My document structure is as follow :
{
"_id" : ObjectId("621ccb5ea46a9e41768e0ba8"),
"cust_name" : "Anuj Kumar",
"product" : [
{
"prod_name" : "Robot",
"price" : 15000
},
{
"prod_name" : "Keyboard",
"price" : 65000
}
],
"order_date" : ISODate("2022-02-22T00:00:00Z"),
"status" : "processed",
"invoice" : {
"invoice_no" : 111,
"invoice_date" : ISODate("2022-02-22T00:00:00Z")
}
}
How to do the following query...
List the details of orders with a value >10000.
I want to display only those objects whose sum of prices is greater than 10000
I try this
db.order.aggregate([{$project : {sumOfPrice : {$sum : "$product.price"} }}])
Output
{ "_id" : ObjectId("621ccb5ea46a9e41768e0ba8"), "sumOfPrice" : 80000 }
{ "_id" : ObjectId("621ccba9a46a9e41768e0ba9"), "sumOfPrice" : 16500 }
{ "_id" : ObjectId("621ccbfaa46a9e41768e0baa"), "sumOfPrice" : 5000 }
I want to check this sumOfPrice is greater than 10000 or not and display those order full object.
You can just add a $match stage right after that checks for this conditions, like so:
db.collection.aggregate([
{
$addFields: {
sumOfPrice: {
$sum: "$product.price"
}
}
},
{
$match: {
sumOfPrice: {
$gt: 10000
}
}
}
])
Mongo Playground
You can also use $expr operator with the find query as:
db.order.find({
$expr: {
$gt: [ {$sum: '$product.price'}, 10000 ]
}
})
Mongo Playground
Related
So I have a collection of users. The user document is a very simple document, and looks like this:
{
username: "player101",
badges: ["score10", "score100"]
}
So how can I query to see how many times each unique value in the badges array occurs across the entire collection?
Use aggregation with $unwind and $group stages, where you can sum badges with $sum arithmetic operator
db.players.aggregate([
{
$unwind: "$badges"
},
{
$group:
{
_id: "$badges",
count: { $sum: 1 }
}
}
]);
on collection players with documents
{ "username" : "player101", "badges" : [ "score10", "score100" ] }
{ "username" : "player102", "badges" : [ "score11", "score100" ] }
{ "username" : "player103", "badges" : [ "score11", "score101" ] }
{ "username" : "player104", "badges" : [ "score12", "score100" ] }
gives you the result
{ "_id" : "score101", "count" : 1 }
{ "_id" : "score11", "count" : 2 }
{ "_id" : "score12", "count" : 1 }
{ "_id" : "score100", "count" : 3 }
{ "_id" : "score10", "count" : 1 }
I have two different collection book and music in JSON .First I give a book collection example:
{
"_id" : ObjectId("b1"),
"author" : [
"Mary",
],
"title" : "Book1",
}
{
"_id" : ObjectId("b2"),
"author" : [
"Joe",
"Tony",
"Mary"
],
"title" : "Book2",
}
{
"_id" : ObjectId("b3"),
"author" : [
"Joe",
"Mary"
],
"title" : "Book3",
}
.......
Mary writes 3 books, Joe write 2 books, Tony writes 1 book. Second I give a music collection example:
{
"_id" : ObjectId("m1"),
"author" : [
"Tony"
],
"title" : "Music1",
}
{
"_id" : ObjectId("m2"),
"author" : [
"Joe",
"Tony"
],
"title" : "Music2",
}
.......
Tony has 2 musics, Joe has 1 music, Mary has 0 music.
I hope to get the number of authors who write more books than music.
Thus, Mary(3 > 0) and Joe(2 > 1) should take into consideration, but not Tony(1 < 2). Thus the final result should be 2(Mary and Joe).
