When i do a find all in a mongodb collection, i want to get maintenanceList sorted by older maintenanceDate to newest.
The sort of maintenanceDate should not affect parents order in a find all query
{
"_id":"507f191e810c19729de860ea",
"color":"black",
"brand":"brandy",
"prixVenteUnitaire":200.5,
"maintenanceList":[
{
"cost":100.40,
"maintenanceDate":"2017-02-07T00:00:00.000+0000"
},
{
"cost":4000.40,
"maintenanceDate":"2019-08-07T00:00:00.000+0000"
},
{
"cost":300.80,
"maintenanceDate":"2018-08-07T00:00:00.000+0000"
}
]
}
Any guess how to do that ?
Thank you
Whatever order the fields are in with the previous pipeline stage, as operations like $project and $group effectively "copy" same position.So, it will not change the order of your fields in your aggregated result.
And the sort of maintenanceDate through aggregation will not affect parents order in a find all query.
So, simply doing this should work.
Assuming my collection name is example.
db.example.aggregate([
{
"$unwind": "$maintenanceList"
},
{
"$sort": {
"_id": 1,
"maintenanceList.maintenanceDate": 1
}
},
{
"$group": {
"_id": "$_id",
"color": {
$first: "$color"
},
"brand": {
$first: "$brand"
},
"prixVenteUnitaire": {
$first: "$prixVenteUnitaire"
},
"maintenanceList": {
"$push": "$maintenanceList"
}
}
}
])
Output:
Related
I have an array of documents that looks like this:
patient: {
conditions: [
{
columnToSortBy: "value",
type: "PRIMARY"
},
{
columnToSortBy: "anotherValue",
type: "SECONDARY"
},
]
}
I need to be able to $sort by columnToSortBy, but using the item in the array where type is equal to PRIMARY. PRIMARY is not guaranteed to be the first item in the array every time.
How do I set my $sort up to accommodate this? Is there something akin to:
// I know this is invalid. It's for illustration purposes
$sort: "columnToSortBy", {$where: {type: "PRIMARY"}}
Is it possible to sort a field, but only when another field matches a query? I do not want the secondary conditions to affect the sort in any way. I am sorting on that one specific element alone.
You need to use aggregation framework
db.collection.aggregate([
{
$unwind: "$patient.conditions" //reshape the data
},
{
"$sort": {
"patient.conditions.columnToSortBy": -1 //sort it
}
},
{
$group: {
"_id": "$_id",
"conditions": { //re group it
"$push": "$patient.conditions"
}
}
},
{
"$project": { //project it
"_id": 1,
"patient.conditions": "$conditions"
}
}
])
Playground
I would like to migrate one of my FireBase projects to Mongo and move the calculations from server side to DB. I already wrote most of the queries but this one is beyond my knowledge.
Player data are saved by week and I need to calculate the sum of donations and points for each players (the rest of the fields should be ignored).
PS: Some of the players are already banned so it would be enough the calculate the fields for a given player set (like: tag in ['playerId1', 'playerId2', ...]). If it's too complex I will do this filtering later on server side.
[
{
"week":"2021-01",
"players":[
{
"donations":20,
"games":3,
"name":"Player1",
"points":258,
"tag":"playerId1"
},
{
"donations":37,
"games":5,
"name":"Player2",
"points":634,
"tag":"playerId2"
},
{ ... }
]
},
{
"week":"2021-02",
"players":[ { ... } ]
}
]
So the result should be something like this:
[
{
"name":"Player1",
"tag":"playerId1",
"donations":90,
"points":980
},
{
"name":"Player2",
"tag":"playerId2",
"donations":80,
"points":1211
}
]
I think the $unwind and the $group operators could be the key but I can't figure out how to use them properly here.
$unwind deconstruct players array
$group by name and get sum of donations and points and get first tag
$project to show required fields
db.collection.aggregate([
{ $unwind: "$players" },
{
$group: {
_id: "$players.name",
donations: { $sum: "$players.donations" },
points: { $sum: "$players.points" },
tag: { $first: "$players.tag" }
}
},
{
$project: {
_id: 0,
name: "$_id",
points: 1,
tag: 1,
donations: 1
}
}
])
Playground
PS: Some of the players are already banned so it would be enough the calculate the fields for a given player set (like: tag in ['playerId1', 'playerId2', ...]).
