I am trying to query multiple sub-documents in MongoDB and return as a single doc.
I think the aggregation framework is the way to go, but, can't see to get it exactly right.
Take the following docs:
{
"board_id": "1",
"hosts":
[{
"name": "bob",
"ip": "10.1.2.3"
},
{
"name": "tom",
"ip": "10.1.2.4"
}]
}
{
"board_id": "2",
"hosts":
[{
"name": "mickey",
"ip": "10.2.2.3"
},
{
"name": "mouse",
"ip": "10.2.2.4"
}]
}
{
"board_id": "3",
"hosts":
[{
"name": "pavel",
"ip": "10.3.2.3"
},
{
"name": "kenrick",
"ip": "10.3.2.4"
}]
}
Trying to get a query result like this:
{
"hosts":
[{
"name": "bob",
"ip": "10.1.2.3"
},
{
"name": "tom",
"ip": "10.1.2.4"
},
{
"name": "mickey",
"ip": "10.2.2.3"
},
{
"name": "mouse",
"ip": "10.2.2.4"
},
{
"name": "pavel",
"ip": "10.3.2.3"
},
{
"name": "kenrick",
"ip": "10.3.2.4"
}]
}
I've tried this:
db.collection.aggregate([ { $unwind: '$hosts' }, { $project : { name: 1, hosts: 1, _id: 0 }} ])
But it's not quite what I want.
You can definitely do this with aggregate. Let's assume your data is in collection named board, so please replace it with whatever your collection name is.
db.board.aggregate([
{$unwind:"$hosts"},
{$group:{_id:null, hosts:{$addToSet:"$hosts"}}},
{$project:{_id:0, hosts:1}}
]).pretty()
it will return
{
"hosts" : [
{
"name" : "kenrick",
"ip" : "10.3.2.4"
},
{
"name" : "pavel",
"ip" : "10.3.2.3"
},
{
"name" : "mouse",
"ip" : "10.2.2.4"
},
{
"name" : "mickey",
"ip" : "10.2.2.3"
},
{
"name" : "tom",
"ip" : "10.1.2.4"
},
{
"name" : "bob",
"ip" : "10.1.2.3"
}
]
}
So your basic problem here is that the arrays are contained in separate documents. So while you are correct to $unwind the array for processing, in order to bring the content into a single array you would need to $group the result across documents, and $push the content to the result array:
db.collection.aggregate([
{ "$unwind": "$hosts" },
{ "$group": {
"_id": null,
"hosts": { "$push": "$hosts" }
}}
])
So just as $unwind will "deconstruct" the array elements, the $push accumulator in $group brings "reconstructs" the array. And since there is no other key to "group" on, this brings all the elements into a single array.
Note that a null grouping key is only really practical when the resulting document would not exceed the BSON limit. Otherwise you are better off leaving the individual elements as documents in themselves.
Optionally remove the _id with an additional $project if required.
Related
I am new to mongodb, I have a requirement and would like to know how to generate custom resultset using Mongo aggregate operator. Any help would be appreciated.
Need to group the collection by "company" and "status" and would need to produce resultset given below.
Collection
[
{
"company": "google",
"status": "active",
"offer": {
"job": "developer",
"salary": 10000.00
},
},
{
"company": "google",
"status": "active",
"offer": {
"job": "designer",
"salary": 500000.00
},
},
{
"company": "amazon",
"status": "inactive",
"offer": {
"job": "designer",
"salary": 500000.00
},
}
]
Expected Result-Set
[
{
"company" : "google",
"report" : [{
"status" : "active",
"totalSalary" : 60000
},
{
"status" : "inactive",
"totalSalary" : 0
}]
},
{
"company" : "amazon",
"report" : [{
"status" : "active",
"totalSalary" : 0
},
{
"status" : "inactive",
"totalSalary" : 500000.00
}]
}
]
You should 100% check the official documentation on aggregates, it's a bit complicated at first but once you get the hang of it they're great. I also recommend you https://mongoplayground.net/, it's a great site for doing this kind of tests.
