I want to groupby the collection and want to pull the comma-separated values.
In the below example, I want to group by "type" and want to pull $sum of "total" and all possible unique values of "value" in a field that should be comma seperated.
collection:
[
{
"type": "1",
"value": "value1",
"total": 10
},
{
"type": "1",
"value": "value3",
"total": 20
},
{
"type": "1",
"value": "value3",
"total": 30
},
{
"type": "2",
"value": "value1",
"total": 10
},
{
"type": "2",
"value": "value2",
"total": 20
}
]
The output that I am expecting:
[
{
"type": "1",
"value": "value1,value3",
"total": 60
},
{
"type": "2",
"value": "value1,value2",
"total": 30
}
]
Please help to provide the approach or code.
This can be achieved just with $group and $project aggregation methods:
https://mongoplayground.net/p/230nt_AMFIm
db.collection.aggregate([
{
$group: {
_id: "$type",
value: {
$addToSet: "$value"
},
total: {
$sum: "$total"
}
},
},
{
$project: {
_id: 0,
type: "$_id",
value: "$value",
total: "$total"
}
},
])
Related
I have a collection where from the backend user can input multiple same name bikes but with different registration number but in front-End I want them to be grouped by matching the same name but as user updates separately display image changes but I want only one display image as it is 1 vehicle
provided there is a node created I will implement it we can sort it by the latest and take the price and image of it
Activa -2 Count
KTM -1 Count
but there is a catch.
Activa 2 bikes but I want only count 2 and the price as it is the same in an array I want only 1 and the same applies to displayimage here display image file path is different but I want the latest one only Sharing data below
Data:
[
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikenumber": "KA05HM2034",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikenumber": "KA05HM2033",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62d7ff7e70b9ab38c6ab0cb1",
"bikename": "Activa",
"bikenumber": "KA05HM2223",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"afterfreekmprice": 22,
"descreption": "Activa",
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658322798414.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658322798414.jpg",
"idx": 0
}
}
]
Expected:
[
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62dba8418ef8f51f454ed757",
"bikename": "Activa",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-v_activa-i-deluxe-1658562557459.jpg",
"file_path": "https://www.example.com/images/upload/bike-v_activa-i-deluxe-1658562557459.jpg",
"idx": 0
}
},
"count": 2
},
{
"_id":{
"price": [
{
"Description": "Hourly",
"Price": "1"
},
{
"Description": "Daily",
"Price": "11"
},
{
"Description": "Monthly",
"Price": "111"
}
],
"_id": "62e69ee3edfe4d0f3cb4994a",
"bikename": "KTM",
"bikebrand": {
"id": 1,
"label": "Honda"
},
"freekm": 234,
"displayimage": {
"file": "bike-2020-honda-city-exterior-8-1659281111883.jpg",
"file_path": "https://www.example.com/images/upload/bike-2020-honda-city-exterior-8-1659281111883.jpg",
"idx": 1
}
}
"count": 1
}
]
You can use the aggregation pipeline,
$sort by _id in descending order
$group by bikename and get the first root document that is latest one in root and count total documents in count
$project to show required documents
db.collection.aggregate([
{ $sort: { _id: -1 } },
{
$group: {
_id: "$bikename",
root: { $first: "$$ROOT" },
count: { $sum: 1 }
}
},
{
$project: {
_id: "$root",
count: 1
}
}
])
Playground
You can use $group for this:
db.collection.aggregate([
{$group: {
_id: "$bikename",
count: {$sum: 1},
data: {$first: "$$ROOT"}
}
},
{$set: {"data.count": "$count"}},
{$replaceRoot: {newRoot: "$data"}}
])
See how it works on the playground example
Example documents
{
"user_id": "1",
"key": "key1",
"percent": 95
},
{
"user_id": "1",
"key": "key1",
"percent": 50
},
{
"user_id": "1",
"key": "key2",
"percent": 50
},
{
"user_id": "1",
"key": "key3",
"percent": 50
},
{
"user_id": "1",
"key": "key3",
"percent": 70
}
I want to extract document that percent less than 95.
