count the number of members before group them - mongodb

here is my document:
member: [
{
name: 'Jack',
group: 0
},
{
name: 'Rose',
group: 0
},
{
name: 'Tom',
group: 1
}
]
first, count the number of members,then group them.
return：
{
count: 3,
member: [
[
{name: 'Jack'},
{name: 'Rose'}
],
[
{name: 'Tom'}
]
]
}
how to do?thank you very much!


First you should calculate member array size using $size in $project, after that unwind member and group them check below query :
db.collectionName.aggregate({
"$project": {
"count": {
"$size": "$member"
},
"member": 1
}
}, {
"$unwind": "$member"
}, {
"$group": {
"_id": "$member.group",
"data": {
"$push": {
"name": "$member.name"
}
},
"count": {
"$first": "$count"
}
}
}, {
"$group": {
"_id": "$count",
"member": {
"$push": "$data"
}
}
}).pretty()

Related

How To Count Values Inside Deeply Nested Array Of Objects in Mongodb Aggregation

I want to sum of values inside array of objects which have another array of objects.
In my case; how can I count 'url' values in all documents inside 'urls' array under 'iocs' array;
Mongo playground: open
Here is document example;
[
{
"_id": {
"$oid": "63b4993d0625ebe8b6f5b06e"
},
"iocs": [
{
"urls": [
{
"url": "7.1.5.2",
}
],
},
{
"urls": [
{
"url": "https://l-ink.me/GeheimeBegierde",
},
{
"url": "GeheimeBegierde.ch",
}
],
},
{
"urls": [
{
"url": "https://l-ink.me/GeheimeBegierde",
}
],
}
],
type: "2"
},
{
"_id": {
"$oid": "63b4993d0624ebe8b6f5b06e"
},
"iocs": [
{
"urls": [
{
"url": "7.1.5.2",
}
],
},
{
"urls": [
{
"url": "https://l-ink.me/GeheimeBegierde",
},
{
"url": "GeheimeBegierde.ch",
}
],
},
{
"urls": [
{
"url": "https://l-ink.me/GeheimeBegierde",
}
],
}
],
type: "3"
},
{
"_id": {
"$oid": "63b4993d0615ebe8b6f5b06e"
},
"iocs": [
{
"urls": [
{
"url": "www.google.com",
}
],
},
{
"urls": [
{
"url": "abc.xyz",
},
{
"url": "GeheimeBegierde.ch",
}
],
},
{
"urls": [
{
"url": "https://123.12",
}
],
}
],
type: "1"
}
]
expected output be like;
url: "7.1.5.2",
count:2,
types:[2,3]
url: "https://l-ink.me/GeheimeBegierde",
count:4,
types:[2,3],
url: "abc.xyz",
count:1,
types:[1],
I tried unwind iocs then project urls but can't figure out how to get this output. I think i must use group but how ? Newbie in mongodb.
Any help would be appreciated. Thanks all.
NOTE: All the answers are working. Thank you all for the contributing.

You could do something like this !
db.collection.aggregate([
{
"$unwind": "$iocs"
},
{
"$unwind": "$iocs.urls"
},
{
"$group": {
"_id": "$iocs.urls.url",
"count": {
"$sum": 1
},
"types": {
"$addToSet": "$type"
}
}
},
{
"$project": {
url: "$_id",
_id: 0,
types: 1,
count: 1
}
},
])
https://mongoplayground.net/p/hhMqh2zI_SX

Here's one way you could do it.
db.collection.aggregate([
{"$unwind": "$iocs"},
{"$unwind": "$iocs.urls"},
{
"$group": {
"_id": "$iocs.urls.url",
"count": {"$count": {}},
"types": {"$addToSet": {"$toInt": "$type"}}
}
},
{
"$set": {
"url": "$_id",
"_id": "$$REMOVE"
}
}
])
Try it on mongoplayground.net.