What you're looking for is something like this
db.collection.aggregate([
{
$group: {
_id: {
company: "$company"
},
report: {
$addToSet: "$offer"
}
}
}
])
You can test it here. You also probably want to rename the resulting _id field that's mandatory in a group aggregate. You can find how to do that here
I have a list of documents like this
[{
"_id": "5dbc95f921d7625303fe2369",
"name": "John",
"itemsPurchased": [{
"offer": "o1",
"items": ["p1"]
},{
"offer": "o1",
"items": ["p1"]
},
{
"offer": "o1",
"items": ["p2"]
},
{
"offer": "o2",
"items": ["p1"]
}, {
"offer": "o7",
"items": ["p1"]
}
]
},
{
"_id": "zbc95f921d7625303fe2363",
"name": "Doe",
"itemsPurchased": [{
"offer": "o1",
"items": ["p11"]
},{
"offer": "o1",
"items": ["p11"]
},
{
"offer": "o2",
"items": ["p13"]
},
{
"offer": "o1",
"items": ["p22"]
},
{
"offer": "o2",
"items": ["p11"]
}, {
"offer": "o3",
"items": ["p11"]
}
]
}
]
And i am trying to compute unique offers on unique products by each customer, expecting the resultant to be like:
[
{
"_id": "5dbc95f921d7625303fe2369",
"name": "John",
"offersAndProducts": {
"o1":2,
"o2":2,
"o3":1
},
{
"_id": "zbc95f921d7625303fe2363",
"name": "Doe",
"offersAndProducts": {
"o1":2,
"o2":1,
"o7":1
}
]
I want to apply aggregations per document, After performing $unwind on itemsPurchased, applied $group on items and then on offer to eliminate the duplication:
{
"$group" : {
"_id" : {
"item" : {
"$arrayElemAt" : [
"$itemsPurchased.item",
0.0
]
},
"count" : {
"$sum" : 1.0
},
"offer" : "$itemsPurchased.offer"
}
}
}
then,
{
"$group" : {
"_id" : "$_id.offer",
"count" : {
"$sum" : 1.0
}
}
}
this gives the array of products and offers for all documents:
[
{o1:4,o2:3,o3:1,o7:1}
]
But i need it at document level.
tried $addFeild, but $unwind and $match operators gives invalid error.
Any other way of achieving this?
Generally speaking, it's an anti-pattern to $unwind an array and then to $group on the original _id since most operations can be done on the array directly, in a single stage. Here is what such a stage would look like:
{$addFields:{
offers:{$arrayToObject:{
$map:{
input:{$setUnion:"$itemsPurchased.offer"},
as:"o",
in:[
"$$o",
{$size:{$setUnion:{$let:{
vars:{items:{$filter:{
input:"$itemsPurchased",
cond:{$eq:["$$this.offer","$$o"]}
}}},
in:{$reduce:{
input:"$$items",
initialValue:[],
in:{$concatArrays:["$$value","$$items.items"]}
}}
}}}
}]
}
}}
}}
What this does is create an array where each element is a two element array (which is a syntax that $arrayToObject can convert to an object where first element is key name and second is value) and the input is a unique set of offers and for each we accumulate an array of products, get rid of duplicates (with $setUnion) and then get the size of the result. What this produces on your input is this:
"offers" : {
"o1" : 2,
"o2" : 2,
"o3" : 1
}
You need to run $unwind and $group twice. To count only unique items you can use $addToSet. To build your keys dynamically you need to use $arrayToObject:
db.collection.aggregate([
{
$unwind: "$itemsPurchased"
},
{
$unwind: "$itemsPurchased.items"
},
{
$group: {
_id: {
_id: "$_id",
offer: "$itemsPurchased.offer"
},
name: { $first: "$name" },
items: { $addToSet: "$itemsPurchased.items" }
}
},
{
$group: {
_id: "$_id._id",
name: { $first: "$name" },
offersAndProducts: { $push: { k: "$_id.offer", v: { $size: "$items" } } }
}
},
{
$project: {
_id: 1,
name: 1,
offersAndProducts: { $arrayToObject: "$offersAndProducts" }
}
}
])
Mongo Playground
This question already has answers here:
Include all existing fields and add new fields to document
(6 answers)
Closed 4 years ago.
