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The _id expression specifies the group key. If you specify an _id value of null, or any other constant value, the $group stage returns a single document that aggregates values across all of the input documents.
The $$ROOT variable contains the source documents for the group. If you'd like to just pass them through unmodified, you can do this by $pushing $$ROOT into the output from the group.
We can group by single as well as multiple field from the collection, we can use $group operator in MongoDB to group fields from the collection and returns the new document as result. We are using $avg, $sum, $max, $min, $push, $last, $first and $addToSet operator with group by in MongoDB.
One of the most efficient ways of grouping the various fields present inside the documents of MongoDB is by using the $group operator, which helps in performing multiple other aggregation functions as well on the grouped data.
If you want to keep the information about the first matching entries for each group, you can try aggregating like:
db.test.aggregate([{
$group: {
_id : '$name',
name : { $first: '$name' },
age : { $first: '$age' },
sex : { $first: '$sex' },
province : { $first: '$province' },
city : { $first: '$city' },
area : { $first: '$area' },
address : { $first: '$address' },
count : { $sum: 1 },
}
}]);
[edited to include comment suggestions]
I came here looking for an answer but wasn't happy with the selected answer (especially given it's age). I found this answer that is a better solution (adapted):
db.test.aggregate({
$group: {
_id: '$name',
person: { "$first": "$$ROOT" },
count: { $sum: 1 }
},
{
"$replaceRoot": { "newRoot": { "$mergeObjects": ["$person", { count: "$count" }]} }
}
}
By the way, if you want to keep not only the first document, you can use$addToSet For example:
db.test.aggregate({
$group: {
_id: '$name',
name : { $addToSet: '$name' }
age : { $addToSet: '$age' },
count: { $sum: 1 }
}
}
You can try out this
db.test.aggregate({
{ $group:
{ _id: '$name',count: { $sum: 1 }, data: { $push: '$$ROOT' } } },
{
$project: {
_id:0,
data:1,
count :1
}
}
}
Use $first with the $$ROOT document and then use $replaceRoot with the first field.
db.test.aggregate([
{ "$group": {
"_id": "$name",
"doc": { "$first": "$$ROOT" }
}},
{ "$replaceRoot": { "newRoot": "$doc" }}
])
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