I need to group the data with set of fields using mongodb aggregate. Here is my data in the collection
[
{
bookName: "aaaa",
bookNo: "1",
registeredDate: "2018-02-01T06:51:16.738Z"
},
{
bookName: "bbbb",
bookNo: "2",
registeredDate: "2018-02-01T06:51:29.244Z"
}
{
bookName: "cccc",
bookNo: "1",
registeredDate: "2018-02-01T06:51:29.244Z"
}
{
bookName: "dddd",
bookNo: "2",
registeredDate: "2018-02-01T06:51:29.244Z"
}
]
I need to aggregate the above data into group by and show the data inside the group. Here is the output that i need,
{
Books: [
{
bookNo: "1",
books: [{
bookName: "aaaa",
registeredDate: "2018-02-01T06:51:16.738Z"
},
{
bookName: "cccc",
registeredDate: ""2018-02-01T06:51:29.244Z"
}]
},
{
bookNo: "2",
books: [{
bookName: "bbbb",
registeredDate: "2018-02-01T06:51:29.244Z"
},
{
bookName: "dddd",
registeredDate: "2018-02-01T06:51:29.244Z"
}]
}
]
}
Someone help me find a solution.
Thanks in Advance,
Try following aggregation:
db.books.aggregate([
{
$group: {
_id: "$bookNo",
books: {
$push: {
bookName: "$bookName",
registeredDate: "$registeredDate"
}
}
}
},
{
$project:{
_id: 0,
bookNo: "$_id",
books: 1
}
}
])
In $group you can specify which fields will be $pushed from each document. Then you can use $project to just rename _id to bookNo field
db.collection.aggregate(
// Pipeline
[
// Stage 1
{
$group: {
_id: '$bookNo',
//Books:{$addToSet:{bookNo:'$bookNo'}},
books: {
$addToSet: {
bookname: '$bookName',
registeredDate: '$registeredDate'
}
},
}
},
// Stage 2
{
$group: {
_id: null,
Books: {
$addToSet: {
books: '$books',
bookNo: '$_id'
}
}
}
},
// Stage 3
{
$project: {
_id: 0,
Books: 1
}
}
]
);
Related
I am using MongoDB 4.2.9 and have the following requirements:
Collection 'A' has multiple documents with a string field 'status' that I need to filter on
Collection 'B' has multiple documents
Collection A
{ _id: "1",
status: "Report",
type: "Academy",
rating: "Excellent",
ReportNo: "A1"
},
{ _id: "2",
status: "Open",
type: "Academy",
rating: "",
ReportNo: ""
},
{ _id: "3",
status: "Draft",
type: "Academy",
rating: "",
ReportNo: ""
},
{ _id: "4",
status: "Report",
type: "Academy",
rating: "Great",
ReportNo: "A4"
}
Collection B
{ _id: "98",
status: "Archived",
type: "Academy",
rating: "So So",
ReportNo: "X2"
},
{ _id: "99",
status: "Archived",
type: "Academy",
rating: "Great",
ReportNo: "X1"
}
Resulting View
{ _id: "1",
status: "Report",
type: "Academy",
rating: "Excellent",
ReportNo: "A1"
},
{ _id: "4",
status: "Report",
type: "Academy",
rating: "Great",
ReportNo: "A4"
},
{ _id: "98",
status: "Archived",
type: "Academy",
rating: "So So",
ReportNo: "X2"
},
{ _id: "99",
status: "Archived",
type: "Academy",
rating: "Great",
ReportNo: "X1"
}
My goal is to create an aggregation view so that I can filter on a status value in Collection 'A' and then merge those results with Collection 'B' and show in the view ?
I can filter on Collection 'A' using the match call, just can't see how to merge resulting documents into Collection 'B'
From my understandings, your "merge" behaviour is actually a union view of filtered view of collection A and collection B.
With MongoDB v4.2, you can use $facet to handle collection A and collection B separately.
simply perform filtering on A
perform uncorrelated $lookup on B
wrangle the result and merge them together to get the union view that you are looking for.
db.createCollection(
"unionView",
{
"viewOn" : "A",
"pipeline" : [
{
"$facet": {
"A": [
{
"$match": {
status: "Report"
}
}
],
"B": [
{
$limit: 1
},
{
"$lookup": {
"from": "B",
"pipeline": [],
"as": "B"
}
},
{
$unwind: "$B"
},
{
"$replaceRoot": {
"newRoot": "$B"
}
}
]
}
},
{
$project: {
all: {
"$setUnion": [
"$A",
"$B"
]
}
}
},
{
$unwind: "$all"
},
{
"$replaceRoot": {
"newRoot": "$all"
}
}
]
}
)
Here is the Mongo Playground for your reference.
With MongoDB v4.4+, you can create a view with $unionWith
db.createCollection(
"unionView",
{
"viewOn" : "A",
"pipeline" : [
{
"$match": {
status: "Report"
}
},
{
"$unionWith": {
"coll": "B"
}
}
]
}
)
Here is the Mongo playground for your reference.
