I'm struggling to query documents now.
I have documents like this:
# reviews
{
_id: 1,
category_id: 1,
comments: [
{
_id: 1,
context: "abc",
is_deleted: false
}
]
}
and I've tried to query like this:
db.reviews.aggregate(
[
{"$match": {"category_id": 1}},
{
"$unwind": {"path": "$comments", "preserveNullAndEmptyArrays": True}
},
{"$match": {"comments.is_deleted": false}},
{
"$group": {
"_id": "$_id",
"category_id": {"$first": "$category_id"},
"comments": {"$push": "$comments"},
}
},
]
)
it worked fine but the one thing is wrong.
when all elements of comments don't match the condition, it doesn't return parent too. but what I exactly want is to return [] when all elements of array don't match.
I'll add an example for your understanding:
what I expected:
[{
_id: 1,
category_id: 1,
comments: []
}]
what I actually got:
[]
I've tried $filter but It excludes the parent object if any of elements of array are not met.
Thank you for your help in advance.
db.collection.aggregate([
{
$match: {
"category_id": 1
}
},
{
$set: {
"comments": {
$filter: {
input: "$comments",
cond: {
$eq: [
"$$this.is_deleted",
false
]
}
}
}
}
}
])
Demo
Related
I have 4 products. I want to know the count of product-4 for users who has product-1 or product-2
Sample data:
[
{
"user_id": 1,
"product_type": "product-1"
},
{
"user_id": 1,
"product_type": "product-4"
},
{
"user_id": 1,
"product_type": "product-4"
},
{
"user_id": 2,
"product_type": "product-1"
}
]
user-1 has two product-4 and one product-1 (that counts 2)
user-2 has only product-1, but no product-4 (hence that does not count)
This is how I tried
db.collection.aggregate([
{
$match: {
product_type: {
$in: [
"product-1",
"product-2",
],
},
},
},
{
$group: {
_id: "$user_id",
},
},
{
$match: {
user_id: { $in: "$_id"}, // I want to use $group's result in here
product_type: "product-4",
},
}
]);
Expected results are:
[
{
"_id": 1,
"count": 2
},
{
"_id": 2,
"count": 0
}
]
Note:
I dont have a backend, I have to this using mongodb only.
Does this answer your question?
db.collection.aggregate([
{$group: {_id: "$user_id", data: {$push: "$product_type"}}},
{$match: {$expr: {$or: [
{$in: ["product-1", "$data"]},
{$in: ["product-2", "$data"]}
]}}},
{$project: {
count: {
$size: {
$filter: {
input: "$data",
cond: {$eq: ["$$this", "product-4"]}
}
}
}
}}
])
See how it works on the playground example
I have following collection
[
{
"_id": ObjectId("57315ba4846dd82425ca2408"),
"myarray": [
{
"point": 5,
"userId": "570ca5e48dbe673802c2d035"
},
{
"point": 2,
"userId": "613ca5e48dbe673802c2d521"
},
{
"point": 4,
"userId": "570ca5e48dbe673802c2d045"
},
{
"point": 4,
"userId": "570ca5e48dbe473802c2d035"
}
]
}
]
I have a collection like above and I want to remove some objects inside array based on userID condition and after removing I have to update one field in mongo with size of array
I'm trying with the below query where removing array elements is working as excepted but array size is not updating properly
db.collection.update({
_id: ObjectId("57315ba4846dd82425ca2408")
},
{
$pull: {
"myarray": {
userId: {
$in: [
"570ca5e48dbe673802c2d035",
"613ca5e48dbe673802c2d521"
]
}
}
},
"$set": {
profilecount: {
$size: "$myarray"
}
}
})
to see result of query please click this link and run query https://mongoplayground.net/p/FtMk7ymacr3
One option is using an update with a pipeline:
db.collection.update({
_id: ObjectId("57315ba4846dd82425ca2408")
},
[{
$set: {
"myarray": {
$filter: {
input: "$myarray",
cond: {
$not: {
$in: [
"$$this.userId",
["570ca5e48dbe673802c2d035", "613ca5e48dbe673802c2d521"]
]
}
}
}
}
}
},
{$set: {profilecount: {$size: "$myarray"}}}
])
See how it works on the playground example
I have a user collection:
[
{"_id": 1,"name": "John", "age": 25, "valid_user": true}
{"_id": 2, "name": "Bob", "age": 40, "valid_user": false}
{"_id": 3, "name": "Jacob","age": 27,"valid_user": null}
{"_id": 4, "name": "Amelia","age": 29,"valid_user": true}
]
I run a '$facet' stage on this collection. Checkout this MongoPlayground.