You can try this query:
Double $unwind to deconstruct the nested array.
Then group by url get the count using $sum nad add the types into a set (to avoid duplicates, otherwise you can use simply $push)
db.collection.aggregate([
{
"$unwind": "$iocs"
},
{
"$unwind": "$iocs.urls"
},
{
"$group": {
"_id": "$iocs.urls.url",
"count": {
"$sum": 1
},
"types": {
"$addToSet": "$type"
}
}
}
])
Example here

Since $unwind is considered an inefficient operation, another option is to use $reduce and only $unwind once:
db.collection.aggregate([
{$project: {
type: 1,
urls: {
$reduce: {
input: "$iocs",
initialValue: [],
in: {$concatArrays: ["$$value", "$$this.urls.url"]}
}
}
}},
{$unwind: "$urls"},
{$group: {
_id: "$urls",
type: {$addToSet: "$type"},
count: {$sum: 1}
}},
{$project: {url: "$_id", count: 1, type: 1}}
])
See how it works on the playground example

Find duplicate name in MongoDB

I'm having a problem in getting the duplicate name in my mongodb to delete duplicates.
{
"users": [
{
"_id": {
"$oid": "61441890a6566a001623b8ed"
},
"name": "Jollibee",
},
{
"_id": {
"$oid": "61441890a6566a001623b8ed"
},
"name": "Jollibee",
},
{
"_id": {
"$oid": "61441890a6566a001623b8ed"
},
"name": "MCDO",
},
{
"_id": {
"$oid": "61441890a6566a001623b8ed"
},
"name": "Burger King",
},
]
}
I want to show in my output only the duplicate names. which is Jollibee.
tried this approach but it only returns me the count of all the users not the duplicated ones. I want to show 2 Jollibee only.
db.collection.aggregate([
{
"$unwind": "$users"
},
{
"$group": {
"_id": "$_id",
"count": {
"$sum": 1
}
}
},
{
"$match": {
"_id": {
"$ne": null
},
"count": {
"$gt": 1
}
}
}
])

Suppose the documents are:
[
{
"_id": {
"$oid": "6226dd742ef592186422ad1d"
},
"name": "Stack test"
},
{
"_id": {
"$oid": "6226dd7d2ef592186422ad1e"
},
"name": "Stack test"
},
{
"_id": {
"$oid": "6226dd912ef592186422ad1f"
},
"name": "Stack test 001"
}
]
Aggreagtion Query:
db.users.aggregate(
[
{
$group: {
_id: "$name",
names: {$push: "$name"}
}
}
]
)
Result:
{
_id: 'Stack test',
names: [ 'Stack test', 'Stack test' ]
},
{
_id: 'Stack test 001',
names: [ 'Stack test 001' ]
}
But a better way to do it will be
Aggregation Query:
db.users.aggregate(
[
{
$group: {
_id: "$name",
count: {$sum: 1}
}
}
]
)
Result:
{
_id: 'Stack test',
count: 2
},
{
_id: 'Stack test 001',
count: 1
}
Now, you can iterate through the count and use the name value in _id

since the $unwind step gives you same _id for all documents grouping by _id is not correct. Instead try grouping by users.name
db.collection.aggregate([
{
"$unwind": "$users"
},
{
"$group": {
"_id": "$users.name",
"count": {
"$sum": 1
}
}
},
{
"$match": {
"_id": {
"$ne": null
},
"count": {
"$gt": 1
}
}
}
])
demo

MongoDB group by and SUM by array

I'm new in mongoDB.
This is one example of record from collection:
{
supplier: 1,
type: "sale",
items: [
{
"_id": ObjectId("60ee82dd2131c5032342070f"),
"itemBuySum": 10
},
{
"_id": ObjectId("60ee82dd2131c50323420710"),
"itemBuySum": 10,
},
{
"_id": ObjectId("60ee82dd2131c50323420713"),
"itemBuySum": 10
},
{
"_id": ObjectId("60ee82dd2131c50323420714"),
"itemBuySum": 20
}
]
}
I need to group by TYPE field and get the SUM. This is output I need:
{
supplier: 1,
sales: 90,
returns: 170
}
please check Mongo playground for better understand. Thank you!

