Develop Reference

Mongo Aggregation pipeline update or push

I have a MongoDB Model which consist of array of members as obejcts.
const guestSchema = new mongoose.Schema({
salutation: {
type: String,
},
members: [membersSchema],
user: {
type: mongoose.Schema.ObjectId,
ref: 'User',
},
});
Members Schema:
const membersSchema = new mongoose.Schema({
name: String,
status: {
type: String,
enum: ['regular', 'helper'],
default: 'regular',
},
});
I want to achieve of doing an update in case documet with given ID exist or push to an array in case ID with document in array does not exist. I use aggregation pipeline, however I am not able to achieve pushing new document to array. Why can't I use push after else statement like this.
const subDocumentToUpsert = { 'name': mem.name, 'status': mem.status, '_id': ObjectId(mem.id)}
const subDocumentNoID = { 'name': mem.name, 'status': mem.status}
await Guest.findOneAndUpdate(
{ "_id": req.params.id },
[
{
$set: {
members: {
$cond: {
if: { $in: [subDocumentToUpsert._id, '$members._id'] },
then: {
$map: {
input: '$members',
as: 'sub_document',
in: {
$cond: {
if: { $eq: ['$$sub_document._id', subDocumentToUpsert._id] },
then: subDocumentToUpsert,
else: '$$sub_document',
},
},
},
},
else: {
$push: {
subDocumentNoID
},
},
},
},
},
},
},
]);
What is the best way of doing so? Thank you

You can do as follow:
db.collection.update({
_id: {
$in: [
1,
2
]
}
},
[
{
$set: {
members: {
$cond: {
if: {
$in: [
5,
"$members._id"
]
},
then: {
$map: {
input: "$members",
as: "sub",
in: {
$cond: {
if: {
$eq: [
"$$sub._id",
5
]
},
then: {
_id: 5,
status: "regular_updated",
name: "Negan_updated"
},
else: "$$sub"
},
},
},
},
else: {
$concatArrays: [
"$members",
[
{
_id: 5,
status: "regular_upserted",
name: "Negan_upserted"
}
]
]
}
}
}
}
}
}
],
{
multi: true
})
Explained:
Check if _id:5 exist in the subobject and update via $map/$cond only the object that has the _id:5.
In case there is no _id:5 add the new object to the array with $concatArrays.
Playground

Add field to each object of an array based on other field

I have the following Array of data:
{
_id: 5f5726ef7d475a61a95c5e0c,
attributes: [
{
values: [
{ name: '1' }
],
},
{
values: [
{ name: '2' }
]
}
],
attr1: [
{ name: "Study Code" },
{ name: "Patient Study" }
]
}
What I need is to add the correspondent value to each on of attr1 objects based on index. So the result would be:
{
_id: 5f5726ef7d475a61a95c5e0c,
attributes: [
{
values: [
{ name: '1' }
],
},
{
values: [
{ name: '2' }
]
},
],
attr1: [
{
name: "Study Code",
values: [{ name: "1" }]
},
{
name: "Patient Study",
values: [{ name: "2" }]
}
],
}
I wonder if that possible using aggregation $addFields in MongoDB

Query
query works if arrays same size
ziparray to make [[member1_1 member2_1], ....]
map to merge member1_1,member2_1 to a document
Playmongo
aggregate(
[{"$set": {"attr1": {"$zip": {"inputs": ["$attributes", "$attr1"]}}}},
{"$set":
{"attr1":
{"$map":
{"input": "$attr1",
"in":
{"$mergeObjects":
[{"$arrayElemAt": ["$$this", 1]},
{"$arrayElemAt": ["$$this", 0]}]}}}}}])

You can use $zip
db.collection.aggregate([
{
"$project": {
attributes: {
"$zip": {
"inputs": [
"$attributes",
"$attr1"
]
}
}
}
}
])
Here is the Mongo playground for your reference.

Mongodb, Verify if my query range is not overlapping a list of Date Ranges in mongodb

I have the following data:
const Availabilities = new Schema({
id: Number,
reservations: [{ from: Date, to: Date }]
});
const Availability = mongoose.model('Availability', Availabilities);
Basically i want to get a list of products in which the dates do not overlap with the query
[
{
id: 1,
reservations: [
{
from: '2022-01-01',
to: '2022-03-31',
},
{
from: '2022-04-01',
to: '2022-06-01',
},
],
},
{
id: 2,
reservations: [
{
from: '2022-01-01',
to: '2022-12-31',
},
],
},
{
id: 3,
reservations: [
{
from: '2022-02-01',
to: '2022-06-30',
},
],
},
]
and I want to filter all those who are not in a specific range.
Example:
query = { from: "2022-07-01", to: "2022-08-31" }
should return
[
{
id: 1,
reservations: [
{
from: '2022-01-01',
to: '2022-03-31',
},
{
from: '2022-04-01',
to: '2022-06-01',
},
],
},
{
id: 3,
reservations: [
{
from: '2022-02-01',
to: '2022-06-30',
},
],
},
]

You can just check if one of two conditions are met, either:
to is smaller than the start of your input ( meaning the session ended before )
from is bigger than the end of your input ( meaning the session started after )
This is how it'll look in code:
db.collection.aggregate([
{
$match: {
reservations: {
$elemMatch: {
$or: [
{
to: {
$lt: "2022-07-01"
},
},
{
from: {
$gt: "2022-08-31"
}
}
]
}
}
}
},
{
$addFields: {
reservations: {
$filter: {
input: "$reservations",
cond: {
$or: [
{
$lt: [
"$$this.to",
"2022-07-01"
]
},
{
$gt: [
"$$this.to",
"2022-08-31"
]
}
]
}
}
}
}
}
])
Mongo Playground

merge and remove elements in nested arrays

i have this array, i want to merge all elements inside the objects in the nested arrays and remove the duplicates..
the array is the output of mongo db populate so answers from there or just js will be amazing :)
"visitors": [
[
{
"name": "matan",
"id": "61793e6a0e08cdcaf213c0b1"
},
{
"name": "shani",
"id": "61793e910e08cdcaf213c0b5"
}
],
[
{
"name": "david",
"id": "6179869cb4944c6b19b05a23"
},
{
"name": "orit",
"id": "617986e535fdf4942ef659bd"
}
],
[
{
"name": "david",
"id": "6179869cb4944c6b19b05a23"
},
{
"name": "orit",
"id": "617986e535fdf4942ef659bd"
}
]
]
would like this output -
"visitors": [
{
"name": "matan",
"id": "61793e6a0e08cdcaf213c0b1"
},
{
"name": "shani",
"id": "61793e910e08cdcaf213c0b5"
},
{
"name": "david",
"id": "6179869cb4944c6b19b05a23"
},
{
"name": "orit",
"id": "617986e535fdf4942ef659bd"
},
]
these are my collections
i need to get all visitors on one solar system,
so > solars > planets > visitors
const solarsModel = new Schema({
planets: [ { type: Schema.Types.ObjectId ,ref:'planet'} ],
starName: { type: String, required: true, default: "" }
})
const planetModel = new Schema({
planetName: { type: String, required: true, default: "" },
system:{type: Schema.Types.ObjectId, ref: 'solar'},
visitors: [{ type: Schema.Types.ObjectId , ref: 'visitor'}]
})
const visitorModel = new Schema({
visitorName:{ type: String, required: true, default: "" },
homePlanet: {type: Schema.Types.ObjectId, ref:"planet" },
visitedPlanets: [{ type: Schema.Types.ObjectId, ref:"planet" }]
})
this is what i did to achieve a result would love to use Aggregate..
const response = await solarModel
.findById({ _id: data.id })
.select({ starName: 1, _id: 0 })
.populate({
path: "planets",
select: { visitors: 1, _id: 0 },
populate: {
path: "visitors",
select: "visitorName",
},
})
.exec();
solved with this
exports.findVisitorSystemHandler = async (data) => {
const systemName = await solarModel.findById({ _id: data.id });
const response = await planetModel.aggregate([
{ $match: { system: makeObjectId(data.id) } },
{
$lookup: {
from: "visitors",
localField: "visitors",
foreignField: "_id",
as: "solarVisitors",
},
},
{
$project: {
solarVisitors: {
visitedPlanets: 0,
homePlanet: 0,
__v: 0,
},
},
},
{ $unwind: "$solarVisitors" },
{
$group: {
_id: null,
system: { $addToSet: systemName.starName },
solarVisitors: {
$addToSet: {
id: "$solarVisitors._id",
name: "$solarVisitors.visitorName",
},
},
},
},
{ $unwind: "$system" },
{
$project: {
_id: 0,
},
},
]);
return response;
};

You can use aggregate() like this:
$unwind twice due to nested array
$group using $addToSet to not get duplicates.
db.collection.aggregate([
{
"$unwind": "$visitors"
},
{
"$unwind": "$visitors"
},
{
"$group": {
"_id": null,
"visitors": {
"$addToSet": {
"id": "$visitors.id",
"name": "$visitors.name"
}
}
}
}
])
Example here

(1) Flatten the array of arrays
visitors = visitors.flat();
Which gives us this:
[
{ name: 'matan', id: '61793e6a0e08cdcaf213c0b1' },
{ name: 'shani', id: '61793e910e08cdcaf213c0b5' },
{ name: 'david', id: '6179869cb4944c6b19b05a23' },
{ name: 'orit', id: '617986e535fdf4942ef659bd' },
{ name: 'david', id: '6179869cb4944c6b19b05a23' },
{ name: 'orit', id: '617986e535fdf4942ef659bd' }
]
(2) Get unique ids
let uniqueIds= [...new Set(visitors.map(v => v.id)]
Which gives us this:
[
'61793e6a0e08cdcaf213c0b1',
'61793e910e08cdcaf213c0b5',
'6179869cb4944c6b19b05a23',
'617986e535fdf4942ef659bd'
]
(3) Get new list of visitors based only on uniqueIds
visitors = uniqueIds.map(id => {
let name = visitors.find(v => v.id === id).name;
return {
id,
name
}
});
Which gives us this:
[
{ name: 'matan', id: '61793e6a0e08cdcaf213c0b1' },
{ name: 'shani', id: '61793e910e08cdcaf213c0b5' },
{ name: 'david', id: '6179869cb4944c6b19b05a23' },
{ name: 'orit', id: '617986e535fdf4942ef659bd' },
]

Query
reduce with concat to flatten
union with an empty array,just to remove duplicates
if you have other fields except visitors they are not affected
PlayMongo
aggregate(
[{"$set":
{"visitors":
{"$setUnion":
[{"$reduce":
{"input": "$visitors",
"initialValue": [],
"in": {"$concatArrays": ["$$value", "$$this"]}}},
[]]}}}])
Results
[{
"visitors": [
{
"name": "david",
"id": "6179869cb4944c6b19b05a23"
},
{
"name": "matan",
"id": "61793e6a0e08cdcaf213c0b1"
},
{
"name": "orit",
"id": "617986e535fdf4942ef659bd"
},
{
"name": "shani",
"id": "61793e910e08cdcaf213c0b5"
}
]
}]

Mongo Query - match and lookup combined

I’ve defined the following query which fetches me all items with an id which is in a given list of ids, and a status of either active or retracted.
const query = {
$and : [
{
$or: [
{
status: ‘active’,
},
{
status: ‘retracted’,
},
],
},
{
id: { $in: ids },
},
],
};
Each of these items has a parent_id field, which can either be null if the item does not have a parent, or can be the id of the parent.
I want my query to fetch all items with the ids I supply, as well as their parent items, if such a parent exists.
For example, if I supply the following IDs
[1,2,3]
and item 2 has a parent with id 5, while item 1 and 3 have parent_id set to null, I want my query to return the following items:
[1,2,3,5].
To achieve this I wrote the following query:
const collection = db.collection(‘myCollection’);
const data = await collection.aggregate([
{$match : query},
{
$lookup: {
from: ‘myCollection’,
let: { parentID: ‘$parent_id’},
pipeline: [
{
$match: {
$expr: {
$eq: [‘$id’, ‘$$parentID’],
},
},
},
as: ‘parent’,
},
},
]).sort(‘created_date’, ‘desc’).toArray();
return data;
However, this always returns null.
Sample Data:
[
{
id: 1,
parent_id: 3,
data: ‘bla bla’
},
{
id: 2,
parent_id: null,
data: ‘bla bla bla’
},
{
id: 3,
parent_id: null,
data: ‘bla’
}
]
Input: [1]
Output:
[
{
id: 1,
parent_id: 3,
data: ‘bla bla’
},
{
id: 3,
parent_id: null,
data: ‘bla’
}
]

The approach with $lookup being run upon same collection should work however it gives you a nested array so you need few additional stages to flatten such array and get all elements as on result set:
db.collection.aggregate([
{
$match: { id: { $in: [1] } }
},
{
$lookup: {
from: "collection",
localField: "parent_id",
foreignField: "id",
as: "parent"
}
},
{
$project: {
all: {
$concatArrays: [
"$parent",
[ "$$ROOT" ]
]
}
}
},
{
$project: {
"all.parent": 0
}
},
{
$unwind: "$all"
},
{
$replaceRoot: {
newRoot: "$all"
}
}
])
Mongo Playground

Your aggregation was malformed and lack some "]" for example closing the pipeline fied.
If you fix that the query works fine for me. Example

You can try this. The input array is [2,3] where 2 has parent id=1 and that is not in the input array. But the output array has the entry.
Working Playground
db.collection.aggregate([
{
$match: {
_id: {
$in: [
2,
3
]
}
}
},
{
$lookup: {
from: "collection",
localField: "p",
foreignField: "_id",
as: "parent"
}
},
{
$project: {
_id: 0,
id: {
$concatArrays: [
[
"$_id"
],
"$parent._id"
]
}
}
},
{
$unwind: "$id"
},
{
$sort: {
id: 1
}
}
])