What I have is a collection of documents in MongoDB that have the structure something like this
[
{
"userid": "user1",
"addresses": [
{
"type": "abc",
"street": "xyz"
},
{
"type": "def",
"street": "www"
},
{
"type": "hhh",
"street": "mmm"
},
]
},
{
"userid": "user2",
"addresses": [
{
"type": "abc",
"street": "ccc"
},
{
"type": "def",
"street": "zzz"
},
{
"type": "hhh",
"street": "yyy"
},
]
}
]
If I can give the "type" and "userid", how can I get the result as
[
{
"userid": "user2",
"type": "abc",
"street": "ccc",
}
]
It would also be great even if I can get the "street" only as the result. The only constraint is I need to get it in the root element itself and not inside an array
Something like this:
db.collection.aggregate([
{
$match: {
userid: "user1" , "address.type":"abc"
}
},
{
$project: {
userid: 1,
address: {
$filter: {
input: "$addresses",
as: "a",
cond: {
$eq: [
"$$a.type",
"abc"
]
}
}
}
}
},
{
$unwind: "$address"
},
{
$project: {
userid: 1,
street: "$address.street",
_id: 0
}
}
])
explained:
Filter only documents with the userid & addresess.type you need
Project/Filter only the addresses elements with the needed type
unwind the address array
project only the needed elements as requested
For best results create index on the { userid:1 } field or compound index on { userid:1 , address.type:1 } fields
playground
You should be able to use unwind, match and project as shown below:
db.collection.aggregate([
{
"$unwind": "$addresses"
},
{
"$match": {
"addresses.type": "abc",
"userid": "user1"
}
},
{
"$project": {
"_id": 0,
"street": "$addresses.street"
}
}
])
You can also duplicate the match step as the first step to reduce the number of documents to unwind.
Here is the playground link.
There is a similar question/answer here.
Related
Need help with mongo db query
Mondo db query - search for parents with state good and children with state bad or missing. output should be an array of all the children with state bad or missing from parents with good state
Below is the JSON list
[
{
"name": "parent-a",
"status": {
"state": "good"
},
"children": [
"child-1",
"child-2"
]
},
{
"name": "child-1",
"state": "good",
"parent": "parent-a"
},
{
"name": "child-2",
"state": {},
"parent": "parent-a"
},
{
"name": "parent-b",
"status": {
"state": "good"
},
"children": [
"child-3",
"child-4"
]
},
{
"name": "child-3",
"state": "good",
"parent": "parent-b"
},
{
"name": "child-4",
"state": "bad",
"parent": "parent-b"
},
{
"name": "parent-c",
"status": {
"state": "bad"
},
"children": [
"child-5",
"child-6"
]
},
{
"name": "child-5",
"state": "good",
"parent": "parent-c"
},
{
"name": "child-6",
"state": "bad",
"parent": "parent-c"
}
]
Expected output
"children": [
{
"name": "child-2",
"state": {}
},
{
"name": "child-4",
"state": "bad"
}
]
Any inputs would be appreciated. Thanks in advance :)
One option is to use $lookup* for this:
db.collection.aggregate([
{$match: {state: {$in: ["bad", {}]}}},
{$lookup: {
from: "collection",
localField: "parent",
foreignField: "name",
pipeline: [
{$match: {"status.state": "good"}}
],
as: "hasGoodParent"
}},
{$match: {"hasGoodParent.0": {$exists: true}}},
{$project: {name: 1, state: 1, _id: 0}}
])
See how it works on the playground example
*If your mongoDB version is lower than 5.0 you need to change the syntax a bit. Drop the localField and foreignField of the $lookup and replace with let and equality match on the pipeline
Here is an approach doing this all without a "$lookup" stage as performance usually suffers when involved. Basically we match all relevant children and parents and we group by the child id. if it has a parent (which means the parent has a "good" state, and a "child" which means the child has a "bad/{}" state then it's matched).
You should make sure you have the appropriate indexes to support the initial query.
Additionally I would personally recommend adding a boolean field on each document to mark wether it's a parent or a child. right now we have to use the field structure based on your input to mark this type but I would consider this a bad practice.