$match - Filter documents.
$group - Group by type and add item into data array which leads to the result like:
[
[/* data 1 */],
[/* data 2 */]
]
$project - Decorate output document.
3.1. First $reduce is used to flatten the nested array to a single array (from Result (2)) via $concatArrays.
3.2. Second $reduce is used to aggregate $sum the itemBuySum.
db.collection.aggregate({
$match: {
supplier: 1
},
},
{
"$group": {
"_id": "$type",
"supplier": {
$first: "$supplier"
},
"data": {
"$push": "$items"
}
}
},
{
"$project": {
_id: 0,
"supplier": "$supplier",
"type": "$_id",
"returns": {
"$reduce": {
"input": {
"$reduce": {
input: "$data",
initialValue: [],
in: {
"$concatArrays": [
"$$value",
"$$this"
]
}
}
},
"initialValue": 0,
"in": {
$sum: [
"$$value",
"$$this.itemBuySum"
]
}
}
}
}
})
Sample Mongo Playground

db.collection.aggregate([
{
$match: {
supplier: 1
},
},
{
"$group": {
"_id": "$ID",
"supplier": {
"$first": "$supplier"
},
"sale": {
"$sum": {
"$cond": {
"if": {
"$eq": [
"$type",
"sale"
]
},
"then": {
"$sum": "$items.itemBuySum"
},
"else": {
"$sum": 0
}
}
}
},
"returns": {
"$sum": {
"$sum": {
"$cond": {
"if": {
"$eq": [
"$type",
"return"
]
},
"then": {
"$sum": "$items.itemBuySum"
},
"else": {
"$sum": 0
}
}
}
}
}
}
},
{
"$project": {
_id: 0,
supplier: 1,
sale: 1,
returns: 1
}
}
])

Group on field while getting the last document for each field with MongoDB

Problem
I'm trying to group a stock inventory by products. At first, my stock entries was fully filled each time so I made this aggregate:
[
{ $sort: { date: 1 } },
{
$group: {
_id: '$userId',
stocks: { $last: '$stocks' },
},
},
{ $unwind: '$stocks' },
{
$group: {
_id: '$stocks.productId',
totalQuantity: { $sum: '$stocks.quantity' },
stocks: { $push: { userId: '$_id', quantity: '$stocks.quantity' } },
},
},
]
Now, it can be possible that a stock entry doesn't contain all the products filled. So I'm stuck while writing the new aggregate.
Basically I need to group every products by productId and have an array of the last entry for each user.
Output
This is my expected output:
[
{
"_id": ObjectId("5e75eae1359fc8159d5b6073"),
"totalQuantity": 33,
"stocks": [
{
"userId": ObjectId("5e75f498359fc8159d5b6075"),
"lastDate": "2020-03-21T11:45:53.077Z",
"quantity": 33
}
]
},
{
"_id": ObjectId("5e75eaea359fc8159d5b6074"),
"totalQuantity": 2,
"stocks": [
{
"userId": ObjectId("5e75f498359fc8159d5b6075"),
"lastDate": "2020-03-21T11:45:53.077Z",
"quantity": 2
}
]
}
]
Documents
Documents (when fully filled):
{
"_id": ObjectId("5e75fe71e4a3e0323ba47e0a"),
"date": "2020-03-21T11:45:53.077Z",
"userId": ObjectId("5e75f498359fc8159d5b6075"),
"stocks": [
{
"productId": ObjectId("5e75eae1359fc8159d5b6073"),
"quantity": 33
},
{
"productId": ObjectId("5e75eaea359fc8159d5b6074"),
"quantity": 2
}
]
}
Sometimes it won't be filled for the whole inventory (that's why I need the lastDate):
{
"_id": ObjectId("5e75fe71e4a3e0323ba47e0a"),
"date": "2020-03-21T11:45:53.077Z",
"userId": ObjectId("5e75f498359fc8159d5b6075"),
"stocks": [
{
"productId": ObjectId("5e75eae1359fc8159d5b6073"),
"quantity": 33
}
]
}

