How to query a mongo collection to return the full document with virtual fields containing calculated values from the sub-document? - mongodb

Question

I'm trying to query a collection for a specific document that contains a sub-document. The sub-document contains values for which I'd like to obtain
the highest and lowest scores from that sub-document and return that result as virtual fields to the original document.
I have the following dataset:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
},
{
"name" : "vallum",
"score" : 100
},
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU"
}
In mongo 2.4, how can I query mongo once to return the following result:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
},
{
"name" : "vallum",
"score" : 100
},
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
"worst_test": {
"name" : "ipsum",
"score" : 65
},
"best_test": {
"name" : "vallum",
"score" : 100
}
}
Where "best_test" and "worst_test" are virtual fields representing the tests with the highest and lowest scores, respectively.
I've tried with many different ways and the closest I've gotten is with this query:
db.students.aggregate([
{ $match: {
'_id': 'd0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e'
}},
{ $unwind: '$tests' },
{ $sort: {'tests.score': 1} },
{ $group: {
_id: '$_id',
student_tests: {$push: "$$ROOT"},
worst_test: {$first: '$tests'},
best_test: { $last: '$tests' }
}}
]);
Which yields this result:
{
"_id" : "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e",
"student_tests" : [
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "ipsum",
"score" : 65
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "lorem",
"score" : 79
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
{
"name" : "Addison Hunt",
"tests" : [
{
"name" : "vallum",
"score" : 100
}
],
"created_at" : 1401488865684,
"class" : "dolor sit amit",
"user_id" : "005G5635231325O4VIAU",
},
],
"worst_test": {
"name" : "ipsum",
"score" : 65
},
"best_test": {
"name" : "vallum",
"score" : 100
}
}

Answer 1

If you are using $$ROOT then in fact you are using MongoDB 2.6 as this is an aggregation variable only introduced in that version.
But while handy for various things, all it does is represent the entire document at the present stage of the pipeline where used. To do what you want and return the original document unmodified but with additional fields, you could use it in $project stage before the $unwind to assign to the _id field, but really you don't have exactly the same document as you would still need to $project at the end in order to get the correct document shape out of those elements.
You best bet is just projecting the fields, but keeping an un-altered copy of the array before any $sort is applied:
db.students.aggregate([
{ "$match": {
"_id": "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e"
}},
{ "$project": {
"name": 1,
"tests": 1,
"created_at": 1,
"class": 1,
"user_id": 1,
"testCopy": "$tests"
}},
{ "$unwind": "$testCopy" },
{ "$sort": { "testCopy.score": 1 } },
{ "$group": {
"_id: "$_id",
"tests": { "$first": "$tests" },
"created_at": { "$first": "$created_at" },
"class": { "$first": "$class" },
"user_id": { "$first": "$user_id" },
"worst_test": { "$first": "$testCopy" },
"best_test": { "$last": "$testCopy" }
}}
]);
Or using $$ROOT as mentioned before, alternately just placing the fields under the _id individually in the $project:
db.students.aggregate([
{ "$match": {
"_id": "d0e78492342f9f-f843ec7-4bd14g3h-bh34j3a9-02d6ah32k8e6b79e"
}},
{ "$project": {
"_id": "$$ROOT",
"tests": 1
}},
{ "$unwind": "$tests" },
{ "$sort": { "tests.score": 1 } },
{ "$group": {
"_id": "$_id",
"aworst_test": { "$first": "$tests" },
"abest_test": { "$last": "$tests" }
}},
{ "$project": {
"_id": "$_id._id",
"tests": "$_id.tests",
"created_at": "$_id.created_at",
"class": "$_id.class",
"user_id": "$_id.user_id",
"worst_test": "$aworst_test",
"best_test": "$abest_test"
}}
]);
But as you see, you are still doing the $project work somewhere in order to get the structure you want, as well as the "renamed fields" to maintain the field order you want as the $project will otherwise "optimize" and "keep" any fields that have not been renamed and "append" new fields after the existing ones.
There really is no simple way to "get all fields" in the same way as you originally found them. Operations like $project and $group are an "all or nothing" affair, where they only explicitly produce what you tell them to.