I want to use this mongoDB collection:
[
{
"_id": {
"$oid": "627c4eb87e7c2b8ba510ac4c"
},
"Contact": [
{
"name": "ABC",
"phone": 5501234,
"mail": "abc#mail.com"
},
{
"name": "DEF",
"phone": 6001234,
"mail": "def#mail.com"
}
],
"nomatter": "trash"
}
]
search for {"name":"ABC"} and return only {"mail":"abc#mail.com"}.
It's possible to use find or it's necessary to use aggregate?
Try this one:
db.collection.aggregate([
{ $match: { "Contact.name": "ABC" } },
{
$project: {
Contact: {
$filter: {
input: "$Contact",
cond: { $eq: [ "$$this.name", "ABC" ] }
}
}
}
},
{ "$replaceWith": { mail: { $first: "$Contact.mail" } } }
])
Mongo Playground
Related
In my example project, I have employees under manager. Db schema is like this;
{
"employees": [
{
"name": "Adam",
"_id": "5ea36b27d7ae560845afb88e",
"bananas": "allowed"
},
{
"name": "Smith",
"_id": "5ea36b27d7ae560845afb88f",
"bananas": "not-allowed"
},
{
"name": "John",
"_id": "5ea36b27d7ae560845afb88g",
"bananas": "not-allowed"
},
{
"name": "Patrick",
"_id": "5ea36b27d7ae560845afb88h",
"bananas": "allowed"
}
]
}
In this case Adam is allowed to eat bananas and Smith is not. If I have to give the permission of eating bananas from Adam to Smith I need to perform update operation twice like this:
db.managers.update(
{ 'employees.name': 'Adam' },
{ $set: { 'employees.$.bananas': 'not-allowed' } }
);
and
db.managers.update(
{ 'employees.name': 'Smith' },
{ $set: { 'employees.$.bananas': 'allowed' } }
);
Is it possible to handle this in a single query?
You can use $map and $cond to perform conditional update to the array entries depending on the name of the employee. A $switch is used for potential extension of cases.
db.collection.update({},
[
{
"$set": {
"employees": {
"$map": {
"input": "$employees",
"as": "e",
"in": {
"$switch": {
"branches": [
{
"case": {
$eq: [
"$$e.name",
"Adam"
]
},
"then": {
"$mergeObjects": [
"$$e",
{
"bananas": "not-allowed"
}
]
}
},
{
"case": {
$eq: [
"$$e.name",
"Smith"
]
},
"then": {
"$mergeObjects": [
"$$e",
{
"bananas": "allowed"
}
]
}
}
],
default: "$$e"
}
}
}
}
}
}
])
Mongo Playground
db.managers.update(
{
$or: [
{"employees.name": "Adam"},
{"employees.name": "Smith"}
]
},
{
$set: {
"employees.$[e].bananas": {
$cond: [{ $eq: ["$e.name", "Adam"] }, "not-allowed", "allowed"]
}
}
},
{
arrayFilters: [{ "e.name": { $in: ["Adam", "Smith"] } }]
}
)
I have a problem with filtering fields(objects) from my document using find function.
Here is my db.collection.find(..) document:
{
"_id": BinData(3, "Uz+QwtoVMt7hjpqMrLxVhQ=="),
"name": "jeorge",
"permissions": [
{
"key": "group.staff",
"value": true,
"context": [
{ "key": "server", "value": "test" }
]
},
{
"key": "group.tester",
"value": true,
"context": [
{ "key": "server", "value": "test" }
]
},
{
"key": "test.test",
"value": true
},
{
"key": "group.default",
"value": true
},
{
"key": "group.helper",
"value": true
}
]
}
How can I filter my document in two ways?
a) Display only fields with nested context object.
b) Display only fields, which hasn't got nested context objects.
Also I need to check if parent's object property key contains string "group" as value (If not - skip it).
