How to group the documents which I just unwinded in mongodb? - mongodb

This is the initial document. I applied $unwind on "colors" and another $unwind on "colors.sizes". But now am not able to group them back. How is that done?
{
"_id": ObjectId("5ef838255c959771c46fc917"),
"created": ISODate("2017-03-04T18:30:00.000Z"),
"updated": ISODate("2020-06-28T00:49:23.000Z"),
"status": "active",
"productId": "3828",
"tags": {
"colors": ["BEIGE", "BLACK", "BLUE", "GREEN", "GREY", "LIGHT BEIGE", "LIGHT PINK", "NAVY", "PINK", "WHITE"],
"size": ["39", "39.5", "40", "40.5", "41", "41.5", "42", "42.5", "43", "43.5", "44", "44.5", "45", "45.5", "46"],
"category": ["SHOES", "SNEAKERS", "MENS SHOES & ACCESSORIES"],
"price": ["3400.00", "1360.00", "2040.00", "1700.00", "45575.00"],
"season": ["SS(2020)", "AW(2019)", "SS(2019)", "AW(2018)", "SS(2017)", "AW(2016)", "SS(2016)", "SS(2015)"],
"hash": ["ACTIVE", "ANKLE", "BLUSH", "COLLAR", "COLOUR", "DESIGNER", "FOIL", "GOLD-TONE", "LACE-UP", "LEATHER", "LOW", "LOW TOP", "ORIGINAL", "PADDED", "ROUND", "ROUND TOE", "RUBBER", "SIGNATURE", "TONAL"]
},
"name": "ORIGINAL ARCHILLES LOW SNEAKER",
"description": "",
"brand": "COMMON PROJECTS",
"styleCode": "220036200",
"availableOnline": true,
"colors": [{
"images": [],
"_id": ObjectId("5ef838255c959771c46fc918"),
"colorId": "1",
"color": "WHITE",
"hexCode": "#FFFFFF",
"status": "active",
"sizes": [{
"extraInfo": [{
"title": "Size And Fit",
"text": "Product measures: Heel 3cm (1.2\"). True to size"
}, {
"title": "Information",
"text": "Designer colour: White. Leather. Rubber outsole. Made in Italy"
}],
"_id": ObjectId("5ef838255c959771c46fc919"),
"sizeId": "1",
"neo": "0220003258813",
"size": "39",
"originalPrice": "3400.00",
"sellingPrice": "3400.00",
"discountPercent": "0.00",
"status": "active"
}, {
"extraInfo": [{
"title": "Size And Fit",
"text": "Product measures: Heel 3cm (1.2\"). True to size"
}, {
"title": "Information",
"text": "Designer colour: White. Leather. Rubber outsole. Made in Italy"
}],
"_id": ObjectId("5ef838255c959771c46fc91a"),
"sizeId": "2",
"neo": "0220111220849",
"size": "39.5",
"originalPrice": "3400.00",
"sellingPrice": "3400.00",
"discountPercent": "0.00",
"status": "active"
}, {
"extraInfo": [{
"title": "Size And Fit",
"text": "Product measures: Heel 3cm (1.2\"). True to size"
}, {
"title": "Information",
"text": "Designer colour: White. Leather. Rubber outsole. Made in Italy"
}],
"_id": ObjectId("5ef838255c959771c46fc91b"),
"sizeId": "3",
"neo": "0220004218564",
"size": "40",
"originalPrice": "3400.00",
"sellingPrice": "3400.00",
"discountPercent": "0.00",
"status": "active"
}]
}, ]
}
db.getCollection('products').aggregate([
{$match:{"productId":"3828"}},
{$unwind:"$colors"},
{$unwind:"$colors.sizes"},
{
$lookup: {
from:"inventories",
localField: "colors.sizes.neo",
foreignField: "sku",
as:"colors.sizes.stores"
}
}
])

