How to flatten a subdocument into root level in MongoDB?
For example, if I have a document like this
{
a: 1,
subdoc: {
b: 2,
c: 3
}
}
How can I convert it into a format like this? (without using project
)
{
a: 1,
b: 2,
c: 3
}
Solution 1:
You can use MongoDB projection i.e $project
aggregation framework pipeline operators as well. (recommended way). If you don't want to use project
check this link
db.collection.aggregation([{$project{ . . }}]);
Below is the example for your case:
db.collectionName.aggregate
([
{ $project: { a: 1, 'b': '$subdoc.b', 'c': '$subdoc.c'} }
]);
Gives you the output as you expected i.e.
{
"a" : 1,
"b" : 2,
"c" : 3
}
Solution 2:
You can use $replaceRoot
with a $addFields
stage as follows:
db.collection.aggregate([
{ "$addFields": { "subdoc.a": "$a" } },
{ "$replaceRoot": { "newRoot": "$subdoc" } }
])
Solution 3:
We can do this with $replaceWith
an alias for $replaceRoot
and the $mergeObjects
operator.
let pipeline = [
{
"$replaceWith": {
"$mergeObjects": [ { "a": "$a" }, "$subdoc" ]
}
}
];
or
let pipeline = [
{
"$replaceRoot": {
"newRoot": {
"$mergeObjects": [{ "a": "$a" }, "$subdoc" ]
}
}
}
];
db.collection.aggregate(pipeline)
Solution 4:
Starting Mongo 4.2
, the $replaceWith
aggregation operator can be used to replace a document by another (in our case by a sub-document) as syntaxic sugar for the $replaceRoot
mentioned by @chridam.
We can thus first include within the sub-document the root field to keep using the $set
operator (also introduced in Mongo 4.2
as an alias for $addFields
) and then replace the whole document by the sub-document using $replaceWith
:
// { a: 1, subdoc: { b: 2, c: 3 } }
db.collection.aggregate([
{ $set: { "subdoc.a": "$a" } },
{ $replaceWith: "$subdoc" }
])
// { b: 2, c: 3, a: 1 }