For situation b, I have tried this function, but it prints result with only the first matched element (based on mongodb's documentation).
db.collection.find(
{
"_id": BinData(3, "Uz+QwtoVMt7hjpqMrLxVhQ==")
},
{
"permissions": {
$elemMatch: {
"key": {
$regex: "group",
},
"value": true
}
}
}
);
Is it possible by single query? Thanks!
db.collection.aggregate([
{
$match: { _id: BinData(3, "Uz+QwtoVMt7hjpqMrLxVhQ==") }
},
{
$set: {
permissions: {
$filter: {
input: "$permissions",
as: "p",
cond: {
$and: [
"$$p.context",
{ $regexMatch: { input: "$$p.key", regex: "group" } }
]
}
}
}
}
}
])
mongoplayground
db.collection.aggregate([
{
$match: {
_id: BinData(3, "Uz+QwtoVMt7hjpqMrLxVhQ==")
}
},
{
$set: {
permissions: {
$filter: {
input: "$permissions",
as: "p",
cond: {
$and: [
{
$not: [ "$$p.context" ]
},
{
$regexMatch: {
input: "$$p.key",
regex: "group"
}
}
]
}
}
}
}
}
])
mongoplayground
I need to fetch distinct nested documents.
Please find the sample document:
{
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z"),
"HList":[
{
"productId": 123,
"name": "Dubai",
"tsh": true
}
],
"PList":[
{
"productId": 123,
"name": "Dubai",
"tsh": false
},
{
"productId": 234,
"name": "India",
"tsh": true
}
],
"CList":[
{
"productId": 234,
"name": "India",
"tsh": false
}
]
}
Expected result is:
{
"produts":[
{
"productId": 123,
"name": "Dubai"
},
{
"productId": 234,
"name": "India"
}
]
}
I tried with this query:
db.property.aggregate([
{
$match: {
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z")
}
},
{
"$project": {
"_id": 0,
"unique": {
"$filter": {
"input": {
"$setDifference": [
{
"$concatArrays": [
"$HList.productId",
"$PList.productId",
"$CList.productId"
]
},
[]
]
},
"cond": {
"$ne": [ "$$this", "" ]
}
}
}
}
}
]);
Is $setDifference aggregation is correct choice here?
My query returns only unique product ids but i need a productId with name.
Could someone help me to solve this?
Thanks in advance
You can use $projectfirst to get rid of tsh field and then run $setUnion which ignores duplicated entries:
db.collection.aggregate([
{
$project: {
"HList.tsh": 0,
"PList.tsh": 0,
"CList.tsh": 0,
}
},
{
$project: {
products: {
$setUnion: [ "$HList", "$PList", "$CList" ]
}
}
}
])
Mongo Playground
The following two aggregations return the expected and same result (you can use any of the two):
db.collection.aggregate( [
{
$project: {
_id: 0,
products: {
$reduce: {
input: { $setUnion: [ "$HList", "$PList", "$CList" ] },
initialValue: [],
in: {
$setUnion: [ "$$value", [ { productId: "$$this.productId", name: "$$this.name" } ] ]
}
}
}
}
}
] )
This one is little verbose:
db.collection.aggregate( [
{
$project: { list: { $setUnion: [ "$HList", "$PList", "$CList" ] } }
},
{
$unwind: "$list"
},
{
$group: {
_id: null,
products: { $addToSet: { "productId": "$list.productId", "name": "$list.name" } }
}
},
{
$project: { _id: 0 }
}
] )
db.collection.aggregate([
{
$match: {
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z")
}
},
{
$project: {
products: {
$filter: {
input: { "$setUnion" : ["$CList", "$HList", "$PList"] },
as: 'product',
cond: {}
}
}
}
},
{
$project: {
"_id":0,
"products.tsh": 1,
"products.name": 1,
}
},
])
I have the following document:
[
{
"callId": "17dac51e-125e-499e-9064-f20bd3b1a9d8",
"caller": {
"firstName": "Test",
"lastName": "Testing",
"phoneNumber": "1231231234"
},
"inquiries": [
{