Try this one:
db.collection.aggregate([
{
$group: {
_id: "$userId",
root: {
$push: "$$ROOT"
}
}
},
{
$addFields: {
root: {
$map: {
input: "$root",
as: "data",
in: {
"stocks": {
$map: {
input: "$$data.stocks",
as: "stock",
in: {
"productId": "$$stock.productId",
"userId": "$$data.userId",
"quantity": "$$stock.quantity",
"lastDate": "$$data.date"
}
}
}
}
}
}
}
},
{
$unwind: "$root"
},
{
$replaceRoot: {
newRoot: "$root"
}
},
{
$unwind: "$stocks"
},
{
$sort: {
"stocks.lastDate": 1
}
},
{
$group: {
_id: "$stocks.productId",
totalQuantity: {
$last: "$stocks.quantity"
},
stocks: {
$last: "$stocks"
}
}
},
{
$addFields: {
stocks: [
{
"lastDate": "$stocks.lastDate",
"quantity": "$stocks.quantity",
"userId": "$stocks.userId"
}
]
}
}
])
MongoPlayground

Aggregate Field Values to Separate Key Names

I have a collection in MongoDB with sample data something like this (simplified):
{
_id: 1,
username: "ted",
content: "4125151",
status: "complete"
}
{
_id: 2,
username: "sam",
content: "4151",
status: "new"
}
{
_id: 3,
username: "ted",
content: "511",
status: "new"
}
{
_id: 4,
username: "ted",
content: "411",
status: "in_progress"
}
{
_id: 5,
username: "pat",
content: "1sds51",
status: "complete"
}
{
_id: 6,
username: "ted",
content: "4151",
status: "in_progress"
}
{
_id: 7,
username: "ted",
content: "4125",
status: "in_progress"
}
I need to aggregate the data such that for each user, I get a count for each status value as well as a total number of records. The result should look like this:
[
{
username: “pat”,
new: 0,
in_progress: 0,
complete: 1,
total: 1
},
{
username: “sam”,
new: 1,
in_progress: 0,
complete: 0,
total: 1
},
{
username: “ted”,
new: 1,
in_progress: 3,
complete: 1,
total: 5
}
]
Or any format that will effectively serve the same purpose which is, I want to be able to use with ngRepeat to display on the front end in this format:
User New In Progress Complete Total
pat 0 0 1 1
sam 1 0 0 1
ted 1 3 1 5
I can perform this aggregation:
{
"$group": {
"_id": {
"username": "$username",
"status": "$status"
},
"count": {
"$sum": 1
}
}
}
This gives me the individual count for each user/status combination that has at least one record. But then I have to piece it together to get it in the format that I can use on the front end. This is not at all ideal.
Is there a way to perform the aggregation to get the data in the format that I need?

What you want is a "conditional" aggregation of the values to produce a distinct field property for each status.
This is pretty simple to do using the $cond operator:
[
{ "$group": {
"_id": "$username",
"new": { "$sum": { "$cond": [{ "$eq": [ "$status", "new" ] },1,0 ] } },
"complete": { "$sum": { "$cond": [{ "$eq": [ "$status", "complete" ] },1,0 ] } },
"in_progress": { "$sum": { "$cond": [{ "$eq": [ "$status", "in_progress" ] },1,0 ] } },
"total": { "$sum": 1 }
}}
]
Presuming of course those are the only "status" values, but if they are not then just add an additional $project to sum the fields you want:
[
{ "$match": { "status": { "$in": [ "new", "complete", "in_progress" ] } } },
{ "$group": {
"_id": "$username",
"new": { "$sum": { "$cond": [{ "$eq": [ "$status", "new" ] },1,0 ] } },
"complete": { "$sum": { "$cond": [{ "$eq": [ "$status", "complete" ] },1,0 ] } },
"in_progress": { "$sum": { "$cond": [{ "$eq": [ "$status", "in_progress" ] },1,0 ] } }
}},
{ "$project": {
"new": 1,
"complete": 1,
"in_progress": 1,
"total": { "$add": [ "$new", "$complete", "$in_progress" ] }
]
Or just include that $add within the $group with the same calculations for the separate fields. But the $match is probably just the best idea if there are indeed other status values you don't want.

Another answer using $group twice and a $push, In this below query you need to compute the final total on UI side.
db.collection.aggregate([
{
"$group": {
"_id": {
"username": "$username",
"status": "$status"
},
"statuscount": {
"$sum": 1
}
}
},
{
"$group": {
"_id": "$_id.username",
"finalstatus": {
"$push": {
"Status": "$_id.status",
"statuscount": "$statuscount"
}
}
}
}
])