CODEWITHSUNDEEP
mongodbmongodb-queryaggregation-framework
samuelluis
samuelluis

Reputation: 1861

Include all existing fields and add new fields to document

I would like to define a $project aggregation stage where I can instruct it to add a new field and include all existing fields, without having to list all the existing fields.

My document looks like this, with many fields:

{
    obj: {
        obj_field1: "hi",
        obj_field2: "hi2"
    },
    field1: "a",
    field2: "b",
    ...
    field26: "z"
}

I want to make an aggregation operation like this:

[
    {
        $project: {
            custom_field: "$obj.obj_field1",
            //the next part is that I don't want to do
            field1: 1,
            field2: 1,
            ...
            field26: 1
        }
    },
    ... //group, match, and whatever...
]

Is there something like an "include all fields" keyword that I can use in this case, or some other way to avoid having to list every field separately?

Upvotes: 186

Views: 138030

Answers (6)

Sede
Sede

Reputation: 61273

In 4.2+, you can use the $set aggregation pipeline operator which is nothing other than an alias to $addFieldsadded in 3.4

The $addFields stage is equivalent to a $project stage that explicitly specifies all existing fields in the input documents and adds the new fields.

db.collection.aggregate([
    { "$addFields": { "custom_field": "$obj.obj_field1" } }
])

Upvotes: 250

Deeksha Sharma
Deeksha Sharma

Reputation: 3359

To add new fields to your document you can use $addFields

from docs

and to all the fields in your document, you can use $$ROOT

db.collection.aggregate([

{ "$addFields": { "custom_field": "$obj.obj_field1" } },
{ "$group": {
        _id : "$field1",
        data: { $push : "$$ROOT" }
    }}
])

Upvotes: 10

mr.byte
mr.byte

Reputation: 21

according to @Deka reply, for c# mongodb driver 2.5 you can get the grouped document with all keys like below;

var group = new BsonDocument
{
 { "_id", "$groupField" },
 { "_document", new BsonDocument { { "$first", "$$ROOT" } } }
};

ProjectionDefinition<BsonDocument> projection = new BsonDocument{{ "document", "$_document"}};
var result = await col.Aggregate().Group(group).Project(projection).ToListAsync();

// For demo first record 
var fistItemAsT = BsonSerializer.Deserialize<T>(result.ToArray()[0]["document"].AsBsonDocument);

Upvotes: 0

Deka
Deka

Reputation: 1343

You can use $$ROOT to references the root document. Keep all fields of this document in a field and try to get it after that (depending on your client system: Java, C++, ...)

 [
    {
        $project: {
            custom_field: "$obj.obj_field1",
            document: "$$ROOT"

        }
    },
    ... //group, match, and whatever...
]

Upvotes: 96

Ghopper21
Ghopper21

Reputation: 10477

As of version 2.6.4, Mongo DB does not have such a feature for the $project aggregation pipeline. From the docs for $project:

Passes along the documents with only the specified fields to the next stage in the pipeline. The specified fields can be existing fields from the input documents or newly computed fields.

and

The _id field is, by default, included in the output documents. To include the other fields from the input documents in the output documents, you must explicitly specify the inclusion in $project.

Upvotes: 3

Victoria Malaya
Victoria Malaya

Reputation: 516

>>> There's something like "include all fields" keyword that I can use in this case or some another solution?

Unfortunaly, there is no operator to "include all fields" in aggregation operation. The only reason, why, because aggregation is mostly created to group/calculate data from collection fields (sum, avg, etc.) and return all the collection's fields is not direct purpose.

Upvotes: 7

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