Sample of my database:
{
name: 'The Perks of Being a Wallflower'
genres: ['Drama', 'Romance']
rating: 8
},
{
name: 'The Edge of Seventeen'
genres: ['Drama', 'Comedy']
rating: 7.3
},
{
name: 'Little Women',
genres: ['Drama', 'Romance']
rating: 7.8
}
I want to project the first element of the genres array if it's Comedy or Romance.
I tried this :
db.shows.find(
{},
{ genres: { $elemMatch: { $or: [{ $eq: 'Drama' }, {$eq: 'Comedy'}] } } }
);
But it gives me this error "unknown top level operator: $eq".
Use $in
The $in operator selects the documents where the value of a field equals any value in the specified array. To specify an $in expression, use the following prototype:
Demo - https://mongoplayground.net/p/MAuWUeP9GIE
db.collection.find({
genres: {
$elemMatch: {
"$in": [ "Drama", "Comedy" ]
}
}
})
Demo - https://mongoplayground.net/p/JJJkoHuz_DZ
db.collection.find({},
{
genres: {
$elemMatch: {
"$in": [ "Drama", "Comedy" ]
}
}
})
Related
I have a database of a the employees of a company that looks like this:
{
_id: 7698,
name: 'Blake',
job: 'manager',
manager: 7839,
hired: ISODate("1981-05-01T00:00:00.000Z"),
salary: 2850,
department: {name: 'Sales', location: 'Chicago'},
missions: [
{company: 'Mac Donald', location: 'Chicago'},
{company: 'IBM', location: 'Chicago'}
]
}
I have an exercise in which I need to write the MongoDb command that returns all them employees who did all their missions in Chicago. I struggle with the all because I cannot find a way to check that all the locations of the missions array are equal to 'Chicago'.
I was thinking about doing it in two time: first find the total number of missions the employee has and then compare it to the number of mission he has in Chicago (that how I would do in SQL I guess). But I cannot found the number of mission the employee did in Chicago. Here is what I tried:
db.employees.aggregate([
{
$match: { "missions": { $exists: true } }
},
{
$project: {
name: 1,
nbMissionsChicago: {
$sum: {
$cond: [
{
$eq: [{
$getField: {
field: { $literal: "$location" },
input: "$missions"
}
}, "Chicago"]
}, 1, 0
]
}
}
}
}
])
Here is the result :
{ _id: 7698, name: 'Blake', nbMissionsChicago: 0 }
{ _id: 7782, name: 'Clark', nbMissionsChicago: 0 }
{ _id: 8000, name: 'Smith', nbMissionsChicago: 0 }
{ _id: 7902, name: 'Ford', nbMissionsChicago: 0 }
{ _id: 7499, name: 'Allen', nbMissionsChicago: 0 }
{ _id: 7654, name: 'Martin', nbMissionsChicago: 0 }
{ _id: 7900, name: 'James', nbMissionsChicago: 0 }
{ _id: 7369, name: 'Smith', nbMissionsChicago: 0 }
First of all, is there a better method to check that all the locations of the missions array respect the condition? And why does this commands returns only 0 ?
Thanks!
If all you need is the agents who had all their missions in "Chicago" then you don't need an aggregation pipeline for it, specifically the approach of filtering the array as part of the aggregation can't utilize an index and will make performance even worse.
A simple query should suffice here:
db.collection.find({
$and: [
{
"missions": {
$exists: true
}
},
{
"missions.location": {
$not: {
$gt: "Chicago"
}
}
},
{
"missions.location": {
$not: {
$lt: "Chicago"
}
}
}
]
})
Mongo Playground
This way we can build an index on the missions field and utilize it properly, any documents with a different value other then "Chigaco" will not match as they will fail the $gt or $lt comparion.
Note that an empty array also matches the condition, you can change the generic "missions" exists condition key into "missions.0": {$exists: true}, this will also require at least one mission.
You are unable to get the correct result as it is not the correct way to iterate the element in an array field.
Instead, you need to work with $size operator to get the size of an array and the $filter operator to filter the document.
Updated: You can directly compare the filtered array with the original array.
db.employees.aggregate([
{
$match: {
"missions": {
$exists: true
}
}
},
{
$project: {
name: 1,
nbMissionsChicago: {
$eq: [
{
$filter: {
input: "$missions",
cond: {
$eq: [
"$$this.location",
"Chicago"
]
}
}
},
"$missions"
]
}
}
}
])
Demo # Mongo Playground
I have the following document structure which contains an array of votes:
{ _id: ObjectId("6350e2c1a15e0e656f4a7472"),
category: 'business',
votes:
[ { voteType: 'like',
userId: ObjectId("62314007da34df3f32f7cfc0") },
{ voteType: 'like',
userId: ObjectId("6356b5cbe2272ebf628451b") } ] }
What I would like to achieve is to add for each document the sum of votes for which voteType = like, while keeping the original document, such as:
[ [{ _id: ObjectId("6350e2c1a15e0e656f4a7472"),
category: 'business',
votes:
[ { voteType: 'like',
userId: ObjectId("62314007da34df3f32f7cfc0") },
{ voteType: 'like',
userId: ObjectId("6356b5cbe2272ebf628451b") } ] }, {sum: 2, voteType: "like"} ], ...]