I am using mongodb and I have a document which return a json like that:
{
"_id": "5ad9a24be78f9d33888d2567",
"tag": [],
"active": 1,
"code": "_CAROT",
"name": [
{
"lang": "uk",
"translation": "carot"
},
{
"lang": "fr",
"translation": "carotte"
}
],
"season": [],
"category": [],
"createdAt": "2018-04-23T07:59:51.261Z",
"updatedAt": "2018-04-23T07:59:51.261Z",
"__v": 0
}
I want to add a filter on the lang, to get only one translation. So I am using aggregate and $filter to do that. This is what I do :
db.products.aggregate(
[ {$match: {'name.lang': "fr"}},
{$project: { name: {$filter: {
input: '$name',
as: 'item',
cond: {$eq: ['$$item.lang', "fr"]}
}}
}}
])
And I get :
{ "_id" : ObjectId("5ad9a24be78f9d33888d2567"), "name" : [ { "lang" : "fr", "translation" : "carotte" } ] }
{ "_id" : ObjectId("5add96fedf3aac3d049196ca"), "name" : [ { "lang" : "fr", "translation" : "tomate" } ] }
However I would like to get the following result :
{
"_id": "5ad9a24be78f9d33888d2567",
"tag": [],
"active": 1,
"code": "_CAROT",
"name": [
{
"lang": "fr",
"translation": "carotte"
}
],
"season": [],
"category": [],
"createdAt": "2018-04-23T07:59:51.261Z",
"updatedAt": "2018-04-23T07:59:51.261Z",
"__v": 0
}
Basically the default result with just the "fr" result on the "name" field.
Is there a way to do it using mongoDB ?
Thanks a lot
You can use the aggregation $unwind to "unwrap" by translation. I means that for each translation value of each document it will create a new document with only this translation value (at the same level, not in a sub document). Then you will have to then filter with $match to only keep the "fr" translations.
Note you will have to copy each field name to have it in the final result.
Example:
db.products.aggregate([
{ $unwind: '$name' }, // scalar product by name
{ $match: { 'name.lang': 'fr' } }, // only keep documents in french
{ $project: { _id: 0, code: 1, 'name.translation': 1 } } // return code + translation (in french)
])
I have a below structure maintained in a sample collection.
{
"_id": "1",
"name": "Stock1",
"description": "Test Stock",
"lines": [
{
"lineNumber": "1",
"priceInfo": {
"buyprice": 10,
"sellprice": 15
},
"item": {
"id": "BAT10001",
"name": "CricketBat",
"description": "Cricket bat"
},
"quantity": 10
},
{
"lineNumber": "2",
"priceInfo": {
"buyprice": 10,
"sellprice": 15
},
"item": {
"id": "BAT10002",
"name": "CricketBall",
"description": "Cricket ball"
},
"quantity": 10
},
{
"lineNumber": "3",
"priceInfo": {
"buyprice": 10,
"sellprice": 15
},
"item": {
"id": "BAT10003",
"name": "CricketStumps",
"description": "Cricket stumps"
},
"quantity": 10
}
]
}
I have a scenario where i will be given lineNumber and item.id, i need to filter the above collection based on lineNumber and item.id and i need to project only selected fields.
Expected output below:
{
"_id": "1",
"lines": [
{
"lineNumber": "1",
"item": {
"id": "BAT10001",
"name": "CricketBat",
"description": "Cricket bat"
},
"quantity": 10
}
]
}
Note: I may not get lineNumber all the times, if lineNumber is null then i should filter for item.id alone and get the above mentioned output.The main purpose is to reduce the number of fields in the output, as the collection is expected to hold huge number of fields.
