I have the following structure to store user's like. For example, _id1 user likes user _id2 & _id3. Where the _id1 user is having a match with user _id2 only.
I need only records with a match.
Document Structure --
[
{'from':'_id1', 'to':'_id2'},
{'from':'_id1', 'to':'_id3'},
{'from':'_id2', 'to':'_id1'},
]
Expected Output --
[
{'from':'_id1', 'to':'_id2'},
{'from':'_id2', 'to':'_id1'},
]
You can $group on the "couple" and only then match those that were summed to be 2. The trick is to make sure the _id your grouping on is the same - I just did it with $cond and create the tuple [ _id1, _id2 ] where the smaller id is always first, like so:
db.collection.aggregate([
{
$group: {
_id: {
$cond: [
{
$gt: [
"$from",
"$to"
]
},
[
"$to",
"$from"
],
[
"$from",
"$to"
]
]
},
sum: {
$sum: 1
},
roots: {
$push: "$$ROOT"
}
}
},
{
$match: {
sum: {
$gt: 1
}
}
},
{
$unwind: "$roots"
},
{
$replaceRoot: {
newRoot: "$roots"
}
}
])
Related
I have two collections - "users" and "follows". "Follows" simply contains documents with a "follower" field and a "followee" field that represent when a user follows another user. What I want to do is to be able to query the users but display the users that I (or whatever user is making the request) follow first. For example if I follow users "5" and "14", when I search the list of users, I want users "5" and "14" to be at the top of the list, followed by the rest of the users in the database.
If I were to first query all the users that I follow from the "Follows" collection and get an array of those userIDs, is there a way that I can sort by using something like {$in: [userIDs]}? I don't want to filter out the users that I do not follow, I simply want to sort the list by showing the users that I do follow first.
I am using nodejs and mongoose for this.
Any help would be greatly appreciated. Thank you!
Answer
db.users.aggregate([
{
$addFields: {
sortBy: {
$cond: {
if: {
$in: [ "$_id", [ 5, 14 ] ]
},
then: 0,
else: 1
}
}
}
},
{
$sort: {
sortBy: 1
}
},
{
$unset: "sortBy"
}
])
Test Here
If you don't want you on the list, then
db.users.aggregate([
{
$addFields: {
sortBy: {
$cond: {
if: {
$in: [ "$_id", [ 5, 14 ] ]
},
then: 0,
else: 1
}
}
}
},
{
$sort: {
sortBy: 1
}
},
{
$unset: "sortBy"
},
{
$match: {
"_id": { $ne: 1 }
}
}
])
Test Here
If you want to sort users first
db.users.aggregate([
{
$sort: {
_id: 1
}
},
{
$addFields: {
sortBy: {
$cond: {
if: {
$in: [
"$_id",
[
5,
14
]
]
},
then: 0,
else: 1
}
}
}
},
{
$sort: {
sortBy: 1,
}
},
{
$unset: "sortBy"
},
{
$match: {
"_id": {
$ne: 1
}
}
}
])
Test Here
Updated:
I have a document on the database that looks like this:
My question is the following:
How can I retrieve the first 10 elements from the friendsArray from database and sort it descending or ascending based on the lastTimestamp value.
I don't want to download all values to my API and then sort them in Python because that is wasting my resources.
I have tried it using this code (Python):
listOfUsers = db.user_relations.find_one({'userId': '123'}, {'friendsArray' : {'$orderBy': {'lastTimestamp': 1}}}).limit(10)
but it just gives me this error pymongo.errors.OperationFailure: Unknown expression $orderBy
Any answer at this point would be really helpful! Thank You!
