How can I get this result in mongodb? - mongodb

Question

I have this data structure:
"_id" : "121212",
"terms" : [
{
"term" : "hi",
"tf" : 2
},
{
"term" : "you",
"tf" : 1
}
]
}
and making this query:
db.foo.aggregate( [
{
$match : { _id : "121212" }
},
{
$project:{ terms:1 }
},
{
$unwind: "$terms"
}
]).pretty();
I have come to get this result in my db:
{
"_id" : "121212",
"terms" : {
"term" : "hi",
"tf" : 2
}
}
{
"_id" : "121212",
"terms" : {
"term" : "you",
"tf" : 1
}
}
but is there any way to get a result like this?:
{
"_id" : "121212",
"term" : "hi",
"tf" : 2
}
{
"_id" : "121212",
"term" : "you",
"tf" : 1
}
I have tried to build the query with $ replaceRoot: {newRoot: "$ terms"}, but after I can't select the _id field anymore.

Answer 1

Well, you can use the $map and $mergeObjects to do this beautifully.
[
{ "$match":{"_id":"121212"}},
{
"$addFields":{
"terms":{
"$map":{
"input":"$terms",
"in":{
"$mergeObjects":[
"$$this",
{
"_id":"$_id"
}
]
}
}
}
}
}
]
If you really need to deconstruct the "terms" array, then add the $unwind: "$terms" to the pipeline.

Answer 2

You can achieve by using $project stage at the end of the pipeline
db.foo.aggregate([
{ "$match" : { "_id": "121212" } },
{ "$unwind": "$terms" },
{ "$project": { "term": "$terms.term", "tf": "$terms.tf" }}
])
Output
[
{
"_id": "121212",
"term": "hi",
"tf": 2
},
{
"_id": "121212",
"term": "you",
"tf": 1
}
]
Check it here

Answer 3

You need to use $mergeObjects inside $replaceRoot:
db.foo.aggregate( [
{
$match : { _id : "121212" }
},
{
$project:{ terms:1 }
},
{
$unwind: "$terms"
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: [ { _id: "$_id" }, "$terms" ]
}
}
}
]).pretty();

Answer 4

Just to complete the range of options:
db.foo.aggregate([
{ "$match" : { "_id": "121212" } }, // filter by "_id"
{ "$addFields": { "terms._id": "$_id" } }, // copy "_id" field into terms
{ "$unwind": "$terms" }, // flatten the "terms" array
{ "$replaceRoot": { "newRoot": "$terms" } } // move the contents of the "terms" field up to the root level
])