The opposite operator of $unwind is $group
Explanation
We $group by productId + colors[*]._id fields grouping root, colors and sizes separately.
Note: You can define each variable separately like this:
created: {$first:"$created"},
updated: {$first:"$updated"},
...
OR
use aggregation variable $$ROOT which contains entire document fields
We "merge" each variables into single object and move into root level.
Note: If we don't use $replaceRoot, the data structure will looks like this:
{
_id: "...",
root: {
_id: "...",
...
colors:{
...
sizes:[...]
}
}
}
Note: Unfortunately, MongoDB doesn't allow merge subkeys like this:
{
$mergeObjects:[
"$root",
{"colors":"$colors"},
{"colors.sizes":"$sizes"} <-- Not allowed
]
}
So we have to use $objectToArray + $map + $arrayToObject operators
$objectToArray - transforms object into array
{"hello": "world"} ---> [{"k":"hello", "v":"world"}]
$map - iterates array and transforms array content.
In this case, we replace `sizes` field with `sizes` from the 1st step
$objectToArray - transforms "special" array into object
[{"k":"hello", "v":"world"}] ---> {"hello": "world"}
Now we group only by productId and group colors and root values separately
We finish merging variables and transform into desired object (same as 2nd step)
Add these steps into your pipeline:
{
$group: {
"_id": {
productId: "$productId",
colors: "$colors._id"
},
"root": {
$first: "$$ROOT"
},
"colors": {
$first: "$colors"
},
"sizes": {
$push: "$colors.sizes"
}
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: [
"$root",
{
"colors": {
$arrayToObject: {
$map: {
input: { $objectToArray: "$colors" },
as: "color",
in: {
k: "$$color.k",
v: {
$cond: [
{ $eq: [ "$$color.k", "sizes" ] },
"$sizes",
"$$color.v"
]
}
}
}
}
}
}
]
}
}
},
{
$group: {
"_id": "$_id",
"root": {
$first: "$$ROOT"
},
"colors": {
$push: "$colors"
}
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: [
"$root",
{ "colors": "$colors" }
]
}
}
}
MongoPlayground

Related

MongoDB Select By Group along with that Count Unique match exclude array and object fields Get data sort by latest objects

I have a collection where from the backend user can input multiple same name bikes but with different registration number but in front-End I want them to be grouped by matching the same name but as user updates separately display image changes but I want only one display image as it is 1 vehicle
provided there is a node created I will implement it we can sort it by the latest and take the price and image of it
Activa -2 Count
KTM -1 Count
but there is a catch.
Activa 2 bikes but I want only count 2 and the price as it is the same in an array I want only 1 and the same applies to displayimage here display image file path is different but I want the latest one only Sharing data below
Data:
[
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikenumber": "KA05HM2034",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikenumber": "KA05HM2033",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62d7ff7e70b9ab38c6ab0cb1",
"bikename": "Activa",
"bikenumber": "KA05HM2223",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"afterfreekmprice": 22,
"descreption": "Activa",
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658322798414.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658322798414.jpg",
"idx": 0
}
}
]
Expected:
[
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
"count": 2
},
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
}
"count": 1
}
]
You can use the aggregation pipeline,
$sort by _id in descending order
$group by bikename and get the first root document that is latest one in root and count total documents in count
$project to show required documents
db.collection.aggregate([
{ $sort: { _id: -1 } },
{
$group: {
_id: "$bikename",
root: { $first: "$$ROOT" },
count: { $sum: 1 }
}
},
{
$project: {
_id: "$root",
count: 1
}
}
])
Playground
You can use $group for this:
db.collection.aggregate([
{$group: {
_id: "$bikename",
count: {$sum: 1},
data: {$first: "$$ROOT"}
}
},
{$set: {"data.count": "$count"}},
{$replaceRoot: {newRoot: "$data"}}
])
See how it works on the playground example

MongoDB Aggregate and Group by Subcategories of products

I have a MongoDB schema that looks like this
const ProductModel = new Schema({
subcategory: {
type : mongoose.Schema.Types.ObjectId,
ref : "Subcategory",
},
product_name: {
type: String
},
description: {
type: String
},
price: {
type: Number
},
});
And a subcategory schema:
const SubcategoryModel = new Schema({
subcategoryName: {
type: String,
}
});
The input query before aggregation looks like this:
[
{
"_id": "111",
"subcategory": {
"_id": "456",
"categoryName": "Sneakers",
},
"product_name": "Modern sneaker",
"description": "Stylish",
"price": 4400
},
{
"_id": "222",
"subcategory": {
"_id": "456",
"categoryName": "Sneakers",
},
"product_name": "Blue shoes",
"description": "Vived colors",
"price": 7500
},
{
"_id": "333",
"subcategory": {
"_id": "123",
"categoryName": "Jackets",
"__v": 0
},
"product_name": "Modern jacket",
"description": "Stylish",
"price": 4400
},
}
]
The final result of the query should look like this:
{
"Sneakers":[
{
"product_name":"Modern sneaker",
"description":"Stylish",
"price":"4400"
},
{
"product_name":"Blue shoes",
"description":"Vived colors",
"price":"7500"
},
"Jackets":{
"...."
}
]
}
Subcategory before aggregation:
"subcategories": [
{
"_id": "123",
"categoryName": "Jackets",
},
{
"_id": "456",
"categoryName": "Sneakers",
}
]
I'm trying to populate the subcategory, And then group the products by their subcategoryName field.
You can use this aggregation query:
First $lookup to do the join between Product and Subcategory creating the array subcategories.
Then deconstructs the array using $unwind.
$group by the name of subproduct adding the entire object using $$ROOT.
The passes the fields you want using $project.
And replaceRoot to get key value into arrays as Sneakers and Jackets.
db.Product.aggregate([
{
"$lookup": {
"from": "Subcategory",
"localField": "subcategory.categoryName",
"foreignField": "categoryName",
"as": "subcategories"
}
},
{
"$unwind": "$subcategories"
},
{
"$group": {
"_id": "$subcategories.categoryName",
"data": {
"$push": "$$ROOT"
}
}
},
{
"$project": {
"data": {
"product_name": 1,
"description": 1,
"price": 1
}
}
},
{
"$replaceRoot": {
"newRoot": {
"$arrayToObject": [
[
{
"k": "$_id",
"v": "$data"
}
]
]
}
}
}
])
Example here
With your provided data, result is:
[
{
"Sneakers": [
{
"description": "Stylish",
"price": 4400,
"product_name": "Modern sneaker"
},
{
"description": "Vived colors",
"price": 7500,
"product_name": "Blue shoes"
}
]
},
{
"Jackets": [
{
"description": "Stylish",
"price": 4400,
"product_name": "Modern jacket"
}
]
}
]

