Dealing with mongodb unique, sparse, compound indexes

Question

Because mongodb will index sparse, compound indexes that contain 1 or more of the indexed fields, it is causing my unique, sparse index to fail because one of those fields

Answer 1

A hash index ended up being sufficient for this

Answer 2

https://docs.mongodb.org/manual/core/index-partial/

As of mongoDB 3.2 you can create partial index to support this as well.

db.users.createIndex(
   { name: 1, email: 1 },
   { unique: true, partialFilterExpression: { email: { $exists: true } } }
)

Answer 3

A sparse index avoids indexing a field that doesn't exist. A unique index avoid documents being inserted that have the same field values. Unfortunately as of MongoDB 2.6.7, the unique constraint is always enforced even when creating a compound index (indexing two or more fields) with the sparse and unique properties.

Example:

db = db.connect("test");
db.a.drop();
db.a.insert([
    {},
    {a : 1},
    {b : 1},
    {a : 1, b : 1}
]);
db.a.ensureIndex({a:1,b:1},  { sparse: true, unique: true } );
db.a.insert({a : 1}); // throws Error but wanted insert to be valid.

However, it works as expected for a single index field with sparse and unique properties. I feel like this is a bug that will get fixed in future releases.

Anyhow, here are two solutions to get around this problem.

1) Add a non-null hash field to each document that is only computed when all the required fields for checking the uniqueness are supplied. Then create a sparse unique index on the hash field.

function createHashForUniqueCheck(obj){
    if( obj.firstName && obj.id){
        return MD5( String( obj.firstName) + String(obj.id) );
    }
    return null;
}

2) On the application side, check for uniqueness before insertion into Mongodb. :-)

• sparse index doc