I am using mongo version 3.4.3 and I have my documents stored in mongo like this -
{
"_id" : ObjectId("5ad5ab8aaf2808b739ba6ab2"),
"ResumeId" : "105839064",
"ResumeDetails" : {
"WorkProfile" : [
{
"Company" : "XXXXXXXXX",
"JobTitle" : "YYYYY",
"JobSkills" : {
"CommonSkills": [],
"OtherSkills": []
}
},
{
"Company" : "XXXXXXXX",
"JobTitle" : "YYYYYY",
"JobSkills" : {
"CommonSkills" : [
ObjectId("5ad5ab860b94c96c738e914a")
],
"OtherSkills" : [
ObjectId("5ad5ab860b94c96c738e9146")
]
}
},
{
"Company" : "XXXXXXX",
"JobTitle" : "YYYY"
}
],
"AdditionalSkills" : {
"CommonSkills" : [
ObjectId("5ad5ab860b94c96c738e9175"),
ObjectId("5ad5ab860b94c96c738e91f0"),
ObjectId("5ad5ab860b94c96c738e9241"),
ObjectId("5ad5ab860b94c96c738e919b")
],
"OtherSkills" : [
ObjectId("5ad5ab860b94c96c738e90e6"),
ObjectId("5ad5ab860b94c96c738e9142"),
ObjectId("5ad5ab860b94c96c738e9211"),
ObjectId("5ad5ab860b94c96c738e9293"),
ObjectId("5ad5ab860b94c96c738e92c8")
]
}
},
"DocId" : "51cb2f49-fcb9-46a0-9040-67e0f986be11"
}
I want to combine all the skills under WorkProfile and AdditionalSkills under 2 separate arrays. I tried the following query
db.ResumeParsedData.aggregate([
{$match: {'DocId': '51cb2f49-fcb9-46a0-9040-67e0f986be11'}},
{$project: {
'JobSkills': {'$concatArrays': [
'$ResumeDetails.WorkProfile.JobSkills.CommonSkills', '$ResumeDetails.WorkProfile.JobSkills.OtherSkills']
},
'AdditionalSkills': {'$setUnion': [
'$ResumeDetails.AdditionalSkills.CommonSkills', '$ResumeDetails.AdditionalSkills.OtherSkills']},
}
}]).pretty()
But I am getting the following output -
{
"_id" : ObjectId("5ad5ab8aaf2808b739ba6ab2"),
"JobSkills" : [
[
ObjectId("5ad5ab860b94c96c738e914a")
],
[
ObjectId("5ad5ab860b94c96c738e9146")
]
],
"AdditionalSkills" : [
ObjectId("5ad5ab860b94c96c738e90e6"),
ObjectId("5ad5ab860b94c96c738e9142"),
ObjectId("5ad5ab860b94c96c738e9175"),
ObjectId("5ad5ab860b94c96c738e919b"),
ObjectId("5ad5ab860b94c96c738e91f0"),
ObjectId("5ad5ab860b94c96c738e9211"),
ObjectId("5ad5ab860b94c96c738e9241"),
ObjectId("5ad5ab860b94c96c738e9293"),
ObjectId("5ad5ab860b94c96c738e92c8")
]
}
How can I fix the JobSkills array field. It currently coming as array of array fields.
I also tried to concatArrays twice as following:
db.ResumeParsedData.aggregate([
{$match: {'DocId': '51cb2f49-fcb9-46a0-9040-67e0f986be11'}},
{$project: {
'JobSkills': {'$concatArrays': { '$concatArrays': [
'$ResumeDetails.WorkProfile.JobSkills.CommonSkills',
'$ResumeDetails.WorkProfile.JobSkills.OtherSkills'
]}},
'AdditionalSkills': {'$setUnion': [
'$ResumeDetails.AdditionalSkills.CommonSkills',
'$ResumeDetails.AdditionalSkills.OtherSkills'
]},
} }
]).pretty()
You can use $reduce (which is available in 3.4) to flatten your array of arrays:
db.ResumeParsedData.aggregate([
{ $match: {"DocId": "51cb2f49-fcb9-46a0-9040-67e0f986be11"} },
{
$project: {
"JobSkills": {
$reduce: {
input: {
$concatArrays: ["$ResumeDetails.WorkProfile.JobSkills.CommonSkills", "$ResumeDetails.WorkProfile.JobSkills.OtherSkills"]
},
initialValue: [],
in: { $setUnion: [ "$$this", "$$value" ] }
}
},
"AdditionalSkills": {"$setUnion": [
"$ResumeDetails.AdditionalSkills.CommonSkills", "$ResumeDetails.AdditionalSkills.OtherSkills"]}
}
}
])
$setUnion guarantees that there will be no duplicates in final array
Related
{
"_id" : ObjectId("5fa919a49bbe481d117506c9"),
"isDeleted" : 0,
"productId" : 31,
"references" : [
{
"_id" : ObjectId("5fa919a49bbe481d117506ca"),
"languageCode" : "en",
"languageId" : 1,
"productId" : ObjectId("5fa919a49bbe481d117506ba")
},
{
"_id" : ObjectId("5fa91cc7d7d52f1e389dee1f"),
"languageCode" : "ar",
"languageId" : 2,
"productId" : ObjectId("5fa91cc7d7d52f1e389dee1e")
}
],
"createdAt" : ISODate("2020-11-09T10:27:48.859Z"),
"updatedAt" : ISODate("2020-11-09T10:27:48.859Z"),
"__v" : 0
},
{
"_id" : ObjectId("5f9aab1d8e475489270ebe3a"),
"isDeleted" : 0,
"productId" : 21,
"references" : [
{
"_id" : ObjectId("5f9aab1d8e475489270ebe3b"),
"languageCode" : "en",
"languageId" : 1,
"productId" : ObjectId("5f9aab1c8e475489270ebe2d")
}
],
"createdAt" : ISODate("2020-10-29T11:44:29.852Z"),
"updatedAt" : ISODate("2020-10-29T11:44:29.852Z"),
"__v" : 0
}
This is my mongoDB collection in which i store the multilingual references to product collection. In productId are the references to product Collection. Now If we have ar in our request, then we will only have the productId of ar languageCode. If that languageCode does not exist then we will have en langCode productId.