I write down following code, but don't know how to compare:
db.book.aggregate([
{ $project:{ _id:0, author:1}},
{ $unwind:"$author" },
{$group:{_id:"$author", count:{$sum:1}}}
]
)
db.music.aggregate([
{ $project:{ _id:0, author:1}},
{ $unwind:"$author" },
{$group:{_id:"$author", count:{$sum:1}}}
]
)
Is it so far right? How to do the following comparison? Thanks.
to solve that problem, we need to use $out phase and store result of both queries in intermediate collection and then use aggregated query to join them ($lookup).
db.books.aggregate([{
$project : {
_id : 0,
author : 1
}
}, {
$unwind : "$author"
}, {
$group : {
_id : "$author",
count : {
$sum : 1
}
}
}, {
$project : {
_id : 0,
author : "$_id",
count : 1
}
}, {
$out : "bookAuthors"
}
])
db.music.aggregate([{
$project : {
_id : 0,
author : 1
}
}, {
$unwind : "$author"
}, {
$group : {
_id : "$author",
count : {
$sum : 1
}
}
}, {
$project : {
_id : 0,
author : "$_id",
count : 1
}
}, {
$out : "musicAuthors"
}
])
db.bookAuthors.aggregate([{
$lookup : {
from : "musicAuthors",
localField : "author",
foreignField : "author",
as : "music"
}
}, {
$unwind : "$music"
}, {
$project : {
_id : "$author",
result : {
$gt : ["$count", "$music.count"]
},
count : 1,
}
}, {
$match : {
result : true
}
}
])
EDIT CHANGES:
used author field instead of _id
added logical statement embeded in document in $project phase
result : { $gt : ["$count", "$music.count"]
Any questions welcome!
Have a fun!
I need to get $sum and $avg of subdocuments, i would like to get $sum and $avg of Channels[0].. and other channels as well.
my data structure looks like this
{
_id : ... Location : 1,
Channels : [
{ _id: ...,
Value: 25
},
{
_id: ... ,
Value: 39
},
{
_id: ..,
Value: 12
}
]
}
In order to get the sum and average of the Channels.Value elements for each document in your collection you will need to use mongodb's Aggregation processing. Further, since Channels is an array you will need to use the $unwind operator to deconstruct the array.
Assuming that your collection is called example, here's how you could get both the document sum and average of the Channels.Values:
db.example.aggregate( [
{
"$unwind" : "$Channels"
},
{
"$group" : {
"_id" : "$_id",
"documentSum" : { "$sum" : "$Channels.Value" },
"documentAvg" : { "$avg" : "$Channels.Value" }
}
}
] )
The output from your post's data would be:
{
"_id" : SomeObjectIdValue,
"documentSum" : 76,
"documentAvg" : 25.333333333333332
}
If you have more than one document in your collection then you will see a result row for each document containing a Channels array.
Solution 1: Using two groups based this example:
previous question
db.records.aggregate(
[
{ $unwind: "$Channels" },
{ $group: {
_id: {
"loc" : "$Location",
"cId" : "$Channels.Id"
},
"value" : {$sum : "$Channels.Value" },
"average" : {$avg : "$Channels.Value"},
"maximun" : {$max : "$Channels.Value"},
"minimum" : {$min : "$Channels.Value"}
}},
{ $group: {
_id : "$_id.loc",
"ChannelsSumary" : { $push :
{ "channelId" : '$_id.cId',
"value" :'$value',
"average" : '$average',
"maximun" : '$maximun',
"minimum" : '$minimum'
}}
}
}
]
)
Solution 2:
there is property i didn't show on my original question that might of help "Channels.Id" independent from "Channels._Id"
db.records.aggregate( [
{
"$unwind" : "$Channels"
},
{
"$group" : {
"_id" : "$Channels.Id",
"documentSum" : { "$sum" : "$Channels.Value" },
"documentAvg" : { "$avg" : "$Channels.Value" }
}
}
] )
Is it possible to find in a nested array the max date and show its price then show the parent field like the actual price.