You can put match condition after $unwind stage,
{ $match: { "players.tag": { $in: ['playerId1', 'playerId2', ..more] } } }
You were right,
play
db.collection.aggregate([
{//Denormalize
"$unwind": "$players"
},
{//Group by name
"$group": {
"_id": "$players.name",
"donations": {
"$sum": "$players.donations"
},
"points": {
"$sum": "$players.points"
},
}
}
])
You can add project stage if you really need name as key than _id
I'm trying to perform nested group, I have an array of documents that has two keys (invoiceIndex, proceduresIndex) I need the documents to be arranged like so
invoices (parent) -> procedures (children)
invoices: [ // Array of invoices
{
.....
"procedures": [{}, ...] // Array of procedures
}
]
Here is a sample document
{
"charges": 226.09000000000003,
"currentBalance": 226.09000000000003,
"insPortion": "",
"currentInsPortion": "",
"claim": "notSent",
"status": "unpaid",
"procedures": {
"providerId": "9vfpjSraHzQFNTtN7",
"procedure": "21111",
"description": "One surface",
"category": "basicRestoration",
"surface": [
"m"
],
"providerName": "B Dentist",
"proceduresIndex": "0"
},
"patientId": "mE5vKveFArqFHhKmE",
"patientName": "Silvia Waterman",
"invoiceIndex": "0",
"proceduresIndex": "0"
}
Here is what I have tried
https://mongoplayground.net/p/AEBGmA32n8P
Can you try the following;
db.collection.aggregate([
{
$group: {
_id: "$invoiceIndex",
procedures: {
$push: "$procedures"
},
invoice: {
$first: "$$ROOT"
}
}
},
{
$addFields: {
"invoice.procedures": "$procedures"
}
},
{
"$replaceRoot": {
"newRoot": "$invoice"
}
}
])
I retain the invoice fields with invoice: { $first: "$$ROOT" }, also keep procedures's $push logic as a separate field. Then with $addFields I move that array of procedures into the new invoice object. Then replace root to that.
You shouldn't use the procedureIndex as a part of _id in $group, for you won't be able to get a set of procedures, per invoiceIndex then. With my $group logic it works pretty well as you see.
Link to mongoplayground
I have the following object with an embedded list of items and I would like to write a query and return all or specific items ordered by date. Is it possible to do it or should I have a different collection for items and keep here their references?
I know that you can match specific element using $elemMatch.
{
"_id": "51cb12857124a215940cf2d4",
"level1":
[
{
"name":"item00",
"description":"item01",
"date": 1238492103
},
{
"name":"item10",
"description":"item11",
"date": 1238492104
}
]
}
If you want these items ordered by date on a more often than not then your best option is to keep the list ordered in the first place. The $push operator has an additional $sort paramter explicitly for this purpose.
db.collection.update(
{ "_id": "51cb12857124a215940cf2d4" },
{ "$push": {
"level1":{
"$each":[{
"name":"item11",
"description":"item11",
"date": 1238492104
}],
"$sort": { "date": 1 }
}
}
)
That actually even adapts so you could just sort your whole collection in one statement:
db.collection.update(
{},
{ "$push": {
"level1":{
"$each":[], "$sort": { "date": 1 }
}
},
{ "multi": true }
)
Without that your ony alternate is to order the results via the .aggregate() method. This really should not be your chosen operation as it requires processing $unwind on the array contents and then $sort operation on the elements within the document. Naturally this comes with some significant overhead on larger selections:
db.collection.aggregate([
{ "$unwind": "$level1" },
{ "$sort": { "_id": 1, "level1.date": 1 } },
{ "$group": {
"_id": "$_id",
"level1": { "$push": "$level1" }
}}
])
I have the following issue:
this query return 1 result which is what I want:
> db.items.aggregate([ {$group: { "_id": "$id", version: { $max: "$version" } } }])
{
"result" : [
{
"_id" : "b91e51e9-6317-4030-a9a6-e7f71d0f2161",
"version" : 1.2000000000000002
}
],
"ok" : 1
}
this query ( I just added projection so I can later query for the entire document) return multiple results. What am I doing wrong?
