How to get Min and Max values from Mongo DB in 1 query? C#

Question

I can get the min and max in 2 Queries, but is it possible with C# to do it with 1?

var max = (await collection.Sort(descendingDateTime).Limit(1).FirstOrDefaultAsync())
                        .GetValue(timeField).ToUniversalTime();

var min = (await collection.Sort(ascendingDateTime).Limit(1).FirstOrDefaultAsync())
                        .GetValue(timeField).ToUniversalTime();

Answer 1

you could do it with a group aggregation like below but it will not be efficient for very large collections as the grouping will cause a collection scan and use up memory (where it might hit the 100mb limit for grouping). it would be better to do it with 2 find queries since it can use an index. as an improvement, you could run both queries somewhat simulaneously by using await Task.WhenAll(minTask, maxTask)

db.collection.aggregate(
[

    {
        $group: {
            _id: null,
            Min: { $min: "$TimeField" },
            Max: { $max: "$TimeField" }

        }
    },
    {
        $project: {
            _id: 0
        }
    }

])

also there's no way afaik to translate the above aggregation pipeline to a strongly-typed c# query as the driver doesn't support grouping by a null id.