I'm very new with mongoose. Can mongoose do something like i want to see how many linkview of James where store is Red only . The expected outcome should be linkview = 2 because 1+1. Please help
So this is User schema
{
"_id": "234",
"name": "James",
"__v": 0,
}
So this is affiliatelink schema
{
"_id": "11",
"store": "Red",
"linkview": "1",
"date": 12/12/12,
"affiliate": "James"
},
{
"_id": "22",
"store": "Red",
"linkview": "1",
"date": 13/12/12,
"affiliate": "James"
},
{
"_id": "33",
"store": "Blue",
"linkview": "1",
"date": 13/12/12,
"affiliate": "James"
}
Here is an aggregation you're looking for:
db.collection.aggregate([
{
$match: {
"affiliate": "James",
store: "Red"
}
},
{
$group: {
_id: null,
"linkview_count": {
$sum: 1
}
}
},
{
$project: {
_id: 0
}
}
])
The result will be:
[
{
"linkview_count": 2
}
]
Playground: https://mongoplayground.net/p/lPPhv9_w4iT
Related
Let's assume we have the following documents in the Mongo db
[
{
"Name": "Jack",
"info": {
"eyes": "brown",
"city": "paris",
"accomodation": "house"
}
},
{
"Name": "Mathew",
"info": {
"eyes": "yellow",
"city": "rome",
"accomodation": "apartment"
}
},
{
"Name": "Peter",
"info": {
"eyes": "brown",
"city": "barcelona",
"accomodation": "house",
"hair_color": "black"
}
}
]
Let's say the input JSON is
{
"eyes": "brown",
"city": "paris",
"accomodation": "house",
"hair_color": "black"
}
I would like to construct a query such that all the key-value pairs of the document should be present in the input json document but not necessarily the other way round.
For the above example - the output should be
{
"Name": "Jack",
"info": {
"eyes": "brown",
"city": "paris",
"accomodation": "house"
}
}
Since all the key value pairs of Jack were met by the input json document even though the input JSON document had hair_color which wasn't present in Jack document
Mongo playground with input : https://mongoplayground.net/p/1T0ZL8yGhPW
All you have to do is convert both your input and info field to an array using $objectToArray and check if their if there's any none matches, like so:
db.collection.aggregate([
{
$addFields: {
rawInput: { // better if you do this in code once and drop this stage.
"$objectToArray": {
"eyes": "brown",
"city": "paris",
"accomodation": "house",
"hair_color": "black"
}
}
}
},
{
$match: {
$expr: {
$eq: [
{
$size: {
$filter: {
input: {
"$objectToArray": "$info"
},
as: "elem",
cond: {
$not: {
"$in": [
"$$elem",
"$rawInput"
]
}
}
}
}
},
0
]
}
}
},
{
"$project": {
Name: 1,
info: 1
}
}
])
Mongo Playground
I have a mongo Database I'll like to "join" two of them and then merge some other fields:
Let's see the schemas:
Students Schema (and data):
{
"_id": ObjectId("5fbd564981b1313de790b580"),
"name": "John Doe",
"age": "21",
"image": "https://XXXX/481.png",
"subjects": [
{
"_id": ObjectId("5fbd4e6881b1313de790b56b"),
"passed": true,
},
{
"_id": ObjectId("5fcb63fa8814d96876c687bf"),
}
],
"__v": NumberInt("1"),
}
and Subject schema:
{
"_id": ObjectId("5fbd4e6881b1313de790b56b"),
"course": 3,
"teacher": "John Smith",
"name": "Math",
},
{
"_id": ObjectId("5fcb63fa8814d96876c687bf"),
"name": "IT",
"course": 8,
"teacher": "John Peter",
}
What I'll like to make a query with the subjects (all info) of a student, also if the student have additional fields in subject like passed add it to the subject subdocument.
Here is my query till now:
db.students.aggregate([
{
$match:
{
_id : ObjectId('5fbd564981b1313de790b580')
}
},
{
$lookup :
{
from : "subjects",
localField : "subjects._id",
foreignField : "_id",
as : "FoundSubject"
}
}
]);
which correctly make the "join" but the merge is still missing, I got as result:
{
"_id": ObjectId("5fbd564981b1313de790b580"),
"name": "John Doe",
"age": "21",
"image": "https://XXXX/481.png",
"subjects": [
{
"_id": ObjectId("5fbd4e6881b1313de790b56b"),
"passed": true,
},
{
"_id": ObjectId("5fcb63fa8814d96876c687bf"),
}
],
"__v": NumberInt("1"),
"FoundSubject": [
{
"_id": ObjectId("5fbd4e6881b1313de790b56b"),
"course": 3,
"teacher": "John Smith",
"name": "Math"
},
{
"_id": ObjectId("5fcb63fa8814d96876c687bf"),
"name": "IT",
"course": 8,
"teacher": "John Peter"
}
]
}
but I'll like to have:
{
"_id": ObjectId("5fbd564981b1313de790b580"),
"name": "John Doe",
"age": "21",
"image": "https://XXXX/481.png",
"subjects": [
{
"_id": ObjectId("5fbd4e6881b1313de790b56b"),
"course": 3,
"teacher": "John Smith",
"name": "Math",
"passed": true,
},
{
"_id": ObjectId("5fcb63fa8814d96876c687bf"),
"name": "IT",
"course": 8,
"teacher": "John Peter"
}
],
"__v": NumberInt("1"),
}
with merged data and field "passed" added. How can accomplish that?