Another thing we did not discuss which doesn't seem possible from the current structure is recursion, can a child have children of it's own? Just some things to consider
db.collection.aggregate([
{
$match: {
$or: [
{
$and: [
{
"status.state": "good"
},
{
parent: {
$exists: false
}
},
{
"children.0": {
$exists: true
}
}
]
},
{
$and: [
{
"state": {
$in: [
"bad",
null,
{}
]
}
},
{
parent: {
$exists: true
}
}
]
}
]
}
},
{
$unwind: {
path: "$children",
preserveNullAndEmptyArrays: true
}
},
{
$addFields: {
isParent: {
$cond: [
{
$eq: [
null,
{
$ifNull: [
"$parent",
null
]
}
]
},
1,
0
]
}
}
},
{
$group: {
_id: {
$cond: [
"$isParent",
"$children",
"$name"
]
},
hasParnet: {
$sum: "$isParent"
},
hasChild: {
$sum: {
$subtract: [
1,
"$isParent"
]
}
},
state: {
"$mergeObjects": {
$cond: [
"$isParent",
{},
{
state: "$state"
}
]
}
}
}
},
{
$match: {
hasChild: {
$gt: 0
},
hasParnet: {
$gt: 0
}
}
},
{
$group: {
_id: null,
children: {
$push: {
name: "$_id",
state: "$state.state"
}
}
}
}
])
Mongo Playground
i want to compare collection with array in aggregate result
i have following two collection.
chat collection
chat.tags is a array value in reference key come from the tags collection.
"chat": [
{
"id": "test1",
"tags": [
"AAA",
"BBB",
"CCC",
"AAA"
]
},
{
"id": "test2",
"tags": [
"AAA",
"BBB",
"CCC"
]
}
]
tag collection
"tag": [
{
"id": "1234",
"key": "AAA",
"name": "a"
},
{
"id": "1235",
"key": "BBB",
"name": "b"
},
{
"id": "1236",
"key": "CCC",
"name": "c"
},
{
"id": "1237",
"key": "DDD",
"name": "d"
},
]
i want to result that id is "test1" and unique tags in chat collection.
i want to following result using with mongo aggregate.
Is it possible with from, let, pipeline when using lookup?
[
{
"chat": [
{
"id": "test1",
"setTags": [
"AAA",
"BBB",
"CCC"
]
}
],
"tag": [
{
"id": "1234",
"key": "AAA",
"name": "a"
},
{
"id": "1235",
"key": "BBB",
"name": "b"
},
{
"id": "1236",
"key": "CCC",
"name": "c"
}
]
}
]
please help me.
This can be achieved with a simple $lookup, like so:
db.chat.aggregate([
{
$match: {
id: "test1"
}
},
{
$lookup: {
from: "tag",
localField: "tags",
foreignField: "key",
as: "tagDocs"
}
},
{
$project: {
chat: [
{
id: "$id",
setTags: "$tags"
}
],
tag: "$tagDocs"
}
}
])
Mongo Playground
I didn't fully understand what the output structure you want is but it can easily be changed via a different $project stage.
--- EDIT ---
With Mongo's v3.6 $lookup syntax the pipeline remains the same, just the $lookup stage changes:
{
$lookup: {
from: "tag",
let: {
tagKeys: "$tags"
},
pipeline: [
{
$match: {
$expr: {
$in: [
"$key",
"$$tagKeys"
]
}
}
}
],
as: "tagDocs"
}
},
Mongo Playground
I have many records in one collection in MongoDB and this is 3 examples to remove only based one QUESTION match criteria.
{
"_id": {
"$oid": "5f0f561256efe82f5082252e"
},
"Item1": false,
"Item2": "",
"Item3": 1,
"Item4": [
{
"Name": "TYPE",
"Value": "QUESTION"
},
{
"Name": "QUESTION",
"Value": "What is your name?"
},
{
"Name": "CORRECT_ANSWER",
"Value": "1"
},
{
"Name": "ANSWER_1",
"Value": "name one"
},
{
"Name": "ANSWER_2",
"Value": "name two"
}
],
"Item5": [
10
],
"Item6": false
}
and another one to compare
{
"_id": {
"$oid": "5f0f561256efe82f5082252c"
},
"Item1": false,
"Item2": "",
"Item3": 2,
"Item4": [
{
"Name": "TYPE",
"Value": "QUESTION"
},
{
"Name": "QUESTION",
"Value": "What is your name?"
