How to sort with the sum of 2 fields in MongoDB

问题:

I have this document:

{ 
 "_id": "59b804e1ee8a4071a5ea3fcc",
 "description" : "description",
 "imagepath" : "https://example.com",
 "type" : "label",
 "downvotes" : -25,
 "upvotes" : 15, 
 "language" : "en", 
 "approoved" : true
},
{ 
 "_id": "59b804e1ee8a4071a5ea4dfd",
 "description" : "description",
 "imagepath" : "https://example.com",
 "type" : "label",
 "downvotes" : 0,
 "upvotes" : 30, 
 "language" : "en", 
 "approoved" : true
}

How to sort these 2 objects by the sum of the upvotes & downvotes values?

(object with id 59b804e1ee8a4071a5ea4dfd should come first and so on)

I've tried to aggregate with

{"$group" : {
    _id:{upvotes:"$upvotes",downvotes:"$downvotes"}, 
    count:{$sum:1}
}} 

but it groups all similar documents. I can't figure out the aggregate process/syntax.

EDIT: as per Veeram answer, here's the working code using mongoose just in case someone drops here:

Labels.aggregate([
        {"$addFields":{ "sort_order":{"$add":["$upvotes", "$downvotes"]}}},
        {"$sort":{"sort_order":-1}},
        {"$project":{"sort_order":0}}
    ]).exec(function (err, labels) {
        if (err) {
            return res.status(404).json({
                error: err
            });
        }
        res.status(200).json(labels);
    });

回答1:

You can try below aggregation in 3.4 for custom sort.

Use $add to sum the upvotes and downvotes in $addFields to keep the computed value as the extra field in the document followed by $sort sort on field.

$project with exclusion to drop the sort field to get expected output.

db.col.aggregate([
 {"$addFields":{ "sort_order":{"$add":["$upvotes", "$downvotes"]}}}, 
 {"$sort":{"sort_order":-1}},
 {"$project":{"sort_order":0}}
])

回答2:

You could create a new field in each document representing the sum of upvotes and downvotes. And then sort documents on this new field:

db.collection.aggregate([
  { $addFields: { "votes": { $sum: [ "$downvotes", "$upvotes"] } } },
  { $sort: { "votes": -1 } }
])