The result I want it to show like this :
{
"_id" : ObjectId("5547e45c97d8b2c816c994c8"),
"actualPrice":19500,
"lastModifDate" :ISODate("2015-05-04T22:53:50.583Z"),
"price":"16000"
}
The data :
db.adds.findOne()
{
"_id" : ObjectId("5547e45c97d8b2c816c994c8"),
"addTitle" : "Clio pack luxe",
"actualPrice" : 19500,
"fistModificationDate" : ISODate("2015-05-03T22:00:00Z"),
"addID" : "1746540",
"history" : [
{
"price" : 18000,
"modifDate" : ISODate("2015-05-04T22:01:47.272Z"),
"_id" : ObjectId("5547ec4bfeb20b0414e8e51b")
},
{
"price" : 16000,
"modifDate" : ISODate("2015-05-04T22:53:50.583Z"),
"_id" : ObjectId("5547f87e83a1dae00bc033fa")
},
{
"price" : 19000,
"modifDate" : ISODate("2015-04-04T22:53:50.583Z"),
"_id" : ObjectId("5547f87e83a1dae00bc033fe")
}
],
"__v" : 1
}
my query
db.adds.aggregate(
[
{ $match:{addID:"1746540"}},
{ $unwind:"$history"},
{ $group:{
_id:0,
lastModifDate:{$max:"$historique.modifDate"}
}
}
])
I dont know how to include other fields I used $project but I get errors
thanks for helping
You could try the following aggregation pipeline which does not need to make use of the $group operator stage as the $project operator takes care of the fields projection:
db.adds.aggregate([
{
"$match": {"addID": "1746540"}
},
{
"$unwind": "$history"
},
{
"$project": {
"actualPrice": 1,
"lastModifDate": "$history.modifDate",
"price": "$history.price"
}
},
{
"$sort": { "lastModifDate": -1 }
},
{
"$limit": 1
}
])
Output
/* 1 */
{
"result" : [
{
"_id" : ObjectId("5547e45c97d8b2c816c994c8"),
"actualPrice" : 19500,
"lastModifDate" : ISODate("2015-05-04T22:53:50.583Z"),
"price" : 16000
}
],
"ok" : 1
}
I have the following MongoDB collection db.students:
/* 0 */
{
"id" : "0000",
"name" : "John"
"subjects" : [
{
"professor" : "Smith",
"day" : "Monday"
},
{
"professor" : "Smith",
"day" : "Tuesday"
}
]
}
/* 1 */
{
"id" : "0001",
"name" : "Mike"
"subjects" : [
{
"professor" : "Smith",
"day" : "Monday"
}
]
}
I want to find the number of subjects for a given student. I have a query:
db.students.find({'id':'0000'})
that will return the student document. How do I find the count for 'subjects'? Is it doable in a simple query?
If query will return just one element :
db.students.find({'id':'0000'})[0].subjects.length;
For multiple elements in cursor :
db.students.find({'id':'0000'}).forEach(function(doc) {
print(doc.subjects.length);
})
Do not forget to check existence of subjects either in query or before check .length
You could use the aggregation framework
db.students.aggregate(
[
{ $match : {'_id': '0000'}},
{ $unwind : "$subjects" },
{ $group : { _id : null, number : { $sum : 1 } } }
]
);
The $match stage will filter based on the student's _id
The $unwind stage will deconstruct your subjects array to multiple documents
The $group stage is when the count is done. _id is null because you are doing the count for only one user and only need to count.
You will have a result like :
{ "result" : [ { "_id" : null, "number" : 187 } ], "ok" : 1 }
Just another nice and simple aggregation solution:
db.students.aggregate([
{ $match : { 'id':'0000' } },
{ $project: {
subjectsCount: { $cond: {
if: { $isArray: "$subjects" },
then: { $size: "$subjects" },
else: 0
}
}
}
}
]).then(result => {
// handle result
}).catch(err => {
throw err;
});
Thanks!