},
{
"Name": "CORRECT_ANSWER",
"Value": "1"
},
{
"Name": "ANSWER_1",
"Value": "name one"
},
{
"Name": "ANSWER_2",
"Value": "name two"
}
],
"Item5": [
10
],
"Item6": false
}
the third one :
{
"_id": {
"$oid": "5f0f561256efe82f5082252d"
},
"Item1": false,
"Item2": "",
"Item3": 3,
"Item4": [
{
"Name": "TYPE",
"Value": "QUESTION"
},
{
"Name": "QUESTION",
"Value": "What is your last name?"
},
{
"Name": "CORRECT_ANSWER",
"Value": "1"
},
{
"Name": "ANSWER_1",
"Value": "name one"
},
{
"Name": "ANSWER_2",
"Value": "name two"
}
],
"Item5": [
10
],
"Item6": false
}
What I'm trying here is to make query with aggregation approach and I only want to focus on Item4 for exactly ("Name": "QUESTION") and the value (the question) for identifying the duplication.
The idea is to looking for duplication in the the question itself only ("What is your name?") in our example here. and I don't want to specify witch question because there are long list of them.
I'm looking just for the duplicated questions no mater what is the question look like.
I used the following approach but still I cannot narrow down the output to be only related to question and its value in order to delete the duplicate in the another step.
db.collections.aggregate([{ $unwind: "$Item4" }, {$group: { _id: { QUESTION: "$Item4.Name.4", Value: "$Item4.Value.4" }}}]).pretty()
I'm executing from mongo shell directly.
The following aggregation will list all the documents (the _ids) which have the duplicates of "Item4.Value" for the condition "Item4.Name": "QUESTION".
db.test.aggregate( [
{
$unwind: "$Item4"
},
{
$match: { "Item4.Name": "QUESTION" }
},
{
$group: {
_id: { "Item4_Value": "$Item4.Value" },
ids: { $push: "$_id" }
}
},
{
$match: { $expr: { $gt: [ { $size: "$ids" }, 1 ] } }
}
] )
It works! thanks a lot. I add it to the rest of code as below :
db.test.find().count()
const duplicatesIds = [];
db.test.aggregate( [
{
$unwind: "$Item4"
},
{
$match: { "Item4.Name": "QUESTION" } //here is the trick...to filter the array to pass only the condition "Item4.Name": "QUESTION".
},
{
$group: {
_id: { "Item4_Value": "$Item4.Value" },
ids: { $push: "$_id" }
}
}
],
{
allowDiskUse: true
}
).forEach(function (doc) {
doc.ids.shift();
doc.ids.forEach(function (dupId) {
duplicatesIds.push(dupId);
})
});
printjson(duplicatesIds);
db.test.remove({_id:{$in:duplicatesIds}})
db.test.find().count()
I got question might be many of you can help me.
so I have data on mongodb.
first data
{
"name" : 'david'
contacts : [
{
"name" : 'john',
"phone" : '123456'
},
{
"name" : 'george',
"phone" : '0987654'
}
]
}
second data
{
"name" : 'anita',
"contacts" : [
{
"name" : 'harry',
"phone" : '123456'
},
{
"name" : 'kurita',
"phone" : '323434'
}
]
}
the problem is,
can I query to find data that have duplicate contacts.phone.
so the result whill show like this.
{
"name" : 'david',
"contacts" : [
{
"name" : 'john',
"phone" : '123456'
}
]
}
{
"name" : 'anita',
"contacts" : [
{
"name" : 'harry',
"phone" : '123456'
}
]
}
data john and anita will show because they have similar data on contacts.phone
sorry for my english btw,
I hope you all understand what I mean.
thank you so much
There are a few steps involved to get the results you need.
We are going to write an aggregate pipeline to get the work done.
