how to group in mongoDB and return all fields in result
Solution 1
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"}}
])
Solution 2
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: '$$ROOT' },
count: {
$sum: 1,
},
},
},
{
$replaceRoot: {
newRoot: { $mergeObjects: [{ count: '$count' }, '$detail'] },
},
}])
Solution 3
When you group data on any database, it means you want to perform accumulated operation on the required field and the other field which will not be include in accumulated operation will be used in group like
db.collection.aggregate([{
$group: {
_id: { field1: "", field1: "" },
acc: { $sum: 1 }
}}]
here in _id object will contains all other fields which you want to hold.
for your data you can try this
db.collection.aggregate([{
$group: {
_id: "$name",
rating: { $first: "$rating" },
tags: { $first: "$tag" },
docid: { $first: "$_id" }
}
},
{
$project: {
_id: "$docid",
name: "$_id",
rating: 1,
tags: 1
}
}])
Solution 4
You can use this query
db.col.aggregate([
{"$group" : {"_id" : "$name","data" : {"$first" : "$$ROOT"}}},
{"$project" : {
"tags" : "$data.tags",
"name" : "$data.name",
"rating" : "$data.rating",
"_id" : "$data._id"
}
}])
Solution 5
I wanted to group my collection by groupById
field and store it as key value pairs having key as groupById
and value as all the items of that group.
db.col.aggregate([{$group :{_id :"$groupById",newfieldname:{$push:"$"}}}]).pretty()
This is working fine for me..
Comments
-
Anukool almost 4 years
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 is not working either. I also tried$first
and it worked but the result data is now in different format.The response format I need looks like:
{ "_id" : ObjectId("5b814b2852d47e00514d6a09"), "tags" : [], "name" : "name here", "rating" : "123456789" }
and after adding $group in my query.response is like this, the value of _id changes. (and the $group is taking only _id, if i try any other keyword it throws an error of accumulator something. please explain this also.)
{ "_id" :"name here" //the value of _id changed to the name field which i used in $group condition }
I have to remove the duplicates in name field, without changing any structure and fields. also I am using nodeJS with mongoose, so please provide the solution that works with it.
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Anukool over 5 yearsi simply want to remove the duplicates. assume i receive 50 documents, 20 of them has same name field, which i want only once. ( eg {{"name" : "name1","rating":"123456"},{"name" : "name2","rating":"123456.004"},{"name" : "name1","rating":"12345614.23"}}). i dont want 3rd document as it has same name as 1st and it is duplicate. what should i do in this case?
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Adnan Ahmed Ansari over 5 yearsYou have different ratings against same name so which one you want to get with group or do you want to all ratings against one name?
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Anukool over 5 yearswhich ever comes first, the rating is not that important, important thing is that documents must have distinct values of name field.
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Anukool over 5 yearsnow i am getting "The field 'name' must be an accumulator object" error. if i replace name with _id, it works but it will change the _id value. {"_id" : "name1","rating" : "0.001","tags" : null,"id" : ObjectId("5b814c9752d47e00514e9fcf")}.. the value of _id is changing which i dont want to. the actual value of _id is receiving in "id" field.
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Adnan Ahmed Ansari over 5 yearsoh sorry my mistake you are right you will use _id for name then use projection. I'm updating my query
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Anukool over 5 yearsby the way is there any way to get all fields without passing all fields. ? i have 30 fields in document.
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Adnan Ahmed Ansari over 5 yearsYes Senthur Deva query is usefull for all fields you can see him answer
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Habib over 4 yearsThank you so much, this question really helps me a lot.
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dd619 over 4 yearsThis is the short and best answer !
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newdeveloper about 4 yearspls help if you can for related questions in mongoDB - stackoverflow.com/questions/61067856/…
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Ashish Saini about 2 yearsits working but converting my array to objects