I want to talk about the first output from the facet stage. The following is the response currently:
{
"user_by_valid_status": [
{
"_id": false,
"count": 1
},
{
"_id": true,
"count": 2
},
{
"_id": null,
"count": 1
}
]
}
However, I want to restructure the output in this way:
"analytics": {
"invalid_user": {
"_id": false
"count": 1
},
"valid_user": {
"_id": true
"count": 2
},
"user_with_unknown_status": {
"_id": null
"count": 1
}
}
The problem with using a '$project' stage along with 'arrayElemAt' is that the order may not be definite for me to associate an index with an attribute like 'valid_users' or others. Also, it gets further complicated because unlike the sample documents that I have shared, my collection may not always contain all the three categories of users.
Is there some way I can do this?
You can use $switch conditional operator,
$project to show value part in v with _id and count field as object, k to put $switch condition
db.collection.aggregate([
{
"$facet": {
"user_by_valid_status": [
{
"$group": {
"_id": "$valid_user",
"count": { "$sum": 1 }
}
},
{
$project: {
_id: 0,
v: { _id: "$_id", count: "$count" },
k: {
$switch: {
branches: [
{ case: { $eq: ["$_id", null] }, then: "user_with_unknown_status" },
{ case: { $eq: ["$_id", false] }, then: "invalid_user" },
{ case: { $eq: ["$_id", true] }, then: "valid_user" }
]
}
}
}
}
],
"users_above_30": [{ "$match": { "age": { "$gt": 30 } } }]
}
},
$project stage in root, convert user_by_valid_status array to object using $arrayToObject
{
$project: {
analytics: { $arrayToObject: "$user_by_valid_status" },
users_above_30: 1
}
}
])
Playground
I am trying to aggregate a batch of documents. There are two fields in the documents I would like to $push. However, lets say they are "_id" and "A" fields, I only want $push "_id" and "A" if "A" is $gt 0.
I tried two approaches.
First one.
db.collection.aggregate([{
"$group":{
"field": {
"$push": {
"$cond":[
{"$gt":["$A", 0]},
{"id": "$_id", "A":"$A"},
null
]
}
},
"secondField":{"$push":"$B"}
}])
But this will push a null value to "field" and I don't want it.
Second one.
db.collection.aggregate([{
"$group":
"field": {
"$cond":[
{"$gt",["$A", 0]},
{"$push": {"id":"$_id", "A":"$A"}},
null
]
},
"secondField":{"$push":"$B"}
}])
The second one simply doesn't work...
Is there a way to skip the $push in else case?
ADDED:
Expected documents:
{
"_id":objectid(1),
"A":2,
"B":"One"
},
{
"_id":objectid(2),
"A":3,
"B":"Two"
},
{
"_id":objectid(3),
"B":"Three"
}
Expected Output:
{
"field":[
{
"A":"2",
"_id":objectid(1)
},
{
"A":"3",
"_id":objectid(2)
},
],
"secondField":["One", "Two", "Three"]
}
You can use "$$REMOVE":
This system variable was added in version 3.6 (mongodb docs)
db.collection.aggregate([{
$group:{
field: {
$push: {
$cond:[
{ $gt: ["$A", 0] },
{ id: "$_id", A:"$A" },
"$$REMOVE"
]
}
},
secondField:{ $push: "$B" }
}
])
In this way you don't have to filter nulls.
