MongoDB: Project to array item with minimum value of field

Question

Suppose my collection consists of items that looks like this:

{
    "items" : [
        {
            "item_id": 1,
            "item_field": 10
        },
        {
            "item_id": 2,
            "item_field": 15
        },
        {
            "item_id": 3,
            "item_field": 3
        },
    ]
}

Can I somehow select the entry of items with the lowest value of item_field, in this case the one with item_id 3?

I'm ok with using the aggregation framework. Bonus point if you can give me the code for the C# driver.

Answer 1

You can use $reduce expression in the following way.

The below query will set the initialValue to the first element of $items.item_field and followed by $lt comparison on the item_field and if true set $$this to $$value, if false keep the previous value and $reduce all the values to find the minimum element and $project to output min item.

db.collection.aggregate([
    {
         $project:  {
            items: {
                $reduce: {
                    input: "$items",
                    initialValue:{ 
                         item_field:{ 
                            $let: { 
                                vars: { obj: { $arrayElemAt: ["$items", 0] } },
                                in: "$$obj.item_field"
                            }
                        }
                    },
                    in: {
                          $cond: [{ $lt: ["$$this.item_field", "$$value.item_field"] }, "$$this", "$$value" ] 
                    }
                }
            }
        }
    }
])