At the moment, the only workaround that I found is through an aggregation although I cannot manage to keep the original documents in the results:
db.getCollection('MyDocument') .aggregate([ {
$unwind: "$votes" }, {
$match: {
"votes.voteType": "like",
} }, {
$group: {
_id: {
name: "$_id",
type: "$votes.voteType"
},
count: {
$sum: 1
}
} },
{ $sort : { "count" : -1 } }, {$limit : 5}
])
which gives me:
{ _id: { name: ObjectId("635004f1b96e494947caaa5e"), type: 'like' },
count: 3 }
{ _id: { name: ObjectId("63500456b96e494947cbd448"), type: 'like' },
count: 3 }
{ _id: { name: ObjectId("63500353b6c7eb0a01df268e"), type: 'like' },
count: 2 }
{ _id: { name: ObjectId("634e315bb7d17339f8077c39"), type: 'like' },
count: 1 }
You can do it like this:
$cond with $isArray - to check if the votes property is of the type array.
$filter - to filter votes based on voteType property.
$size - to get the sized of the filtered array.
db.collection.aggregate([
{
"$set": {
"count": {
"$cond": {
"if": {
"$isArray": "$votes"
},
"then": {
"$size": {
"$filter": {
"input": "$votes",
"cond": {
"$eq": [
"$$this.voteType",
"like"
]
}
}
}
},
"else": 0
}
}
}
}
])
Working example
My model :
const scheduleTaskSchema = new Schema({
activity: { type: Object, required: true },
date: { type: Date, required: true },
crew: Object,
vehicle: Object,
pickups: Array,
details: String,
});
const ScheduleTaskModel = mongoose.model("schedule_task", scheduleTaskSchema),
and this aggregation pipeline :
let aggregation = [
{
$sort: {
"pickups.0.time": 1,
},
},
{
$group: {
_id: "$date",
tasks: { $push: "$$ROOT" },
},
},
{ $sort: { _id: -1 } },
];
if (hasDateQuery) {
aggregation.unshift({
$match: {
date: { $gte: new Date(start_date), $lte: new Date(end_date) },
},
});
} else {
aggregation.push({ $limit: 2 });
}
const scheduledTasksGroups = await ScheduleTaskModel.aggregate(aggregation);
the crew object can have arbitrary number of keys with this structure :
crew : {
drivers: [
{
_id: "656b1e9cf5b894a4f2v643bc",
name: "john"
},
{
_id: "567b1e9cf5b954a4f2c643bhh",
name: "bill"
}
],
officers: [
{
_id: "655b1e9cf5b6632a4f2c643jk",
name: "mark"
},
{
_id: "876b1e9af5b664a4f2c234bb",
name: "jane"
}
],
//...any number of keys that contain an array of objects that all have an _id
}
I'm looking for a way to return all documents (before sorting/grouping) that contain a given _id anywhere within the crew object without knowing which key to search,it can be many different keys that all contain an array of objects that all have an _id
Any ideas ?
You can use $objectToArray for this:
db.collection.aggregate([
{$addFields: {crewFilter: {$objectToArray: "$crew"}}},
{$set: {
crewFilter: {$size: {
$reduce: {
input: "$crewFilter",
initialValue: [],
in: {$concatArrays: [
"$$value",
{$filter: {
input: "$$this.v",
as: "member",
cond: {$eq: ["$$member._id", _id]}
}
}
]
}
}
}}
}},
{$match: {crewFilter: {$gt: 0}}}
])
See how it works on the playground example
My data looks like this:
[
{
"_id":"61717cafd351f3ae8b6d205a",
"restaurant":"Hogwarts",
"purchasedAt":"2021-10-20T17:47:40.166Z",
"products":[
{
"name":"Meat Samosa",
"price":3.95,
"quantity":1,
"_id":"61717cafd351f3ae8b6d205b"
},
{
"name":"Pilau Rice",
"price":2.95,
"quantity":1,
"_id":"61717cafd351f3ae8b6d205f"
}
]
},
{
"_id":"61717cb2d351f3ae8b6dd05b",
"restaurant":"Hogwarts",
"purchasedAt":"2021-10-20T03:14:11.111Z",
"products":[
{
"name":"Pilau Rice",
"price":2.95,
"quantity":1,
"_id":"61717cb2d351f3ae8b6dd05d"
}
]
},
]
I am trying to find a query that will get me all the products (no duplicates) and their quantities added up. Notice that the products id are different even when they are the same(same name) Ideally my response would look like this
[
{
name: "Meat Samosa",
price: 3.95,
quantity: 1
},
{
name: "Pilau Rice",
price: 2.95,
quantity: 2
}
]
$project to show required fields
$unwind deconstruct the products array
$group by name and get the first price and count the quantity sum
$project to show required fields
db.collection.aggregate([
{
$project: {
_id: 0,
products: 1
}
},
{ $unwind: "$products" },
{
$group: {
_id: "$products.name",
price: { $first: "$products.price" },
quantity: { $sum: "$products.quantity" }
}
},
{
$project: {
_id: 0,
name: "$_id",
price: 1,
quantity: 1
}
}
])
Playground
For example, with this data:
{id: 1, fname: "Barry", lname: "Sullivan"}
{id: 2, fname: "Sarah", lname: "Bailey"}
{id: 3, fname: "Drake", lname: "Barry"}
Is there a way, with a single query, that I could check to see if anyone had the same lname as id: 1 fname?
You can use $facet to run two separate queries and get the result as one document. This will give you two separate arrays: 1-element with id:1 and the other documents. Then you can simply run $filter to get matching lnames:
db.collection.aggregate([
{
$facet: {
first: [ { $match: { id: 1 } } ],
others: [ { $match: { $expr: { $ne: [ "$id", 1 ] } } } ]
}
},
{
$unwind: "$first"
},
{
$project: {
matches: {
$filter: {
input: "$others",
cond: { $eq: [ "$$this.lname", "$first.lname" ] }
}
}
}
}
])
Mongo Playground