This is my answer to the question after reading the post suggested by #Veeram
db.collection.aggregate([{
"$group":{
"field": {
"$push": {
"$cond":[
{"$gt":["$A", 0]},
{"id": "$_id", "A":"$A"},
null
]
}
},
"secondField":{"$push":"$B"}
},
{
"$project": {
"A":{"$setDifference":["$A", [null]]},
"B":"$B"
}
}])
One more option is to use $filter operator:
db.collection.aggregate([
{
$group : {
_id: null,
field: { $push: { id: "$_id", A : "$A"}},
secondField:{ $push: "$B" }
}
},
{
$project: {
field: {
$filter: {
input: "$field",
as: "item",
cond: { $gt: [ "$$item.A", 0 ] }
}
},
secondField: "$secondField"
}
}])
On first step you combine your array and filter them on second step
$group: {
_id: '$_id',
tasks: {
$addToSet: {
$cond: {
if: {
$eq: [
{
$ifNull: ['$tasks.id', ''],
},
'',
],
},
then: '$$REMOVE',
else: {
id: '$tasks.id',
description: '$tasks.description',
assignee: {
$cond: {
if: {
$eq: [
{
$ifNull: ['$tasks.assignee._id', ''],
},
'',
],
},
then: undefined,
else: {
id: '$tasks.assignee._id',
name: '$tasks.assignee.name',
thumbnail: '$tasks.assignee.thumbnail',
status: '$tasks.assignee.status',
},
},
},
},
},
},
},
}
Using the example from the Mongo docs:
{ _id: 1, results: [ { product: "abc", score: 10 }, { product: "xyz", score: 5 } ] }
{ _id: 2, results: [ { product: "abc", score: 8 }, { product: "xyz", score: 7 } ] }
{ _id: 3, results: [ { product: "abc", score: 7 }, { product: "xyz", score: 8 } ] }
db.survey.find(
{ id: 12345, results: { $elemMatch: { product: "xyz", score: { $gte: 6 } } } }
)
How do I return survey 12345 (regardless of even if it HAS surveys or not) but only return surveys with a score greater than 6? In other words I don't want the document disqualified from the results based on the subdocument, I want the document but only a subset of subdocuments.
What you are asking for is not so much a "query" but is basically just a filtering of content from the array in each document.
You do this with .aggregate() and $project:
db.survey.aggregate([
{ "$project": {
"results": {
"$setDifference": [
{ "$map": {
"input": "$results",
"as": "el",
"in": {
"$cond": [
{ "$and": [
{ "$eq": [ "$$el.product", "xyz" ] },
{ "$gte": [ "$$el.score", 6 ] }
]}
]
}
}},
[false]
]
}
}}
])
So rather than "contrain" results to documents that have an array member matching the condition, all this is doing is "filtering" the array members out that do not match the condition, but returns the document with an empty array if need be.
The fastest present way to do this is with $map to inspect all elements and $setDifference to filter out any values of false returned from that inspection. The possible downside is a "set" must contain unique elements, so this is fine as long as the elements themselves are unique.
Future releases will have a $filter method, which is similar to $map in structure, but directly removes non-matching results where as $map just returns them ( via the $cond and either the matching element or false ) and is then better suited.
Otherwise if not unique or the MongoDB server version is less than 2.6, you are doing this using $unwind, in a non performant way:
db.survey.aggregate([
{ "$unwind": "$results" },
{ "$group": {
"_id": "$_id",
"results": { "$push": "$results" },
"matched": {
"$sum": {
"$cond": [
{ "$and": [
{ "$eq": [ "$results.product", "xyz" ] },
{ "$gte": [ "$results.score", 6 ] }
]},
1,
0
]
}
}
}},
{ "$unwind": "$results" },
{ "$match": {
"$or": [
{
"results.product": "xyz",
"results.score": { "$gte": 6 }
},
{ "matched": 0 }
}},
{ "$group": {
"_id": "$_id",
"results": { "$push": "$results" },
"matched": { "$first": "$matched" }
}},
{ "$project": {
"results": {
"$cond": [
{ "$ne": [ "$matched", 0 ] },
"$results",
[]
]
}
}}
])
Which is pretty horrible in both design and perfomance. As such you are probably better off doing the filtering per document in client code instead.
You can use $filter in mongoDB 3.2
db.survey.aggregate([{
$match: {
{ id: 12345}
}
}, {
$project: {
results: {
$filter: {
input: "$results",
as: "results",
cond:{$gt: ['$$results.score', 6]}
}
}
}
}]);
It will return all the sub document that have score greater than 6. If you want to return only first matched document than you can use '$' operator.
You can use $redact in this way:
db.survey.aggregate( [
{ $match : { _id : 12345 }},
{ $redact: {
$cond: {
if: {
$or: [
{ $eq: [ "$_id", 12345 ] },
{ $and: [
{ $eq: [ "$product", "xyz" ] },
{ $gte: [ "$score", 6 ] }
]}
]
},
then: "$$DESCEND",
else: "$$PRUNE"
}
}
}
] );
It will $match by _id: 12345 first and then it will "$$PRUNE" all the subdocuments that don't have "product":"xyz" and don't have score greater or equal 6. I added the condition ($cond) { $eq: [ "$_id", 12345 ] } so that it wouldn't prune the whole document before it reaches the subdocuments.