Consider I have the following data below:
{
"id":123,
"name":"apple",
"codes":["ABC", "DEF", "EFG"]
}
{
"id":234,
"name":"pineapple",
"codes":["DEF"]
}
{
"id":345,
"name":"banana",
"codes":["HIJ","KLM"]
}
If I didn't want to search by a specific code, is there a way to find all fruits in my mongodb collection which shares the same code?
db.collection.aggregate([
{ $unwind: '$codes' },
{ $group: { _id: '$codes', count: {$sum:1}, fruits: {$push: '$name'}}},
{ $match: {'count': {$gt:1}}},
{ $group:{_id:null, total:{$sum:1}, data:{$push:{fruits: '$fruits', code:'$_id'}}}}
])
result:
{ "_id" : null, "total" : 1, "data" : [ { "fruits" : [ "apple", "pineapple" ], "code" : "DEF" } ] }
Related
Inward collections
{"ord" : 1,
"products" : [
{
"name" : "apple",
"qty" : "10",
"batch" : "jun-2021"
},
{
"name" : "banana",
"qty" : 20,
"batch" : "jan-2021"
}
]
}
outward collections
{
"_id" : ObjectId("5edde5487957d9efea972a74"),
"inv" : 1,
"products" : [
{
"name" : "apple",
"qty" : 13,
"batch" : "jun-2021"
}
]
}
Now, I would like to perform actual stock quantity check for particular product and batch (grouping together) both the collections
You may try this way:
We join them with inward.ord = outward.inv condition.
Flatten products field.
Group by product's name and batch to sum qty value.
db.inward.aggregate([
{
$lookup: {
from: "outward",
let: {
ord: "$ord",
products: "$products"
},
pipeline: [
{
$match: {
$expr: {
$eq: [ "$$ord", "$inv" ]
}
}
},
{
$project: {
products: {
$concatArrays: [
"$$products",
"$products"
]
}
}
},
{
$unwind: "$products"
},
{
$replaceWith: "$products"
}
],
as: "products"
}
},
{
$unwind: "$products"
},
{
$group: {
_id: {
batch: "$products.batch",
name: "$products.name"
},
qty: {
$sum: "$products.qty"
}
}
}
])
MongoPlayground
Note: You need to have MongoDB v4.2
Below are the sample collection.
col1:
"_id" : ObjectId("5ec293782bc00b43b463b67c")
"status" : ["running"],
"name" : "name1 ",
"dcode" : "dc001",
"address" : "address1",
"city" : "city1"
col2:
"_id" : ObjectId("5ec296182bc00b43b463b68f"),
"scode" : ObjectId("5ec2933df6079743c0a2a1f8"),
"ycode" : ObjectId("5ec293782bc00b43b463b67c"),
"city" : "city1",
"lockedDate" : ISODate("2020-05-20T00:00:00Z"),
"_id" : ObjectId("5ec296182bc00b43b463688b"),
"scode" : ObjectId("5ec2933df6079743c0a2a1ff"),
"ycode" : ObjectId("5ec293782bc00b43b463b67c"),
"city" : "city1",
"lockedDate" : ISODate("2020-05-20T00:00:00Z"),
"_id" : ObjectId("5ec296182bc00b43b44fc6cb"),
"scode" :null,
"ycode" : ObjectId("5ec293782bc00b43b463b67c"),
"city" : "city1",
"lockedDate" : ISODate("2020-05-20T00:00:00Z"),
problemStatement:
I want to display name from col1 & count of documents from col2 according to ycode where scode is != null
Tried attempt:
db.col1.aggregate([
{'$match':{
city:'city1'
}
},
{
$lookup:
{
from: "col2",
let: {
ycode: "$_id",city:'$city'
},
pipeline: [
{
$match: {
scode:{'$ne':null},
lockedDate:ISODate("2020-05-20T00:00:00Z"),
$expr: {
$and: [
{
$eq: [
"$ycode",
"$$ycode"
]
},
{
$eq: [
"$city",
"$$city"
]
}
]
},
},
},
], as: "col2"
}
},
{'$unwind':'$col2'},
{'$count':'ycode'},
{
$project: {
name: 1,
status: 1,
}
},
])
now problem with this query is it either displays the count or project the name & status i.e if i run this query in the current format it gives {} if I remove {'$count':'ycode'} then it project the values but doesn't give the count and if I remove $project then i do get the count {ycode:2} but then project doesn't work but I want to achieve both in the result. Any suggestions
ORM: mongoose v>5, mongodb v 4.0
You can try below query :
db.col1.aggregate([