use aggregate
first unwind
then sort according timestap
group by _id to create sorted array
use addfields and filter for getting first 10 item of array
db.collection.aggregate([
{ $match:{userId:"123"}},
{
"$unwind": "$friendsArray"
},
{
$sort: {
"friendsArray.lastTimeStamp": 1
}
},
{
$group: {
_id: "$_id",
friendsArray: {
$push: "$friendsArray"
}
},
},
{
$addFields: {
friendsArray: {
$filter: {
input: "$friendsArray",
as: "z",
cond: {
$lt: [
{
$indexOfArray: [
"$friendsArray",
"$$z"
]
},
10
]
}// 10 is n first item
}
}
},
}
])
https://mongoplayground.net/p/2Usk5sRY2L2
and for pagination use this
db.collection.aggregate([
{ $match:{userId:"123"}},
{
"$unwind": "$friendsArray"
},
{
$sort: {
"friendsArray.lastTimeStamp": 1
}
},
{
$group: {
_id: "$_id",
friendsArray: {
$push: "$friendsArray"
}
},
},
{
$addFields: {
friendsArray: {
$filter: {
input: "$friendsArray",
as: "z",
cond: {
$and: [
{
$gt: [
{
$indexOfArray: [
"$friendsArray",
"$$z"
]
},
10
]
},
{
$lt: [
{
$indexOfArray: [
"$friendsArray",
"$$z"
]
},
20
]
},
]
}// 10 is n first item
}
}
},
}
])
The translation of your find to aggregation(we need unwind that why aggregation is used) would be like the bellow query.
Test code here
Query (for descending replace 1 with -1)
db.collection.aggregate([
{
"$match": {
"userId": "123"
}
},
{
"$unwind": {
"path": "$friendsArray"
}
},
{
"$sort": {
"friendsArray.lastTimeStamp": 1
}
},
{
"$limit": 10
},
{
"$replaceRoot": {
"newRoot": "$friendsArray"
}
}
])
If you want to skip some before limit add one stage also
{
"$skip" : 10
}
To take the 10-20 messages for example.
I have a document like this:
{
value: "field2",
field1: [ ... ],
field2: [ ... ],
...
}
Where value will be the value of one of the fields in the document. and many different fields are possible for one document.
I want to match a document. fetch the relevant field only and them do some calculations on it.
For example I want to do:
{
$unwind: "$value"
}
And get the results of field2 unwinded.
How can I do this?
It's a little bit "hacky" but you can achieve this using operators like $objectToArray and $filter like so:
db.collection.aggregate([
{
$addFields: {
"values": {
$arrayElemAt: [
{
$filter: {
input: {
$objectToArray: "$$ROOT"
},
as: "field",
cond: {
$eq: [
"$$field.k",
"$value"
]
}
}
},
0
]
}
}
},
{
$unwind: "$values.v"
},
{
$replaceRoot: {
newRoot: "$values.v"
}
},
])
MongoPlayground
As part of an aggregate I need to run this transformation:
let inheritances = await db.collection('inheritance').aggregate([
{ $match: { status: 1 }}, // inheritance active
{ $project: { "_id":1, "name": 1, "time_trigger": 1, "signers": 1, "tree": 1, "creatorId": 1, "redeem": 1, "p2sh": 1 } },
{ $lookup:
{
from: "user",
let: { creatorId: { $concat: [ "secretkey", { $toString: "$creatorId" } ] }, time_trigger: "$time_trigger"},
pipeline: [
{ $match:
{ $expr:
{ $and:
[
{ $eq: [ "$_id", sha256( { $toString: "$$creatorId" } ) ] },
{ $gt: [ new Date(), { $add: [ { $multiply: [ "$$time_trigger", 24*60*60*1000 ] }, "$last_access" ] } ] },
]
}
}
},
],
as: "user"
},
},
{ $unwind: "$user" }
]).toArray()
creatorId comes from a lookup, and in order to compare it to _id I first need to do a sha256.
How can I do it?
Thanks.
External functions will not work with the aggregation framework. Everything is parsed to BSON by default. It is all basically processed from BSON operators to native C++ code implementation, This is by design for performance.
Basically in short, you can't do this. I recommend just storing the hashed value on every document as a new field, otherwise you'll have to do it in code just before the pipeline.