When there are multiple documents with a specific key, that one is greater than 95 and one is less than 95, the document with that key must not be displayed. (In the above example, key1 corresponds. Because one is greater than 95 and the other is less than 95.)
And If all of multiple documents have a percentage below 95, the document with the highest percentage must be extracted.
Current query that I used
db.test.aggregate([
{
"$match": {
"percent": {
"$lte": 95
},
"user_id": {
"$eq": user_id,
},
}
},
{
"$group": {
"_id": "$key",
"max": {"$max": "$percent"}
}
},
{
"$project": {
"_id": 1,
"percent": "$max"
}
},
])
Result
{"key": "key1", "percent": 50}
{"key": "key2", "percent": 50}
{"key": "key3", "percent": 70}
According to the conditions described above, key1 should be excluded from the results because there are documents with a percentage of 95 or higher.
Expected result
{"key": "key2", "percent": 50}
{"key": "key3", "percent": 70}
Is this a problem that can be solved with MongoDB's query?
Thanks!
db.test.aggregate([
{
"$match": {
"user_id": {
"$eq": user_id,
},
}
},
{
"$group": {
"_id": "$key",
"max": {"$max": "$percent"}
}
},
{
"$match": {
"max": {
"$lte": 95
}
}
},
{
"$project": {
"_id": 1,
"percent": "$max"
}
},
])
After changing order of match query, it works perfectly.
I would like to independently group the results of an or clause, including overlap. The data set is rather large so running 2 queries sequentially will result in an undesirable wait time. I am hoping I can somehow project which clause returned the corresponding data. Given this data set:
[
{
"_id": 1,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("2"),
"date": ISODate("2014-03-01T08:00:00Z")
},
{
"_id": 2,
"item": "jkl",
"name": "Toby",
"price": NumberDecimal("20"),
"quantity": NumberInt("1"),
"date": ISODate("2014-03-01T09:00:00Z")
},
{
"_id": 3,
"item": "xyz",
"name": "Keith",
"price": NumberDecimal("5"),
"quantity": NumberInt("10"),
"date": ISODate("2014-03-15T09:00:00Z")
},
{
"_id": 4,
"item": "abc",
"name": "Dwight",
"price": NumberDecimal("5"),
"quantity": NumberInt("20"),
"date": ISODate("2014-04-04T11:21:39.736Z")
},
{
"_id": 5,
"item": "abc",
"name": "Ryan",
"price": NumberDecimal("10"),
"quantity": NumberInt("10"),
"date": ISODate("2014-04-04T21:23:13.331Z")
},
{
"_id": 6,
"item": "def",
"name": "Jim",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("5"),
"date": ISODate("2015-06-04T05:08:13Z")
},
{
"_id": 7,
"item": "abc",
"name": "Keith",
"price": NumberDecimal("7.5"),
"quantity": NumberInt("10"),
"date": ISODate("2015-09-10T08:43:00Z")
},
{
"_id": 8,
"item": "abc",
"name": "Michael",
"price": NumberDecimal("10"),
"quantity": NumberInt("5"),
"date": ISODate("2016-02-06T20:20:13Z")
},
]
I would like to receive this result:
[{
"_id": {
"name": "Keith"
},
"count": 2
},
{
"_id": {
"item": "abc",
},
"count": 5
}]
Here is what I have tried so far:
db.collection.aggregate([
{
$match: {
$or: [
{
item: "abc"
},
{
name: "Keith"
}
]
}
},
{
$group: {
_id: {
item: "$item",
name: "$name"
},
count: {
$sum: 1
}
}
}
])
You can use $facet to get multiple aggregation pipelines into the same stage in this way:
Using $facet there are two "outputs" one group by name and other by item.
In each one there are multiple stages:
First $match to process only documents you want.