Let's say I have the input docs below:
[
{
"_id": "6225ca4052e7c226e2dd836d",
"data": [
"07",
"07",
"12",
"19",
"07",
"32"
]
},
{
"_id": "6225ca4052e7c226e2dd888f",
"data": [
"99",
"97",
"52",
"99",
"58",
"92"
]
}
]
I want to count the occurrences of every element in data string array per document. In JS, I can use countBy. How can I achieve the same using MongoDB Aggregation Framework?
I have tried to $reduce but MongoDB seems to not support assigning dynamic field to object.
{
$reduce: {
input: '$data',
initialValue: {},
in: { // assign `$$this` with count to `$$value`, but failed! }
}
}
Below is the desired output.
[
{
"_id": "6225ca4052e7c226e2dd836d",
"freqs": {
"12": 1,
"19": 1,
"32": 1,
"07": 3
}
},
{
"_id": "6225ca4052e7c226e2dd888f",
"freqs": {
"52": 1,
"58": 1,
"92": 1,
"97": 1,
"99": 2
}
}
]
db.collection.aggregate([
{
$match: {}
},
{
$unwind: "$data"
},
{
$group: {
_id: "$data",
c: { $sum: 1 },
id: { $first: "$_id" }
}
},
{
$group: {
_id: "$id",
data: { $push: { k: "$_id", v: "$c" } }
}
},
{
$set: {
data: { $arrayToObject: "$data" }
}
}
])
mongoplayground
db.collection.aggregate([
{
$set: {
data: {
$function: {
body: "function(d) {let obj = {}; d.forEach(e => {if(obj[e]==null) { obj[e]=1; }else{ obj[e]++; }}); return obj;}",
args: [
"$data"
],
lang: "js"
}
}
}
}
])
mongoplayground
I have an array field (contains objects) in multiple documents, I want to merge the arrays into one array and group the array by object key. I have manage to group the array but I dont know how to group the data. See the code I tried below
const test = await salesModel.aggregate([
{ $unwind: "$items" },
{
$group: {
_id: 0,
data: { $addToSet: '$items' }
},
}
])
Result of the query:
{
_id: 0,
data: [
{
_id: 61435b3c0f773abaf77a367e,
price: 3000,
type: 'service',
sellerId: 61307abca667678553be81cb,
},
{
_id: 613115808330be818abaa613,
price: 788,
type: 'product',
sellerId: 61307abca667678553be81cb,
},
{
_id: 61307c1ea667676078be81cc,
price: 1200,
type: 'product',
sellerId: 61307abca667678553be81cb,
}
]
}
Now I want to group the data array by object key data.sellerId and sum price
Desired Output:
{
data: [
{
sumPrice: 788,
sellerId: 613115808330be818abaa613,
},
{
sumPrice: 1200,
sellerId: 61307abca667678553be81cb,
}
]
}
Extend with the current query and result with:
$unwind: Deconstruct the array field to multiple documents.
$group: Group by data.sellerId to sum ($sum) for data.price.
$group: Group by 0 with $addToSet to combine multiple documents into one document with data.
MongoDB aggregation query
db.collection.aggregate([
{
$unwind: "$data"
},
{
$group: {
_id: {
sellerId: "$data.sellerId"
},
"sumPrice": {
$sum: "$data.price"
}
}
},
{
"$group": {
"_id": 0,
"data": {
$addToSet: {
"sellerId": "$_id.sellerId",
"sumPrice": "$sumPrice"
}
}
}
}
])
Sample Mongo Playground
Output
[
{
"_id": 0,
"data": [
{
"sellerId": ObjectId("61307abca667678553be81cb"),
"sumPrice": 4988
}
]
}
]
If you want to re-write the query, here are the query with sample input.