mongodb distinct query values

I have the following mongodb documents:
{
"_id": "",
"name": "example1",
"colors": [
{
"id": 1000000,
"properties": [
{
"id": "1000",
"name": "",
"value": "green"
},
{
"id": "2000",
"name": "",
"value": "circle"
}
]
} ]
}
{
"_id": "",
"name": "example2",
"colors": [
{
"id": 1000000,
"properties": [
{
"id": "1000",
"name": "",
"value": "red"
},
{
"id": "4000",
"name": "",
"value": "box"
}
]
} ]
}
I would like to get distinct queries on the value field in the array where id=1000
db.getCollection('product').distinct('colors.properties.value', {'colors.properties.id':{'$eq': 1000}})
but it returns all values in the array.
The expected Result would be:
["green", "red"]
There are a lot of way to do.
$match eliminates unwanted data
$unwind de-structure the array
$addToSet in $group gives the distinct data
The mongo script :
db.collection.aggregate([
{
$match: {
"colors.properties.id": "1000"
}
},
{
"$unwind": "$colors"
},
{
"$unwind": "$colors.properties"
},
{
$match: {
"colors.properties.id": "1000"
}
},
{
$group: {
_id: null,
distinctData: {
$addToSet: "$colors.properties.value"
}
}
}
])
Working Mongo playground

