indexed key."

So is it impossible to create a unique index on an optional field? Should I create a compound index with say a userId as well to solve this? In my specific case I have a user collection that has an optional embedded oauth object. eg

>db.users.ensureIndex( { "name":1, "oauthConnections.provider" : 1, "oauthConnections.providerId" : 1 } );

My sample user

{  name: "Bob"
   ,pwd: "myPwd"
   ,oauthConnections [
      {
         "provider":"Facebook",
         "providerId" : "12345",
         "key":"blah"
      }
     ,{
         "provider":"Twitter",
         "providerId" : "67890",
         "key":"foo"
      }
     ]
}

Answer 1

I believe that this is possible: You can have an index that is sparse and unique . This way, non-existant values never make it to the index, hence they can't be duplicate.

Caveat : This is not possible with compound indexes. I'm not quite sure about your question. Your citing a part of the documentation that concerns compound indexes -- there, missing values will be inserted, but from your question I guess you're not looking for a solution w/ compound indexes?

Here's a sample:

> db.Test.insert({"myId" : "1234", "string": "foo"});
> show collections
Test
system.indexes
>
> db.Test.find();
{ "_id" : ObjectId("4e56e5260c191958ad9c7cb1"), "myId" : "1234", "string" : "foo" }
>

> db.Test.ensureIndex({"myId" : 1}, {sparse: true, unique: true});
>
> db.Test.insert({"myId" : "1234", "string": "Bla"});
E11000 duplicate key error index: test.Test.$myId_1  dup key: { : "1234" }
>
> db.Test.insert({"string": "Foo"});
> db.Test.insert({"string": "Bar"});
> db.Test.find();
{ "_id" : ObjectId("4e56e5260c191958ad9c7cb1"), "myId" : "1234", "string" : "foo" }
{ "_id" : ObjectId("4e56e5c30c191958ad9c7cb4"), "string" : "Foo" }
{ "_id" : ObjectId("4e56e5c70c191958ad9c7cb5"), "string" : "Bar" }

Also note that compound indexes can't be sparse

Answer 2

It is not impossible to index an optional field. The docs are talking about a unique index. Once you've specified a unique index, you can only insert one document per value for that field, even if that value is null.

If you want a unique index on an optional field but still allow multiple nulls, you could try making the index both unique and sparse, although I have no idea if that's possible. I couldn't find an answer in the documentation.

Answer 3

There's no good way to uniquely index an optional field. You can either fill it with a default (the _id on the user would work), let your access layer enforce uniqueness, or change your "schema" a bit.

We have a separate collection for oauth login tokens, partially for this reason. We never really need to access those in a context where having them as embedded docs is an obvious win. If this is a relatively easy change to make, it's probably your best bet.

----edit----

As the other answers points, you can achieve this with a sparse index. It's even a documented use. You should probably accept one of those answers instead of mine.

http://www.mongodb.org/display/DOCS/Indexes#Indexes-SparseIndexes