Then $group with _id name or item, and $count to get the total.
db.collection.aggregate([
{
"$facet": {
"groupByName": [
{
"$match": {"name": "Keith"}
},
{
"$group": {"_id": "$name","count": {"$sum": 1}}
}
],
"groupByItem": [
{
"$match": {"item": "abc"}
},
{
"$group": {"_id": "$item","count": {"$sum": 1}}
}
]
}
}
])
Example here
The output is:
{
"groupByItem": [
{
"_id": "abc",
"count": 5
}
],
"groupByName": [
{
"_id": "Keith",
"count": 2
}
]
}
Here it is:
mongos> db.n.aggregate([ { $facet:{ names:[ {$match:{name:"Keith"}} , {$group:{_id:{name:"$name"}, count:{$sum:1}}} ] , items:[ {$match:{item:"abc"}},{ $group:{_id:{item:"$item"}, count:{$sum:1}} } ] } } , {$project:{ "namesANDitems":{$concatArrays:[ "$names","$items" ]} }} ,{$unwind:"$namesANDitems"} ,{$replaceRoot:{newRoot:"$namesANDitems"} } ]).pretty()
{ "_id" : { "name" : "Keith" }, "count" : 2 }
{ "_id" : { "item" : "abc" }, "count" : 5 }
mongos>
explained:
You create two pipes via $facet
Match in every facet pipe what you need to group pipe1=names , pipe2=items
Join the arrays from the two pipes in single array named "namesANDitems"
Convert the array to object with $unwind
Remove the temporary object name namesANDitems so you have only the two objects as requested
I would like to get the unique elements of all arrays in a collection. Consider the following collection
[
{
"collection": "collection",
"myArray": [
{
"name": "ABC",
"code": "AB"
},
{
"name": "DEF",
"code": "DE"
}
]
},
{
"collection": "collection",
"myArray": [
{
"name": "GHI",
"code": "GH"
},
{
"name": "DEF",
"code": "DE"
}
]
}
]
I can achieve this by using $unwind and $group like this:
db.collection.aggregate([
{
$unwind: "$myArray"
},
{
$group: {
_id: null,
data: {
$addToSet: "$myArray"
}
}
}
])
And get the output:
[
{
"_id": null,
"data": [
{
"code": "GH",
"name": "GHI"
},
{
"code": "DE",
"name": "DEF"
},
{
"code": "AB",
"name": "ABC"
}
]
}
]
However, the array "myArray" will have a lot of elements (about 6) and the number of documents passed into this stage of the pipeline will be about 600. So unwinding the array would give me a total of 3600 documents being processed. I would like to know if there's a way for me to achieve the same result without unwinding
You can use below aggregation
db.collection.aggregate([
{ "$group": {
"_id": null,
"data": { "$push": "$myArray" }
}},
{ "$project": {
"data": {
"$reduce": {
"input": "$data",
"initialValue": [],
"in": { "$setUnion": ["$$this", "$$value"] }
}
}
}}
])
Output
[
{
"_id": null,
"data": [
{
"code": "AB",
"name": "ABC"
},
{
"code": "DE",
"name": "DEF"
},
{
"code": "GH",
"name": "GHI"
}
]
}
]
I've got collection that looks like:
[{
"org": "A",
"type": "simple",
"payFor": 3,
"price": 100
},
{
"org": "A",
"type": "custom",
"payFor": 2,
"price": 115
},
{
"org": "B",
"type": "simple",
"payFor": 1,
"price": 110
},
{
"org": "B",
"type": "custom",
"payFor": 2,
"price": 200
},
{
"org": "B",
"type": "custom",
"payFor": 4,
"price": 220
}]
And need to produce result with query to perform group by "org" where payments appears for only first "payFor" prices in "type".
I'm trying to use expression result by $slice operator in $add but this is not works.
pipeline:
[{
"$group": {
"_id": {
"org": "$org",
"type": "$type"
},
"payFor": {
"$max": "$payFor"
},
"count": {
"$sum": 1
},
"prices": {
"$push": "$price"
}
}
},
{
"$group": {
"_id": "$_id.org",
"payments": {
"$push": {
"type": "$_id.type",
"forFirst": "$payFor",
"sum": {
"$cond": [
{
"$gte": [
"$payFor",
"$count"
]
},
{
"$add": {
"$prices": {
"$slice": "$count"
}
}
},
{
"$add": "$prices"
}
]
}
}
}
}
}]
I know that it is possible to traverse unwinded prices and pick only "payFor" count of them. but result collections are more rich than in example above and this operation will produce some unecessary overheads.
Need some advice from community. Please. Thanks.