MongoDb Distinct with query C# driver

Question

I am trying to use the db.collection.distinct(field, query) command, documented here. I am trying to call this with the C# driver, documented here.

Currently I am using the code:

_repository.Search(item.searchCriteria).Select(i => i.messageId).Distinct().ToList()

where messageId is a string and the Search function does:

//Create search accross all properties of type.
public IQueryable<SearchType> Search(SearchType entity)
    {
        Type entityType = entity.GetType();
        var propertiesToSearch = entityType.GetProperties(BindingFlags.GetProperty | BindingFlags.Instance | BindingFlags.Public);
        query = _collection.AsQueryable();
        query = query.WhereAnd(
            query.ElementType,
            propertiesToSearch.Select(p => new SearchCriteria()
                {
                    Column = p.Name,
                    Value = p.GetValue(entity),
                    Operation = WhereOperation.Equal
                }).ToArray());
        return query;
    }

So this should get converted to:

db.collection.distinct("messageId", { $and: [ { prop1: "" }, { prop2: "" } ]  })

I am getting the following error when this is run though:

"Distinct is only supported for a single field. Projections used with Distinct must resolve to a single field in the document."

I am using Mongo 2.4.9 and the official C# driver 1.8.3

did you try aggregate/group. that should give you what you are looking for. — HaBo
– HaBo, Commented Mar 24, 2014 at 0:41
Have you tried trying without your search function experimentally? — WiredPrairie
– WiredPrairie, Commented Mar 24, 2014 at 0:55

Neil Lunn · Accepted Answer · 2014-03-24 00:55:13Z

4

The .distinct() method is an older implementation that is more of a convenience method wrapping mapReduce. For anything more involved that simple operations you should use .aggregate().

So the shell equivalent:

db.collection.aggregate([
    { "$match": { "$and": [ { "prop1": "" }, { "prop2": "" } ] } },
    { "$group": { "_id": "$messageId" } }  
])

The documents are just formed as a chain of BSON documents. There are various examples here.

edited Mar 24, 2014 at 0:55

answered Mar 24, 2014 at 0:05

Neil Lunn

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3 Comments

Neil Lunn Over a year ago

@Andrew does the change here better suit your purpose? You really should be using aggregate here. Which is the point.

Andrew Poland Over a year ago

Thanks, that's what I wanted to achieve :)

Stennie Over a year ago

@NeilLunn The distinct command pre-dates the Aggregation Framework .. however it is implemented in C++ and unrelated to MapReduce (ref: db/commands/distinct.cpp). The distinct() helper in the shell is a wrapper for the distinct command.

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