First you need to unwind your array values with the following:
{
$unwind: "$contacts"
}
Doc: https://docs.mongodb.com/manual/reference/operator/aggregation/unwind/
This would result into:
[
{
"_id": ObjectId("5a934e000102030405000000"),
"contacts": {
"name": "john",
"phone": "123456"
},
"name": "david"
},
{
"_id": ObjectId("5a934e000102030405000000"),
"contacts": {
"name": "george",
"phone": "0987654"
},
"name": "david"
},
{
"_id": ObjectId("5a934e000102030405000001"),
"contacts": {
"name": "harry",
"phone": "123456"
},
"name": "anita"
},
{
"_id": ObjectId("5a934e000102030405000001"),
"contacts": {
"name": "kurita",
"phone": "323434"
},
"name": "anita"
}
]
This would be much easier for us to group by field.
Doc: https://docs.mongodb.com/manual/reference/operator/aggregation/group/
{
$group: {
_id: {
phone: "$contacts.phone"
},
name: {
$addToSet: "$name"
},
contacts: {
$addToSet: "$contacts.name"
},
count: {
$sum: 1
}
}
}
That gives the following output:
[
{
"_id": {
"phone": "323434"
},
"contacts": [
"kurita"
],
"count": 1,
"name": [
"anita"
]
},
{
"_id": {
"phone": "123456"
},
"contacts": [
"john",
"harry"
],
"count": 2,
"name": [
"david",
"anita"
]
},
{
"_id": {
"phone": "0987654"
},
"contacts": [
"george"
],
"count": 1,
"name": [
"david"
]
}
]
Based on the output we need to match the count greater then 1 like:
Doc: https://docs.mongodb.com/manual/reference/operator/aggregation/match/
{
$match: {
count: {
"$gt": 1
}
}
}
Result is:
[
{
"_id": {
"phone": "123456"
},
"contacts": [
"john",
"harry"
],
"count": 2,
"name": [
"david",
"anita"
]
}
]
The query would look like:
Doc: https://docs.mongodb.com/manual/reference/operator/aggregation-pipeline/
db.collection.aggregate([
{
$unwind: "$contacts"
},
{
$group: {
_id: {
phone: "$contacts.phone"
},
name: {
$addToSet: "$name"
},
contacts: {
$addToSet: "$contacts.name"
},
count: {
$sum: 1
}
}
},
{
$match: {
count: {
"$gt": 1
}
}
}
])
MongoPlayground: https://mongoplayground.net/p/qSvhcYyAcQO
I hope this gives you a small idea what is possible with the aggregation pipeline.
Update / fix
According to your requirements you wish to have 2 objects foreach name that has duplicate contacts then you could use unwind again after the match.
[
{
"_id": {
"phone": "123456"
},
"contacts": [
"harry",
"john"
],
"count": 2,
"name": "david"
},
{
"_id": {
"phone": "123456"
},
"contacts": [
"harry",
"john"
],
"count": 2,
"name": "anita"
}
]
Cheers, Kevin
I have bunch of documents in the following format in my mongodb, I am using moongoose as ORM.
Can some one help me make a query to get all the contents having the name=abc inside data-items.content
Document 1:
{
"title": "Some title",
"addresse": "data",
"data-items": {
"content": [
{
"name": "abc",
"age": "poster5.jpg"
},
{
"name": "def",
"age": "poster5.jpg"
},
{
"name": "hij",
"age": "poster5.jpg"
}]
}
}
Document 2:
{
"title": "another title",
"addresse": "data",
"data-items": {
"content": [
{
"name": "abc",
"age": "poster7.jpg"
},
{
"name": "def",
"age": "poster5.jpg"
},
{
"name": "hij",
"age": "poster5.jpg"
}]
}
}
Any help is appreciated
You can simply use the dot notation to query an array of nested documents:
Model.find({"data-items.content.name": "abc"})
EDIT: to get only subdocuments matching your condition you can use below aggregation:
Model.aggregate([
{
$match: {
"data-items.content.name": "abc"
}
},
{
$unwind: "$data-items.content"
},
{
$match: {
"data-items.content.name": "abc"
}
},
{
$replaceRoot: {
newRoot: "$data-items.content"
}
}
])
$unwind will give you single document per content and $replaceRoot will promote it to the root level.