{ "$match": { city: "city1" } },
{
$lookup: {
from: "col2",
let: { id: "$_id", city: "$city" }, /** Create local variables from fields of `col1` but not from `col2` */
pipeline: [
{
$match: { scode: { "$ne": null }, lockedDate: ISODate("2020-05-20T00:00:00Z"),
$expr: { $and: [ { $eq: [ "$ycode", "$$id" ] }, { $eq: [ "$city", "$$city" ] } ] }
}
},
{ $project: { _id: 1 } } // Optional, But as we just need count but not the entire doc, holding just `_id` helps in reduce size of doc
],
as: "col2" // will be an array either empty (If no match found) or array of objects
}
},
{
$project: { _id: 0, name: 1, countOfCol2: { $size: "$col2" } }
}
])
Test : mongoplayground
I have a dataset in mongodb collection named visitorsSession like
{ip : 192.2.1.1,country : 'US', type : 'Visitors',date : '2019-12-15T00:00:00.359Z'},
{ip : 192.3.1.8,country : 'UK', type : 'Visitors',date : '2019-12-15T00:00:00.359Z'},
{ip : 192.5.1.4,country : 'UK', type : 'Visitors',date : '2019-12-15T00:00:00.359Z'},
{ip : 192.8.1.7,country : 'US', type : 'Visitors',date : '2019-12-15T00:00:00.359Z'},
{ip : 192.1.1.3,country : 'US', type : 'Visitors',date : '2019-12-15T00:00:00.359Z'}
I am using this mongodb aggregation
[{$match: {
nsp : "/hrm.sbtjapan.com",
creationDate : {
$gte: "2019-12-15T00:00:00.359Z",
$lte: "2019-12-20T23:00:00.359Z"
},
type : "Visitors"
}}, {$group: {
_id : "$country",
totalSessions : {
$sum: 1
}
}}, {$project: {
_id : 0,
country : "$_id",
totalSessions : 1
}}, {$sort: {
country: -1
}}]
using above aggregation i am getting results like this
[{country : 'US',totalSessions : 3},{country : 'UK',totalSessions : 2}]
But i also total visitors also along with result like totalVisitors : 5
How can i do this in mongodb aggregation ?
You can use $facet aggregation stage to calculate total visitors as well as visitors by country in a single pass:
db.visitorsSession.aggregate( [
{
$match: {
nsp : "/hrm.sbtjapan.com",
creationDate : {
$gte: "2019-12-15T00:00:00.359Z",
$lte: "2019-12-20T23:00:00.359Z"
},
type : "Visitors"
}
},
{
$facet: {
totalVisitors: [
{
$count: "count"
}
],
countrySessions: [
{
$group: {
_id : "$country",
sessions : { $sum: 1 }
}
},
{
$project: {
country: "$_id",
_id: 0,
sessions: 1
}
}
],
}
},
{
$addFields: {
totalVisitors: { $arrayElemAt: [ "$totalVisitors.count" , 0 ] },
}
}
] )
The output:
{
"totalVisitors" : 5,
"countrySessions" : [
{
"sessions" : 2,
"country" : "UK"
},
{
"sessions" : 3,
"country" : "US"
}
]
}
You could be better off with two queries to do this.
To save the two db round trips following aggregation can be used which IMO is kinda verbose (and might be little expensive if documents are very large) to just count the documents.
Idea: Is to have a $group at the top to count documents and preserve the original documents using $push and $$ROOT. And then before other matches/filter ops $unwind the created array of original docs.
db.collection.aggregate([
{
$group: {
_id: null,
docsCount: {
$sum: 1
},
originals: {
$push: "$$ROOT"
}
}
},
{
$unwind: "$originals"
},
{ $match: "..." }, //and other stages on `originals` which contains the source documents
{
$group: {
_id: "$originals.country",
totalSessions: {
$sum: 1
},
totalVisitors: {
$first: "$docsCount"
}
}
}
]);
Sample O/P: Playground Link
[
{
"_id": "UK",
"totalSessions": 2,
"totalVisitors": 5
},
{
"_id": "US",
"totalSessions": 3,
"totalVisitors": 5
}
]
given below is my data in mongo db.I want to fetch all the unique ids from the field articles ,which is nested under the jnlc_subjects index .The result should contain only the articles array with distinct object Ids.