Using the example from the Mongo docs:
{ _id: 1, results: [ { product: "abc", score: 10 }, { product: "xyz", score: 5 } ] }
{ _id: 2, results: [ { product: "abc", score: 8 }, { product: "xyz", score: 7 } ] }
{ _id: 3, results: [ { product: "abc", score: 7 }, { product: "xyz", score: 8 } ] }
db.survey.find(
{ id: 12345, results: { $elemMatch: { product: "xyz", score: { $gte: 6 } } } }
)
How do I return survey 12345 (regardless of even if it HAS surveys or not) but only return surveys with a score greater than 6? In other words I don't want the document disqualified from the results based on the subdocument, I want the document but only a subset of subdocuments.
What you are asking for is not so much a "query" but is basically just a filtering of content from the array in each document.
You do this with .aggregate() and $project:
db.survey.aggregate([
{ "$project": {
"results": {
"$setDifference": [
{ "$map": {
"input": "$results",
"as": "el",
"in": {
"$cond": [
{ "$and": [
{ "$eq": [ "$$el.product", "xyz" ] },
{ "$gte": [ "$$el.score", 6 ] }
]}
]
}
}},
[false]
]
}
}}
])
So rather than "contrain" results to documents that have an array member matching the condition, all this is doing is "filtering" the array members out that do not match the condition, but returns the document with an empty array if need be.
The fastest present way to do this is with $map to inspect all elements and $setDifference to filter out any values of false returned from that inspection. The possible downside is a "set" must contain unique elements, so this is fine as long as the elements themselves are unique.
Future releases will have a $filter method, which is similar to $map in structure, but directly removes non-matching results where as $map just returns them ( via the $cond and either the matching element or false ) and is then better suited.
Otherwise if not unique or the MongoDB server version is less than 2.6, you are doing this using $unwind, in a non performant way:
db.survey.aggregate([
{ "$unwind": "$results" },
{ "$group": {
"_id": "$_id",
"results": { "$push": "$results" },
"matched": {
"$sum": {
"$cond": [
{ "$and": [
{ "$eq": [ "$results.product", "xyz" ] },
{ "$gte": [ "$results.score", 6 ] }
]},
1,
0
]
}
}
}},
{ "$unwind": "$results" },
{ "$match": {
"$or": [
{
"results.product": "xyz",
"results.score": { "$gte": 6 }
},
{ "matched": 0 }
}},
{ "$group": {
"_id": "$_id",
"results": { "$push": "$results" },
"matched": { "$first": "$matched" }
}},
{ "$project": {
"results": {
"$cond": [
{ "$ne": [ "$matched", 0 ] },
"$results",
[]
]
}
}}
])
Which is pretty horrible in both design and perfomance. As such you are probably better off doing the filtering per document in client code instead.
You can use $filter in mongoDB 3.2
db.survey.aggregate([{
$match: {
{ id: 12345}
}
}, {
$project: {
results: {
$filter: {
input: "$results",
as: "results",
cond:{$gt: ['$$results.score', 6]}
}
}
}
}]);
It will return all the sub document that have score greater than 6. If you want to return only first matched document than you can use '$' operator.
You can use $redact in this way:
db.survey.aggregate( [
{ $match : { _id : 12345 }},
{ $redact: {
$cond: {
if: {
$or: [
{ $eq: [ "$_id", 12345 ] },
{ $and: [
{ $eq: [ "$product", "xyz" ] },
{ $gte: [ "$score", 6 ] }
]}
]
},
then: "$$DESCEND",
else: "$$PRUNE"
}
}
}
] );
It will $match by _id: 12345 first and then it will "$$PRUNE" all the subdocuments that don't have "product":"xyz" and don't have score greater or equal 6. I added the condition ($cond) { $eq: [ "$_id", 12345 ] } so that it wouldn't prune the whole document before it reaches the subdocuments.