> db.items.aggregate([ {$group: { "_id": "$id", version: { $max: "$version" } }, $project: { _id : 1 } }])
{
"result" : [
{
"_id" : ObjectId("5139310a3899d457ee000003")
},
{
"_id" : ObjectId("513931053899d457ee000002")
},
{
"_id" : ObjectId("513930fd3899d457ee000001")
}
],
"ok" : 1
}
found the answer
1. first I need to get all the _ids
db.items.aggregate( [
{ '$match': { 'owner.id': '9e748c81-0f71-4eda-a710-576314ef3fa' } },
{ '$group': { _id: '$item.id', dbid: { $max: "$_id" } } }
]);
2. then i need to query the documents
db.items.find({ _id: { '$in': "IDs returned from aggregate" } });
which will look like this:
db.items.find({ _id: { '$in': [ '1', '2', '3' ] } });
( I know its late but still answering it so that other people don't have to go search for the right answer somewhere else )
See to the answer of Deka, this will do your job.
Not all accumulators are available in $project stage. We need to consider what we can do in project with respect to accumulators and what we can do in group. Let's take a look at this:
db.companies.aggregate([{
$match: {
funding_rounds: {
$ne: []
}
}
}, {
$unwind: "$funding_rounds"
}, {
$sort: {
"funding_rounds.funded_year": 1,
"funding_rounds.funded_month": 1,
"funding_rounds.funded_day": 1
}
}, {
$group: {
_id: {
company: "$name"
},
funding: {
$push: {
amount: "$funding_rounds.raised_amount",
year: "$funding_rounds.funded_year"
}
}
}
}, ]).pretty()
Where we're checking if any of the funding_rounds is not empty. Then it's unwind-ed to $sort and to later stages. We'll see one document for each element of the funding_rounds array for every company. So, the first thing we're going to do here is to $sort based on:
funding_rounds.funded_year
funding_rounds.funded_month
funding_rounds.funded_day
In the group stage by company name, the array is getting built using $push. $push is supposed to be part of a document specified as the value for a field we name in a group stage. We can push on any valid expression. In this case, we're pushing on documents to this array and for every document that we push it's being added to the end of the array that we're accumulating. In this case, we're pushing on documents that are built from the raised_amount and funded_year. So, the $group stage is a stream of documents that have an _id where we're specifying the company name.
Notice that $push is available in $group stages but not in $project stage. This is because $group stages are designed to take a sequence of documents and accumulate values based on that stream of documents.
$project on the other hand, works with one document at a time. So, we can calculate an average on an array within an individual document inside a project stage. But doing something like this where one at a time, we're seeing documents and for every document, it passes through the group stage pushing on a new value, well that's something that the $project stage is just not designed to do. For that type of operation we want to use $group.
Let's take a look at another example:
db.companies.aggregate([{
$match: {
funding_rounds: {
$exists: true,
$ne: []
}
}
}, {
$unwind: "$funding_rounds"
}, {
$sort: {
"funding_rounds.funded_year": 1,
"funding_rounds.funded_month": 1,
"funding_rounds.funded_day": 1
}
}, {
$group: {
_id: {
company: "$name"
},
first_round: {
$first: "$funding_rounds"
},
last_round: {
$last: "$funding_rounds"
},
num_rounds: {
$sum: 1
},
total_raised: {
$sum: "$funding_rounds.raised_amount"
}
}
}, {
$project: {
_id: 0,
company: "$_id.company",
first_round: {
amount: "$first_round.raised_amount",
article: "$first_round.source_url",
year: "$first_round.funded_year"
},
last_round: {
amount: "$last_round.raised_amount",
article: "$last_round.source_url",
year: "$last_round.funded_year"
},
num_rounds: 1,
total_raised: 1,
}
}, {
$sort: {
total_raised: -1
}
}]).pretty()
In the $group stage, we're using $first and $last accumulators. Right, again we can see that as with $push - we can't use $first and $last in project stages. Because again, project stages are not designed to accumulate values based on multiple documents. Rather they're designed to reshape documents one at a time. Total number of rounds is calculated using the $sum operator. The value 1 simply counts the number of documents passed through that group together with each document that matches or is grouped under a given _id value. The project may seem complex, but it's just making the output pretty. It's just that it's including num_rounds and total_raised from the previous document.