I'm new to MongoDB coming from MySQL.
Thanks
You need to merge both objects, add below stage after $lookup,
MongoDB Version From 3.4
$map to iterate loop of students array
$reduce to iterate loop of FoundSubject array, check condition if condition match then return required fields otherwise return initial value
$project to remove FoundSubject from result
{
$addFields: {
subjects: {
$map: {
input: "$subjects",
as: "s",
in: {
$reduce: {
input: "$FoundSubject",
initialValue: {},
in: {
$cond: [
{ $eq: ["$$s._id", "$$this._id"] },
{
_id: "$$this._id",
course: "$$this.course",
name: "$$this.name",
teacher: "$$this.teacher",
passed: "$$s.passed"
},
"$$value"
]
}
}
}
}
}
}
},
{ $project: { FoundSubject: 0 } }
Playground
MongoDB Version From 4.4
$map to iterate loop of students array,
$filter to get matching document from FoundSubject array and $first to get first object from array returned by filter
$mergeObjects to merge current objects with found result object from filter
remove FoundSubject using $$REMOVE
// skipping your stages
{
$addFields: {
FoundSubject: "$$REMOVE",
subjects: {
$map: {
input: "$subjects",
as: "s",
in: {
$mergeObjects: [
"$$s",
{
$first: {
$filter: {
input: "$FoundSubject",
cond: { $eq: ["$$s._id", "$$this._id"] }
}
}
}
]
}
}
}
}
}
Playground
I am struggling to find some examples of using the mongo aggregation framework to process documents which has an array of items where each item also has an array of other obejects (array containing an array)
In the example document below what I would really like is an example that sums the itemValue in the results array of all cases in the document and accross the collection where the result.decision was 'accepted'and group by the document locationCode
However, even an example that found all documents where the result.decision was 'accepted' to show or that summmed the itemValue for the same would help
Many thanks
{
"_id": "333212",
"data": {
"locationCode": "UK-555-5566",
"mode": "retail",
"caseHandler": "A N Other",
"cases": [{
"caseId": "CSE525666",
"items": [{
"id": "333212-CSE525666-1",
"type": "hardware",
"subType": "print cartridge",
"targetDate": "2020-06-15",
"itemDetail": {
"description": "acme print cartridge",
"quantity": 2,
"weight": "1.5"
},
"result": {
"decision": "rejected",
"decisionDate": "2019-02-02"
},
"isPriority": true
},
{
"id": "333212-CSE525666-2",
"type": "Stationery",
"subType": "other",
"targetDate": "2020-06-15",
"itemDetail": {
"description": "staples box",
"quantity": 3,
"weight": "1.66"
},
"result": {
"decision": "accepted",
"decisionDate": "2020-03-03",
"itemValue": "23.01"
},
"isPriority": true
}
]
},
{
"caseId": "CSE885655",
"items": [{
"id": "333212-CSE885655-1",
"type": "marine goods",
"subType": "fish food",
"targetDate": "2020-06-04",
"itemDetail": {
"description": "fish bait",
"quantity": 5,
"weight": "0.65"
},
"result": {
"decision": "accepted",
"decisionDate": "2020-03-02"
},
"isPriority": false
},
{
"id": "333212-CSE885655-4",
"type": "tobacco products",
"subType": "cigarettes",
"deadlineDate": "2020-06-15",
"itemDetail": {
"description": "rolling tobbaco",
"quantity": 42,
"weight": "2.25"
},
"result": {
"decision": "accepted",
"decisionDate": "2020-02-02",
"itemValue": "48.15"
},
"isPriority": true
}
]
}
]
},
"state": "open"
}
You're probably looking for $unwind. It takes an array within a document and creates a separate document for each array member.