Mongo Data
{
"_id" : ObjectId("5c9216f1a21a4a31e0c7fa56"),
"jnlc_journal_category" : "Biology",
"jnlc_subjects" : [
{
"subject" : "Conservation Biology",
"views" : "123",
"articles" : [
ObjectId("5c4e93d0135edb6812200d5f"),
ObjectId("5c4e9365135edb6a12200d60"),
ObjectId("5c4e93a8135edb6912200d61")
]
},
{
"subject" : "Micro Biology",
"views" : "20",
"articles" : [
ObjectId("5c4e9365135edb6a12200d60"),
ObjectId("5c4e93d0135edb6812200d5f"),
ObjectId("5c76323fbaaccf5e0bae7600"),
ObjectId("5ca33ce19d677bf780fc4995")
]
},
{
"subject" : "Marine Biology",
"views" : "8",
"articles" : [
ObjectId("5c4e93d0135edb6812200d5f")
]
}
]
}
Required result
I want to get output in following format
articles : [
ObjectId("5c4e9365135edb6a12200d60"),
ObjectId("5c4e93a8135edb6912200d61"),
ObjectId("5c76323fbaaccf5e0bae7600"),
ObjectId("5ca33ce19d677bf780fc4995"),
ObjectId("5c4e93d0135edb6812200d5f")
]
Try as below:
db.collection.aggregate([
{
$unwind: "$jnlc_subjects"
},
{
$unwind: "$jnlc_subjects.articles"
},
{ $group: {_id: null, uniqueValues: { $addToSet: "$jnlc_subjects.articles"}} }
])
Result:
{
"_id" : null,
"uniqueValues" : [
ObjectId("5ca33ce19d677bf780fc4995"),
ObjectId("5c4e9365135edb6a12200d60"),
ObjectId("5c4e93a8135edb6912200d61"),
ObjectId("5c4e93d0135edb6812200d5f"),
ObjectId("5c76323fbaaccf5e0bae7600")
]
}
Try with this
db.collection.aggregate([
{
$unwind:{
path:"$jnlc_subjects",
preserveNullAndEmptyArrays:true
}
},
{
$unwind:{
path:"$jnlc_subjects.articles",
preserveNullAndEmptyArrays:true
}
},
{
$group:{
_id:"$_id",
articles:{
$addToSet:"$jnlc_subjects.articles"
}
}
}
])
If you don't want to $group with _id ypu can use null instead of $_id
According to description as mentioned into above question,as a solution to it please try executing following aggregate operation.
db.collection.aggregate(
// Pipeline
[
// Stage 1
{
$match: {
"_id": ObjectId("5c9216f1a21a4a31e0c7fa56")
}
},
// Stage 2
{
$unwind: {
path: "$jnlc_subjects",
}
},
// Stage 3
{
$unwind: {
path: "$jnlc_subjects.articles"
}
},
// Stage 4
{
$group: {
_id: null,
articles: {
$addToSet: '$jnlc_subjects.articles'
}
}
},
// Stage 5
{
$project: {
articles: 1,
_id: 0
}
},
]
);
I am using MongoDB version v3.4. I have a documents collection and sample datas are like this:
{
"mlVoters" : [
{"email" : "a#b.com", "isApproved" : false}
],
"egVoters" : [
{"email" : "a#b.com", "isApproved" : false},
{"email" : "c#d.com", "isApproved" : true}
]
},{
"mlVoters" : [
{"email" : "a#b.com", "isApproved" : false},
{"email" : "e#f.com", "isApproved" : true}
],
"egVoters" : [
{"email" : "e#f.com", "isApproved" : true}
]
}
Now if i want the count of distinct email addresses for mlVoters:
db.documents.aggregate([
{$project: { mlVoters: 1 } },
{$unwind: "$mlVoters" },
{$group: { _id: "$mlVoters.email", mlCount: { $sum: 1 } }},
{$project: { _id: 0, email: "$_id", mlCount: 1 } },
{$sort: { mlCount: -1 } }
])
Result of the query is:
{"mlCount" : 2.0,"email" : "a#b.com"}
{"mlCount" : 1.0,"email" : "e#f.com"}
And if i want the count of distinct email addresses for egVoters i do the same for egVoters field. And the result of that query would be:
{"egCount" : 1.0,"email" : "a#b.com"}
{"egCount" : 1.0,"email" : "c#d.com"}
{"egCount" : 1.0,"email" : "e#f.com"}
So, I want to combine these two aggregation and get the result as following (sorted by totalCount):
{"email" : "a#b.com", "mlCount" : 2, "egCount" : 1, "totalCount":3}
{"email" : "e#f.com", "mlCount" : 1, "egCount" : 1, "totalCount":2}
{"email" : "c#d.com", "mlCount" : 0, "egCount" : 1, "totalCount":1}
How can I do this? How should the query be like? Thanks.
First you add a field voteType in each vote. This field indicates its type. Having this field, you don't need to keep the votes in two separate arrays mlVoters and egVoters; you can instead concatenate those arrays into a single array per document, and unwind afterwards.
At this point you have one document per vote, with a field that indicates which type it is. Now you simply need to group by email and, in the group stage, perform two conditional sums to count how many votes of each type there are for every email.
Finally you add a field totalCount as the sum of the other two counts.
db.documents.aggregate([
{
$addFields: {
mlVoters: {
$ifNull: [ "$mlVoters", []]
},
egVoters: {
$ifNull: [ "$egVoters", []]
}
}
},
{
$addFields: {
"mlVoters.voteType": "ml",
"egVoters.voteType": "eg"
}
},
{
$project: {
voters: { $concatArrays: ["$mlVoters", "$egVoters"] }
}
},
{
$unwind: "$voters"
},
{
$project: {
email: "$voters.email",
voteType: "$voters.voteType"
}
},
{
$group: {
_id: "$email",
mlCount: {
$sum: {
$cond: {
"if": { $eq: ["$voteType", "ml"] },
"then": 1,
"else": 0
}
}
},
egCount: {
$sum: {
$cond: {
"if": { $eq: ["$voteType", "eg"] },
"then": 1,
"else": 0
}
}
}
}
},
{
$addFields: {
totalCount: {
$sum: ["$mlCount", "$egCount"]
}
}
}
])