Unique index in mongoDB 3.2 ignoring null values - mongodb

Question

I want to add the unique index to a field ignoring null values in the unique indexed field and ignoring the documents that are filtered based on partialFilterExpression.
The problem is Sparse indexes can't be used with the Partial index.
Also, adding unique indexes, adds the null value to the index key field and hence the documents can't be ignored based on $exist criteria in the PartialFilterExpression.
Is it possible in MongoDB 3.2 to get around this situation?

Answer 1

I am adding this answer as I was looking for a solution and didn't find one. This may not answer exactly this question or may be, but will help lot of others out there like me.
Example. If the field with null is houseName and it is of type string, the solution can be like this
db.collectionName.createIndex(
{name: 1, houseName: 1},
{unique: true, partialFilterExpression: {houseName: {$type: "string"}}}
);
This will ignore the null values in the field houseName and still be unique.

Answer 2

Yes, you can create partial index in MongoDB 3.2
Please see https://docs.mongodb.org/manual/core/index-partial/#index-type-partial
MongoDB recommend usage of partial index over sparse index. I'll suggest you to drop your sparse index in favor of partial index.

Answer 3

You can create partial index in mongo:3.2.
Example, if ipaddress can be "", but "127.0.0.1" should be unique. The solution can be like this:
db.collectionName.createIndex(
{"ipaddress":1},
{"unique":true, "partialIndexExpression":{"ipaddress":{"$gt":""}}})
This will ignore "" values in ipaddress filed and still be unique

Answer 4

{
"YourField" : {
"$exists" : true,
"$gt" : "0",
"$type" : "string"
}
}

Answer 5

To create at mongodbCompass you must write it as JSON:
for find other types wich supports see this link.

Answer 6

Yes, that can be a kind of a problem that the partial filter expression cannot contain any 'not' filters.
For those who can be interested in a C# solution for an index like this, here is an example.
We have a 'User' entity, which has one-to-one 'relation' to a 'Doctor' entity.
This relation is represented by the not required, nullable field 'DoctorId' in the 'User' entity. In other words, there is a requirement that a given 'Doctor' can be linked to only single 'User' at a time.
So we need an unique index which can fire an exception when something attempts to set DoctorId to the same Guid which already set for any other 'User' entity. At the same time multiple 'null' entries must be allowed for the 'DoctorId' field, since many users do not have any doctor attached to them.
The solution to build this kind of an index looks like:
var uniqueDoctorIdIndexDefinition = new IndexKeysDefinitionBuilder<User>()
.Ascending(o => o.DoctorId);
var existsFilter = Builders<User>.Filter.Exists(o => o.DoctorId);
var notNullFilter = Builders<User>.Filter.Type(o => o.DoctorId, BsonType.String);
var andFilter = Builders<User>.Filter.And(existsFilter, notNullFilter);
var createIndexOptions = new CreateIndexOptions<User>
{
Unique = true,
Name = UniqueDoctorIdIndexName,
PartialFilterExpression = andFilter,
};
var uniqueDoctorIdIndex = new CreateIndexModel<User>(
uniqueDoctorIdIndexDefinition,
createIndexOptions);
users.Indexes.CreateOne(uniqueDoctorIdIndex);
Probably in your description of a 'User' entity you must directly specify the BsonType of the 'DoctorId' field, by using an attribute, for example in our case it was:
[BsonRepresentation(BsonType.String)]
public Guid? DoctorId { get; set; }
I am more than sure that there is a more proficient and compact solution for this problem, so would be happy if somebody suggests it here.

Answer 7

Here is an example that I modified from the mongoDB partial index documentation:
db.contacts.createIndex(
{ email: 1 },
{ unique: true, partialFilterExpression: { email: { $exists: true } } }
)
IMPORTANT
To use the partial index, a query must contain the filter expression (or a modified filter expression that specifies a subset of the filter expression) as part of its query condition.
You can see that queries such as:
db.contacts.find({'email':'name#email.com'}).explain()
will indicate that they doing an index scan, even if you don't specify {$exists: true} because you're implicitly specifying a subset of the partialFilterExpression by specifying an email in your filter.
On the other hand, the following query will do a collection scan:
db.contacts.find({email: {$exists: false}})
WARNING
mythicalcoder's answer (currently the highest voted answer) is very misleading because it successfully creates a unique index, but the query planner will not generally be able to use the index you've created unless you add houseName: {$type: "string"} into your filter expression. This can have performance costs which you might not be aware of and can cause problems down the road.