{ foos: [1, 2] } -> { foos: 1 }, { foos: 2}
With that you can create a flat document structure and match & group as normal.
db.collection.aggregate([
{
$unwind: "$data.cases"
},
{
$unwind: "$data.cases.items"
},
{
$match: {
"data.cases.items.result.decision": "accepted"
}
},
{
$group: {
_id: "$data.locationCode",
value: {
$sum: {
$toDecimal: "$data.cases.items.result.itemValue"
}
}
}
},
{
$project: {
_id: 0,
locationCode: "$_id",
value: "$value"
}
}
])
https://mongoplayground.net/p/Xr2WfFyPZS3
Alternative solution...
We group by data.locationCode and sum all items with this condition:
cases[*].items[*].result.decision" == "accepted"
db.collection.aggregate([
{
$group: {
_id: "$data.locationCode",
itemValue: {
$sum: {
$reduce: {
input: "$data.cases",
initialValue: 0,
in: {
$sum: {
$concatArrays: [
[ "$$value" ],
{
$map: {
input: {
$filter: {
input: "$$this.items",
as: "f",
cond: {
$eq: [ "$$f.result.decision", "accepted" ]
}
}
},
as: "item",
in: {
$toDouble: {
$ifNull: [ "$$item.result.itemValue", 0 ]
}
}
}
}
]
}
}
}
}
}
}
}
])
MongoPlayground
I have an array as below:
const test = [{
"_id": 1,
"name": "apple",
"car": "ford"
},{
"_id": 2,
"name": "melon",
"car": "ferrari"
},{
"_id": 3,
"name": "perl",
"car": "Renaut"
}]
And there is are documents of Mongodb as below:
[{
"name": "perl", "company": "A"
},{
"name": "melon", "company": "B"
},{
"name": "apple", "company": "C"
},{
"name": "apple", "company": "D"
},{
"name": "perl", "company": "E"
},{
"name": "apple", "company": "F"
}]
And I want to get this result using mongodb aggregate:
[{
"name": "perl", "company": "A", testInform: { "_id": 3, "name": "perl", "car": "Renaut"}
},{
"name": "melon", "company": "B", testInform: { "_id": 2, "name": "melon", "car": "ferrari"}
},{
"name": "apple", "company": "C", testInform: { "_id": 1, "name": "apple", "car": "ford"}
},{
"name": "apple", "company": "D", testInform: { "_id": 1, "name": "apple", "car": "ford"}
},{
"name": "perl", "company": "E", testInform: { "_id": 3, "name": "perl", "car": "Renaut"}
},{
"name": "apple", "company": "F", testInform: { "_id": 1, "name": "apple", "car": "ford"}
}]
I think to use aggregate with $match and $facet, etc., but I don't know exactly how to do this. Could you recommend a solution for this?
Thank you so much for reading this.
$lookup with pipeline keyword
db.demo2.aggregate(
{
$lookup:
{
from: "demo1",
let: { recordName: "$name"},
pipeline: [
{ $match:
{ $expr:
{ $and:
[
{ $eq: [ "$$recordName", "$name" ] },
]
}
}
},
],
as: "testInform"
}
}
)
If the test array data is stored in a collection then acheiving O/P is pretty straightforward $lookup with $project aggregation
$arrayElemAt Why? because the lookup would fetch the joined documents in an array as testInform
db.maindocs.aggregate([
{
$lookup: {
from: "testdocs",
localField: "name",
foreignField: "name",
as: "testInform"
}
},
{
$project: {
_id: 0,
name: 1,
company: 1,
testInform: { $arrayElemAt: ["$testInform", 0] }
}
}
]);
Update based on comments:
The idea is to iterate the cursor from the documents stored in mongodb Array.prototype.find() the object from test which matches the name field, add it to result.
const test = [
{
_id: 1,
name: "apple",
car: "ford"
},
{
_id: 2,
name: "melon",
car: "ferrari"
},
{
_id: 3,
name: "perl",
car: "Renaut"
}
];
const cursor = db.collection("maindocs").find();
const result = [];
while (await cursor.hasNext()) {
const doc = await cursor.next();
const found = test.find(e => e.name === doc.name);
if (found) {
doc["testInform"] = found;
}
result.push(doc);
}
console.info("RESULT::", result);
The aggregation has one stage: Iterate over the test array and get the array element as an object which matches the name field in both the document and the array (using the $reduce operator).
const test = [ { ... }, ... ]
db.test_coll.aggregate( [
{
$addFields: {
testInform: {
$reduce: {
input: test,
initialValue: { },
in: {
$cond: [ { $eq: [ "$$this.name", "$name" ] },
{ $mergeObjects: [ "$$this", "$$value" ] },
"$$value"
]
}
}
}
}
}
] )
I have collection with documents like this one:
{
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}
My query is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })
This query returns:
{
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}
But I need the result like:
{ "names": [ "Bob", "Alice" ] }
Which parameters should I add to the .aggregate() function?
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })