Develop Reference

Mongoose - renaming object key within array

I have this one schema
{
_id: "123456",
id: "123",
inventory: [
{
id: "foo",
count: 0
},
{
id: "bar",
count: 3
}
]
}
I wanted every "count" keys in the inventory array to be "price" which will look like this at the end:
{
_id: "123456",
id: "123",
inventory: [
{
id: "foo",
price: 0
},
{
id: "bar",
price: 3
}
]
}
And I've tried this
Model.updateOne({ id: "123" }, { $unset: { inventory: [{ count: 1 }] } } )
But it seems to be deleting the "inventory" field itself

The first thing here is to try to use $rename but how the docs explain:
$rename does not work if these fields are in array elements.
So is necessary to look for another method. So you can use this update with aggregation query:
This query uses mainly $map, $arrayToObject and $objectToArray. The trick here is:
Create a new field called inventory (overwrite existing one)
Iterate over every value of the array with $map, and then for each object in the array use $objectToArray to create an array and also iterate over that second array using again $map.
Into this second iteration create fields k and v. Field v will be the same (you don't want to change the value, only the key). And for field k you have to change only the one whose match with your condition, i.e. only change from count to price. If this condition is not matched then the key remain.
db.collection.update({},
[
{
$set: {
inventory: {
$map: {
input: "$inventory",
in: {
$arrayToObject: {
$map: {
input: {$objectToArray: "$$this"},
in: {
k: {
$cond: [
{
$eq: ["$$this.k","count"]
},
"price",
"$$this.k"
]
},
v: "$$this.v"
}
}
}
}
}
}
}
}
])
Example here

MongoDB: aggregation, array of objects to string value

My question is pretty similar to: MongoDB Aggregation join array of strings to single string, but instead of pure Array, like: ['Batman', 'Robin'] I have Array of objects:
_id: 1,
field_value: [
{
_id: 2,
name: "Batman"
},
{
_id: 3,
name: "Robin"
}
]
I am trying to use $reduce but got error instead.
I want to receive the following result:
_id: 1,
field_value: "Batman, Robin" /** <= String value */
or at least array of property values:
_id: 1,
field_value: ["Batman", "Robin"] /** <= Array of strings (name property) */
My MongoPlayground data example

You need the same approach with $reduce, $$this represents a single field_value entity so you need $$this.name:
db.collection.aggregate([
{
$project: {
field_value: {
$reduce: {
input: "$field_value",
initialValue: "",
in: {
$concat: [
"$$value",
{ $cond: [ { $eq: [ "$$value", "" ] }, "", "," ] },
{ $toString: "$$this.name" }
]
}
}
}
}
}
])
Mongo Playground

Run a specific $match query only if string is not empty

I am trying to implement a search of documents in a collection. One of the queries in the $match stage checks if anything in a list matches another list from my JS (not within the document).
However, what I want to do is only run that query if the list has contents. If it is empty then just ignore it.
Essentially like choosing options in a filter. If you choose something, then it will search for that, if you choose nothing then it doesn't limit based on that.
Person documents:
{
name: String,
dob: Date,
gender: String,
favoriteColors: [{
color: String,
otherInfo: String
}]
}
My pipeline:
let colors = ["red", "blue", "green"];
collection.aggregate([
{$match: {
gender: "male",
//run below only if colors array is not empty
favoriteColors: {
$elemMatch: {
color: {
$in: colors
}
}
}
}}
])
How can I basically just run that part of the aggregate only if the array is not empty? I really don't want to use an if statement and write that same aggregate twice based on the two different conditions. Is it even possible to do this?

You can use something like this using $cond like so:
let colors = [ ... ]
db.collection.aggregate([
{
$match: {
$expr: {
$and: [
{
$eq: [
"$gender",
"male"
]
},
{
$cond: [
{
$gt: [
colors.length,
0
]
},
{
$gt: [
{
$size: {
$setIntersection: [
{
$map: {
input: "$favoriteColors",
as: "color",
in: "$$color.color"
}
},
colors
]
}
},
0
]
},
{}
]
}
]
}
}
}
])
But if you're already using code why not move the same logic into js?:
let colors = [ ... ];
let conds = [
{
"gender": "Male"
}
];
if (colors.length) {
conds.push({
"favoriteColors.color": {$in: colors}
})
}
db.collection.aggregate([
{
$match: {
$and: conds
}
}
])

mongoDB project array element after filter

I have the following (simplified) aggregation:
Model.aggregate([
{
$lookup: {
from: 'orders',
localField: '_id',
foreignField: 'customer',
as: 'orders',
},
},
{
$project: {
openOrders: {
$filter: {
input: '$orders',
as: 'order',
cond: { $eq: ['$$order.status', 'open'] },
},
},
},
},
])
which returns the following:
{
_id: ...,
openOrders: [
[Object], [Object]
],
}
Those [Object]'s are simply the returned objects, persisted in the database, with all their fields.
I don't find a way to project/filter out those objects' fields and instead return only their _id's:
{
_id: ...,
openOrders: [
_id: ...,
_id: ....
],
}
EDIT: I'd rather prefer the following expected output:
{
_id: ...,
openOrders: [
{ _id: ... },
{ _id: ... }
],
}
I tried adding a new $project stage at various points of the aggregation with no success. Can someone help me?

You should add a $project stage like below:
{
$project: {
openOrders: 'openOrders._id'
}
}
This will give the output like:
{
_id: ...,
openOrders: [
_id1,
_id2,
...
],
}
instead of
{
_id: ...,
openOrders: [
_id: ...,
_id: ....
],
}
I suggest this type of querying because, if you actually see openOrders, it's just the array of _ids, so adding only one _id field inside the array doesn't make sence.
If you still want the output to be like the array of object, then you can use the below:
{
$project: {
'openOrders._id': 1
}
}

As you need an array of _id's like this
openOrders: [ _id: ..., _id: .... ]
but not an array of _id's in objects :
openOrders: [ {_id: ...}, {_id: ....} ]
You need to use $reduce instead of $filter :
Try below query :
db.collection.aggregate([
{
$project: {
openOrders: {
$reduce: {
input: "$orders", // Same like `$filter` use reduce to iterate on array
initialValue: [], // consider an initial value
in: { // If condition is met, push value to array else return holding array as is.
$cond: [ { $eq: [ "$$this.status", "open" ] },
{ $concatArrays: [ "$$value", [ "$$this._id" ] ] },
"$$value"
]
}
}
}
}
}
])
Test : mongoplayground
Note : In javaScript - if you're printing a JSON with objects, you need to print it with JSON.stringify(yourJSON) - which makes it a string, So that you don't see [Object], [Object] in console rather you would see actual objects.
Update :
If you need an array of objects with _id field just add another $project stage at the end, but I would highly suggest to use $reduce and get an array for your scenario :
{ $project: { "openOrders._id": 1 } } // which would just extract `_id` fields in each objects
Test : mongoplayground

MongoDB: Project only the items that was queried for in the array?

I have a user document, each user has an array of objects
Given an array of item tags, I need to find the user whose item array has the item-tag, and return the entire user object except the items array, in which I only want to return the first item tags that existed in the tagArray that was used for the intial query.
//user document
{
user: 'John',
items: [ObjectId('ABC'), ObjectId('123') ...]
}
//item document
{
_id: ObjectId('ABC'),
tag: 'some-unique-id'
},
{
_id: ObjectId('DEF'),
tag: 'some-unique-tag'
}
Users have a 1-to-N relationship with items, the items may repeat within the User's items array.
This is what I current have, which returns the entire user object, but also all the items within the array.
const tagArray = [ 'some-unique-id', 'some-unique-tag']
items.aggregate([
{ $match: { 'tag': { $in: tagArray } }},
{ $lookup: {
from: "users",
localField: "tag",
foreignField: '_id',
as: 'userInfo'
}
},
{
$project: {??} //<--- I'm pretty sure I'm missing something in the project
])
Outcome that I have now:
{
_id: ObjectId('ABC'),
tag: 'some-unique-id'
userInfo : [ {user: 'John', items: [ObjectId('ABC'), ObjectId('123') ...] }]
}
What I want to achieve:
{
_id: ObjectId('ABC'),
tag: 'some-unique-id'
userInfo : [ {user: 'John', items: [ObjectId('ABC')]} ]
}
Edit:
There is a similar question here : Retrieve only the queried element in an object array in MongoDB collection
However in my case, I need the filter condition to be "one of the the tags that is in the tagArray.
Any suggestion or pointers would be appreciated, thank you!

I don't know if I understood well what you need, but I think this is a good start (maybe you can modify it by yourself):
Test data:
// users collection
[
{
user: "John",
items: [
ObjectId("5a934e000102030405000002"),
ObjectId("5a934e000102030405000003")
]
}
]
// items collection
[
{
_id: ObjectId("5a934e000102030405000002"),
tag: "some-unique-id"
},
{
_id: ObjectId("5a934e000102030405000009"),
tag: "some-unique-tag"
}
]
}
Query:
db.users.aggregate([
{
$lookup: {
from: "items",
localField: "items",
foreignField: "_id",
as: "userInfo"
}
},
// create new fields inside the userInfo array
{
$project: {
"userInfo.user": "$user",
"userInfo.items": "$items",
"tag": {
$arrayElemAt: ["$userInfo.tag", 0]
}
}
},
// filter the userInfo.items field, based on _id field
// it's important to use $arrayElemAt here
{
$addFields: {
"userInfo.items": {
$filter: {
input: {
$arrayElemAt: [
"$userInfo.items",
0
]
},
as: "i",
cond: {
$in: [
"$$i",
[
"$_id"
]
]
}
}
}
}
}
])
Result:
[
{
"_id": ObjectId("5a934e000102030405000002"),
"tag": "some-unique-id",
"userInfo": [
{
"items": [
ObjectId("5a934e000102030405000002")
],
"user": "John"
}
]
}
]