I am using aggregate method in mongoDB to group but when I use $group
it returns the only field which I used to group. I have tried $project
but it
user2683814's solution worked for me but in my case, I have a counter accumulator when we replace the newRoot object, the count field is missing in the final stage so I've used $mergeObjects
operator to get my count field back.
db.collection.aggregate([
{
$group: {
_id: '$product',
detail: { $first: '$$ROOT' },
count: {
$sum: 1,
},
},
},
{
$replaceRoot: {
newRoot: { $mergeObjects: [{ count: '$count' }, '$detail'] },
},
}])
You can use this query
db.col.aggregate([
{"$group" : {"_id" : "$name","data" : {"$first" : "$$ROOT"}}},
{"$project" : {
"tags" : "$data.tags",
"name" : "$data.name",
"rating" : "$data.rating",
"_id" : "$data._id"
}
}])
You can use below aggregation query.
$$ROOT
to keep the whole document per each name followed by $replaceRoot
to promote the document to the top.
db.col.aggregate([
{"$group":{"_id":"$name","doc":{"$first":"$$ROOT"}}},
{"$replaceRoot":{"newRoot":"$doc"}}
])
I wanted to group my collection by groupById
field and store it as key value pairs having key as groupById
and value as all the items of that group.
db.col.aggregate([{$group :{_id :"$groupById",newfieldname:{$push:"$"}}}]).pretty()
This is working fine for me..
When you group data on any database, it means you want to perform accumulated operation on the required field and the other field which will not be include in accumulated operation will be used in group like
db.collection.aggregate([{
$group: {
_id: { field1: "", field1: "" },
acc: { $sum: 1 }
}}]
here in _id object will contains all other fields which you want to hold.
for your data you can try this
db.collection.aggregate([{
$group: {
_id: "$name",
rating: { $first: "$rating" },
tags: { $first: "$tag" },
docid: { $first: "$_id" }
}
},
{
$project: {
_id: "$docid",
name: "$_id",
rating: 1,
tags: 1
}
}])