"inquiryId": "b0d14381-ce75-49aa-a66a-c36ae20b72a8",
"routeHistory": [
{
"assignedUserId": "cfa0ffe9-c77d-4eec-87d7-4430f7772e81",
"routeDate": "2020-01-01T06:00:00.000Z",
"status": "routed"
},
{
"assignedUserId": "cfa0ffe9-c77d-4eec-87d7-4430f7772e81",
"routeDate": "2020-01-03T06:00:00.000Z",
"status": "routed"
}
]
},
{
"inquiryId": "9d743be9-7613-46d7-8f9b-a04b4b899b56",
"routeHistory": [
{
"assignedUserId": "cfa0ffe9-c77d-4eec-87d7-4430f7772e81",
"routeDate": "2020-01-01T06:00:00.000Z",
"status": "ended"
},
{
"assignedUserId": "cfa0ffe9-c77d-4eec-87d7-4430f7772e81",
"routeDate": "2020-01-03T06:00:00.000Z",
"status": "ended"
}
]
}
]
}
]
And I'm using the following aggregate:
{
$unwind: '$inquiries',
},
{
$addFields: {
'inquiries.routeHistory': {
$filter: {
input: '$inquiries.routeHistory',
cond: {
$eq: [{ $max: '$inquiries.routeHistory.routeDate' }, '$$this.routeDate'],
},
},
},
},
},
{
$group: {
_id: '$_id',
callId: { $first: '$callId' },
caller: { $first: '$caller' },
inquiries: { $push: '$inquiries' },
},
}
I would like to expand this query to be able to further filter at the inquiry grain, so that I am returning only the inquiry that contains my specified criteria. E.g. if I wanted to find where inquiry.routeHistory.status = ended, I would expect the following results:
[
{
"callId": "17dac51e-125e-499e-9064-f20bd3b1a9d8",
"caller": {
"firstName": "Test",
"lastName": "Testing",
"phoneNumber": "1231231234"
},
"inquiries": [
{
"inquiryId": "9d743be9-7613-46d7-8f9b-a04b4b899b56",
"routeHistory": [
{
"assignedUserId": "cfa0ffe9-c77d-4eec-87d7-4430f7772e81",
"routeDate": "2020-01-03T06:00:00.000Z",
"status": "ended"
}
]
}
]
}
]
Is there a way to do nested $addField or is there another route I could take?
Since you're using $unwind you can do that easily by adding $match since expression: "inquiries.routeHistory.status": "ended" will return true if there's any document in routeHistory having such status:
db.collection.aggregate([
{
$unwind: "$inquiries"
},
{
$match: {
"inquiries.routeHistory.status": "ended"
}
},
{
$addFields: {
"inquiries.routeHistory": {
$filter: {
input: "$inquiries.routeHistory",
cond: {
$eq: [ { $max: "$inquiries.routeHistory.routeDate" }, "$$this.routeDate" ]
}
}
}
}
},
{
$group: {
_id: "$_id",
callId: { $first: "$callId" },
caller: { $first: "$caller" },
inquiries: { $push: "$inquiries" }
}
}
])
Mongo Playground
let's say I have docs such as
{
"nickname": "my nickname",
"comments": [
{
"id": 1
},
{
"id": 1
}
]
}
how do I update it to look like
{
"nickname": "my nickname",
"comments": [
{
"id": 1,
"nickname": "my nickname"
},
{
"id": 1,
"nickname": "my nickname"
}
]
}
This does not seem to be working
db.getCollection('users').update(
{
"comments.nickname": null
},
{ "$set": { "comments.$.nickname": "$nickname" } });
This is just an example to represent my problem.
I would not like to hear about re-structuring and optimizing the fields.
Thanks!
Try this (v4.2):
db.users.updateMany(
{"comments.nickname":null},
[
{"$set": {"comments.nickname": "$nickname"}}
]
)
Note: It will override if any comments.nickname already exists
db.users.aggregate([
{
$match: {
"comments.nickname": null
}
},
{
$addFields: {
comments: {
$map: {
input: "$comments",
in: {
id: "$$this.id",
nickname: {
$cond: [
"$$this.nickname",
"$$this.nickname",
"$nickname"
]
}
}
}
}
}
},
{
$out: "users"
}
])
Note: It will keep already existing values