I tried the below query,
db.sample.aggregate([
{ "$match" : { "_id" : "1"} ,
{ "$project" : { "lines" : { "$filter" : { "input" : "$lines" , "as" : "line" , "cond" :
{ "$and" : [ { "$eq" : [ "$$line.lineNumber" , "3"]} , { "$eq" : [ "$$line.item.id" , "BAT10001"]}]}}}}}
])
But i got all the fields, i'm not able to exclude or include the required fields.
I tried the below query and it worked for me,
db.Collection.aggregate([
{ $match: { _id: '1' } },
{
$project: {
lines: {
$map: {
input: {
$filter: {
input: '$lines',
as: 'line',
cond: {
$and: [
{ $eq: ['$$line.lineNumber', '3'] },
{ $eq: ['$$line.item.id', 'BAT10001'] },
],
},
},
},
as: 'line',
in: {
lineNumber: '$$line.lineNumber',
item: '$$line.item',
quantity: '$$line.quantity',
},
},
},
},
},
])
You can achieve it with $unwind and $group aggregation stages:
db.collection.aggregate([
{$match: {"_id": "1"}},
{$unwind: "$lines"},
{$match: {
$or: [
{"lines.lineNumber":{$exists: true, $eq: "1"}},
{"item.id": "BAT10001"}
]
}},
{$group: {
_id: "$_id",
lines: { $push: {
"lineNumber": "$lines.lineNumber",
"item": "$lines.item",
"quantity": "$lines.quantity"
}}
}}
])
$match - sets the criterias for the documents filter. The first stage is takes document with _id = "1", the second takes only documents which have lines.lineNumber equal to "1" or item.id equal to "BAT10001".
$unwind - splits the lines array into seperated documents.
$group - merges the documents by the _id element and puts the generated object with lineNumber, item and quantity elements into the lines array.
This question has previously been marked as a duplicate of this question I can with certainty confirm that it is not.
This is not a duplicate of the linked question because the elements in question are not an array but embedded in individual objects of an array as fields. I am fully aware of how the query in the linked question should work, however that scenario is different from mine.
I have a question regarding the $lookup query of MongoDb. My data structure looks as follows:
My "Event" collection contains this single document:
{
"_id": ObjectId("mongodbobjectid..."),
"name": "Some Event",
"attendees": [
{
"type": 1,
"status": 2,
"contact": ObjectId("mongodbobjectidHEX1")
},
{
"type": 7,
"status": 4,
"contact": ObjectId("mongodbobjectidHEX2")
}
]
}
My "Contact" collection contains these documents:
{
"_id": ObjectId("mongodbobjectidHEX1"),
"name": "John Doe",
"age": 35
},
{
"_id": ObjectId("mongodbobjectidHEX2"),
"name": "Peter Pan",
"age": 60
}
What I want to do is perform an aggregate query with the $lookup operator on the "Event" collection and get the following result with full "contact" data:
{
"_id": ObjectId("mongodbobjectid..."),
"name": "Some Event",
"attendees": [
{
"type": 1,
"status": 2,
"contact": {
"_id": ObjectId("mongodbobjectidHEX1"),
"name": "John Doe",
"age": 35
}
},
{
"type": 7,
"status": 4,
"contact": {
"_id": ObjectId("mongodbobjectidHEX2"),
"name": "Peter Pan",
"age": 60
}
}
]
}
I have done the same with single elements of "Contact" referenced in another document but never when embedded in an array. I am unsure of which pipeline arguments to pass to get the above shown result?
I also want to add a $match query to the pipeline to filter the data, but that is not really part of my question.
Try this one
db.getCollection('Event').aggregate([{ "$unwind": "$attendees" },
{ "$lookup" : { "from" : "Contact", "localField" : "attendees.contact", "foreignField": "_id", "as" : "contactlist" } },
{ "$unwind": "$contactlist" },
{ "$project" :{
"attendees.type" : 1,
"attendees.status" : 1,
"attendees.contact" : "$contactlist",
"name": 1, "_id": 1
}
},
{
"$group" : {
_id : "$_id" ,
"name" : { $first : "$name" },
"attendees" : { $push : "$attendees" }
}
}
])