Input
[
{
items: [
{
_id: ObjectId("61435b3c0f773abaf77a367e"),
price: 3000,
type: "service",
sellerId: ObjectId("61307abca667678553be81cb"),
},
{
_id: ObjectId("613115808330be818abaa613"),
price: 788,
type: "product",
sellerId: ObjectId("61307abca667678553be81cb"),
},
{
_id: ObjectId("61307c1ea667676078be81cc"),
price: 1200,
type: "product",
sellerId: ObjectId("61307abca667678553be81cb"),
}
]
}
]
Mongo aggregation query
db.collection.aggregate([
{
$unwind: "$items"
},
{
$group: {
_id: {
sellerId: "$items.sellerId"
},
"sumPrice": {
$sum: "$items.price"
}
}
},
{
"$group": {
"_id": 0,
"data": {
$addToSet: {
"sellerId": "$_id.sellerId",
"sumPrice": "$sumPrice"
}
}
}
}
])
Sample 2 on Mongo Playground
Output
[
{
"_id": 0,
"data": [
{
"sellerId": ObjectId("61307abca667678553be81cb"),
"sumPrice": 4988
}
]
}
]
db.setting.aggregate([
{
$match: {
status: true,
deleted_at: 0,
_id: {
$in: [
ObjectId("5c4ee7eea4affa32face874b"),
ObjectId("5ebf891245aa27c290672325")
]
}
}
},
{
$lookup: {
from: "site",
localField: "_id",
foreignField: "admin_id",
as: "data"
}
},
{
$project: {
name: 1,
status: 1,
price: 1,
currency: 1,
numberOfRecord: {
$size: "$data"
}
}
},
{
$sort: {
numberOfRecord: 1
}
}
])
how to push the currency into price object using project please guide thanks a lot, also eager to know what is difference between $addtoSet and $push, what is good option to opt it from project or fix it from $addField
https://mongoplayground.net/p/RiWnnRtksb4
Output should be like this:
[
{
"_id": ObjectId("5ebf891245aa27c290672325"),
"currency": "USD",
"name": "Menz",
"numberOfRecord": 0,
"price": {
"numberDecimal": "20",
"currency": "USD",
},
"status": true
},
{
"_id": ObjectId("5c4ee7eea4affa32face874b"),
"currency": "USD",
"name": "Dave",
"numberOfRecord": 2,
"price": {
"numberDecimal": "10",
"currency": "USD"
},
"status": true
}
]
You can insert a field into an object with project directly, like this (field price):
$project: {
name: 1,
status: 1,
price: {
numberDecimal: "$price.numberDecimal",
currency: "$currency"
},
numberOfRecord: {
$size: "$data"
}
}
By doing it with project, there is no need to use $addField.
For the difference between $addToSet and $push, read this great answer.
You can just set the object structure while projecting, so in this case there's no need for either $push or $addToSet.
{
$project: {
name: "1",
status: 1,
price: {
currency: "$currency",
numberDecimal: "$price.numberDecimal"
},
currency: 1,
numberOfRecord: {
$size: "$data",
}
}
}
Now the difference between $push and $addToSet is pretty trivial and derived from the name, $push saves all items while $addToSet will just create a set of them, for example:
input:
[
//doc1
{
item: 1
},
//doc2
{
item: 2
},
//doc3
{
item: 1
}
]
Now this:
{
$group: {
_id: null,
items: {$push: "$item"}
}
}
Will result in:
{_id: null, items: [1, 2, 1]}
While:
{
$group: {
_id: null,
items: {$addToSet: "$item"}
}
}
Will result in:
{_id: null, items: [1, 2]}
db.customerOrder.insert({
firstName: "Andrew",
lastName: "Lee",
DOB: ISODate("1974-10-28T00:00:00Z"),
phone: "+1 (959) 567-3312",
email: "mark#gmail.com",
address: {
street: "Cornish Street, Victoria",
houseNumber: "68",
postalCode: "3024",
country: "Australia",
},
language: ["English", "Mandarin"],
balance: 0,
orders: [
{
orderNumber: "ord003",
orderDate: ISODate("2020-01-10T00:00:00Z"),
staffNumber: "stf789"
}
]
});
Given the document above, and other documents which contain other orders and order number, how do i specify an aggregation so that it will only list all orderNumbers that's handled by a staffNumber x?
Example, orderNumber ord004 and ord005 is handled by staffNumber stf890
I tried doing
db.customerOrder.aggregate([ {"$match":{"orders.staffNumber":"stf890"}}, {"$project":{"orders.orderNumber":1, "_id":0}} ])
but the result was
{
"orders" : [
{
"orderNumber" : "ord003"
},
{
"orderNumber" : "ord003"
},
{
"orderNumber" : "ord005"
}
]
}
{
"orders" : [
{
"orderNumber" : "ord001"
},
{
"orderNumber" : "ord005"
}
]
}
{
"orders" : [
{
"orderNumber" : "ord003"
},
{
"orderNumber" : "ord004"
}
]
}
I expect the result to output only ord004 and ord005
How do i achieve this?
Thank you for your help
Try this! your query is almost correct but you're missing the case of matching orderNumber.
db.customerOrder.aggregate([
{
"$match":{
"orders.staffNumber":"stf890"
}
},
{
$unwind:{
"path":"$orders"
}
},
{
"$match":{
"orders.orderNumber":{$in:["ord004","ord005"]}
}
},
{
"$project":{
"orders.orderNumber":1,
"_id":0
}
}
])
If you don't care about the structure you can just $unwind and then match. otherwise you need to use something like $filter
Option 1:
db.customerOrder.aggregate([
{
"$match": {
"orders.staffNumber": "stf890"
}
},
{
"$unwind": "$orders"
},
{
"$match": {
"orders.staffNumber": "stf890"
}
},
{
"$project": {"orders.orderNumber": 1, "_id": 0}
}])
Option 2:
db.customerOrder.aggregate([
{
"$match": {
"orders.staffNumber": "stf890"
}
},
{
$project: {
orders: {
$filter: {
input: "$orders",
as: "order",
cond: {
$eq: ["$$order.staffNumber", "stf890"]
}
}
}
}
},
{
"$project": {
"orders.orderNumber": 1,
"_id": 0
}
}
])