How to have multiple nested arrays in an array

I have the collection data from a csv file with header. When i run my query
db.ties.aggregate(
[
{
$group:
{
_id: { "SHOP": "$SHOP" },
isLinkedTo: { $push: { "PERSON": "$PERSON", "CITY": "$CITY", "ROOM": "$ROOM", "STYLE": "$STYLE", "hasDonated": {"DATE": "$DATE", "OBJECT": "$OBJECT", "COST": "$COST", "COLOR": "$COLOR", "PAYMENT": "$PAYMENT"}}}
}
},
{ $out: "ties"}
],
{ allowDiskUse: true }
)
I have like result:
{
"_id": {
"Shop": "FirstShopNameCovered"
},
"isLinkedTo": [{
"PERSON": "Carleen",
"CITY": "Rome",
"ROOM": "Kitchen",
"STYLEPREFERED": "Modern",
"hasDonated": {
"DATE": "2019-10-11",
"OBJECT": "Set of dishes",
"COST": 72,
"COLOR": "White",
"PAYMENT": "Credit card"
}
}, {
"PERSON": "Carleen",
"CITY": "Rome",
"ROOM": "Kitcher",
"STYLEPREFERED": "Modern",
"hasDonated": {
"DATE": "2018-10-26",
"OBJECT": "Set of chairs",
"COST": 353,
"COLOR": "Grey",
"PAYMENT": "Coupon"
}
}, {
"PERSON": "Pernick",
"CITY": "Venezia",
"ROOM": "Bathroom",
"STYLE": "Minimalist",
"hasDonated": {
"DATE": "2018-09-18",
"OBJECT": "Mirror",
"COST": 68,
"COLOR": "Brown",
"PAYMENT": "Credit card"
}
}
You can see that there is replicated the Person "PERSON": "Carleen" with all data with 2 different arrays hasDonated.
I wish have something like this result, with person not replicated that contains all hasDonated arrays where he is present:
"_id": {
"Shop": "NameCovered"
},
"isLinkedTo": [{
"PERSON": "Carleen",
"CITY": "Rome",
"ROOM": "Kitchen",
"STYLE": "RetrĂ²",
"hasDonated": {
"DATE": "2019-10-11",
"OBJECT": "Set of dishes",
"COST": 72,
"COLOR": "White",
"PAYMENT": "Credit card"
},
{
"DATE": "2018-10-26",
"OBJECT": "Chair",
"COST": 53,
"COLOR": "Grey",
"PAYMENT": "Coupon"
}
}, {
"PERSON": "Pernick",
"CITY": "Venezia",
"ROOM": "Bathroom",
"STYLE": "Minimalist",
"hasDonated": {
"DATE": "2018-09-18",
"OBJECT": "Mirror",
"COST": 68,
"COLOR": "Brown",
"PAYMENT": "Credit card"
}
How can I do to have the result like this?
First we need to $unwind to flat the array. Then group the hasDonated using $group where unique is found by combination of "_id" and "PERSON" as you mentioned.
[
{
"$unwind": "$isLinkedTo"
},
{
$group: {
_id: {
_id: "$_id",
per: "$isLinkedTo.PERSON"
},
isLinkedTo: {
$first: {
PERSON: "$isLinkedTo.PERSON",
CITY: "$isLinkedTo.CITY",
ROOM: "$isLinkedTo.ROOM",
STYLEPREFERED: "$isLinkedTo.STYLEPREFERED"
}
},
hasDonated: {
$addToSet: "$isLinkedTo.hasDonated"
}
}
},
{
$addFields: {
_id: "$_id._id",
"isLinkedTo.hasDonated": "$hasDonated"
}
},
{
$project: {
hasDonated: 0
}
},
{
$group: {
_id: "$_id",
isLinkedTo: {
$push: "$isLinkedTo"
}
}
}
]
Working Mongo playground

Aggregate nested arrays

I have multiple documents, and I'm trying to aggregate all documents with companyId = xxx and return one array with all the statuses.
So it will look like this:
[
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
},
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
},
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
},
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
}
]
The document look like this:
[
{
"companyId": "xxx",
"position": "",
"section": "",
"comment": "",
"items": [
{
"any": "111",
"name": "some name",
"description": "some description",
"version": "3",
"status": [
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
}
]
},
{
"any": "222",
"name": "some name",
"description": "some description",
"version": "3",
"status": [
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
}
]
}
]
},
{
"companyId": "xxx",
"position": "",
"section": "",
"comment": "",
"items": [
{
"any": "111",
"name": "some name",
"description": "some description",
"version": "3",
"status": [
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
}
]
},
{
"any": "222",
"name": "some name",
"description": "some description",
"version": "3",
"status": [
{
"status": "created",
"date": "2019-03-16T10:59:59.200Z"
},
{
"status": "completed",
"date": "2019-03-16T11:00:37.750Z"
}
]
}
]
}
]
Any suggestion, how to implement this?
Then I want to loop over the array (in code) and count how many items in status created, and completed. maybe it could be done with the query?
Thanks in advance
You can use below aggregation:
db.col.aggregate([
{
$match: { companyId: "xxx" }
},
{
$unwind: "$items"
},
{
$unwind: "$items.status"
},
{
$replaceRoot: {
newRoot: "$items.status"
}
},
{
$group: {
_id: "$status",
count: { $sum: 1 }
}
}
])
Double $unwind will return single status per document and then you can use $replaceRoot to promote each status to root level of your document.
Additionally you can add $group stage to count documents by status.
In addition to the #mickl answer, you can add $project pipeline to get the result as a flat list of status and count.
db.collectionName.aggregate([
{
$match: { companyId: "xxx" }
},
{
$unwind: "$items"
},
{
$unwind: "$items.status"
},
{
$replaceRoot: {
newRoot: "$items.status"
}
},
{
$group: {
_id: "$status",
count: { $sum: 1 }
}
},
{
$project: {
"status":"$_id",
"count":1,
_id:0
}
}
])
If the number of documents on which you are executing the above query is too much then you should avoid using $unwind in the initial stage of aggregation pipeline.
Either you should use $project after $match to reduce the selection of fields or you can use below query:
db.col.aggregate([
{
$match: {
companyId: "xxx"
}
},
{
$project: {
_id: 0,
data: {
$reduce: {
input: "$items.status",
initialValue: [
],
in: {
$concatArrays: [
"$$this",
"$$value"
]
}
}
}
}
},
{
$unwind: "$data"
},
{
$replaceRoot: {
newRoot: "$data"
}
}
])