For Example if the user pass ar then the query should return
"productId" : ObjectId("5fa91cc7d7d52f1e389dee1e")
"productId" : ObjectId("5f9aab1c8e475489270ebe2d")
I have tried using $or with $elemMatch but I am not able to get the desired result. Also i am thinking of using $cond. can anyone help me construct the query.
We can acheive
$facet helps to categorized the incoming documents
In the arArray, we get all documents which has"references.languageCode": "ar" (This document may or may not have en), then de-structure the references array, then selecting the "references.languageCode": "ar" only using $match. $group helps to get all productIds which belong to "references.languageCode": "ar"
In the enArray, we only get documents which have only "references.languageCode": "en". Others are same like arArray.
$concatArrays helps to concept both arArray,enArray arrays
$unwind helps to de-structure the array.
$replaceRoot helps to make the Object goes to root
Here is the mongo script.
db.collection.aggregate([
{
$facet: {
arAarray: [
{
$match: {
"references.languageCode": "ar"
}
},
{
$unwind: "$references"
},
{
$match: {
"references.languageCode": "ar"
}
},
{
$group: {
_id: "$_id",
productId: {
$addToSet: "$references.productId"
}
}
}
],
enArray: [
{
$match: {
$and: [
{
"references.languageCode": "en"
},
{
"references.languageCode": {
$ne: "ar"
}
}
]
}
},
{
$unwind: "$references"
},
{
$group: {
_id: "$_id",
productId: {
$addToSet: "$references.productId"
}
}
}
]
}
},
{
$project: {
combined: {
"$concatArrays": [
"$arAarray",
"$enArray"
]
}
}
},
{
$unwind: "$combined"
},
{
"$replaceRoot": {
"newRoot": "$combined"
}
}
])
Working Mongo playground
You can test this solution to see if it is useful for you question:
db.collection.aggregate([
{
$addFields: {
foundResults:
{
$cond: {
if: { $in: ["ar", "$references.languageCode"] }, then:
{
$filter: {
input: "$references",
as: "item",
cond: {
$and: [{ $eq: ["$$item.languageCode", 'ar'] },
]
}
}
}
, else:
{
$filter: {
input: "$references",
as: "item",
cond: {
$and: [{ $eq: ["$$item.languageCode", 'en'] },
]
}
}
}
}
}
}
},
{ $unwind: "$foundResults" },
{ $replaceRoot: { newRoot: { $mergeObjects: ["$foundResults"] } } },
{ $project: { _id: 0, "productId": 1 } }
])
In my mongodb collection, I have sometimes two, sometimes one and sometimes null arrays on a document. Now I'd like to get one array over the whole collection with the values of these arrays.
The document looks like this:
{
"title" : "myDocument",
"listOne" : [
"valueOne",
"valueTwo"
],
"listTwo" : [
"abc",
"qwer"
]
},
{
"title" : "myDocumentTwo",
"listTwo" : [
"321"
]
},
{
"title" : "myDocumentAlpha",
"listOne" : [
"alpha",
"beta"
]
},
{
"title" : "myDocumentbeta"
}
And I expect the following output:
"combinedList" : [
"valueOne",
"valueTwo",
"abc",
"qwer",
"321",
"alpha",
"beta"
]
It's like every possible value from these twos array out of every document in this collection.
You can do this using aggregate and $concatArrays
db.collection.aggregate([
{
$project: {
combinedList: {
$concatArrays: [{$ifNull: ["$listOne", []]}, {$ifNull: ["$listTwo", []]}]
}
}
},
{ $unwind: "$combinedList" },
{ $group: { _id: null, combinedList: { $addToSet: "$combinedList"}}},
{ $project: { _id: 0, combinedList: 1 }}
])
I am having the following collection:
{
"price" : [
55800000,
62800000
],
"surface" : [
81.05,
97.4
],
}
I would want to calculate the price/m2. I have tried the following but got the following error: "$divide only supports numeric types, not array and array".
db.entries.aggregate(
[
{ $project: { url: 1, pricePerSquareMeter: { $divide: [ "$price", "$surfaces" ] } } }
]
)
Would you know how to solve this? Ultimately would want to have an array like this:
{
"price" : [
55800000,
62800000
],
"surface" : [
81.05,
97.4
],
"pricePerSquareMeter" : [
688463.91,
644763.86
]
}
Important: The price and surface should also be ordered so that the calculation is valid.
You can use below aggregation
db.collection.aggregate([
{ "$addFields": {
"pricePerSquareMeter": {
"$map": {
"input": { "$range": [0, { "$size": "$price" }] },
"in": {
"$divide": [
{ "$arrayElemAt": ["$price", "$$this"] },
{ "$arrayElemAt": ["$surface", "$$this"] }
]
}
}
}
}}
])
MongoPlayground
You can use below aggregation
db.entries.aggregate([
{ $addFields: {
pricePerSquareMeter: {
$map: {
input: '$price',
as: 'item',
in: {
$divide: [
"$$item",
{ $arrayElemAt: [
"$surface",
{ "$indexOfArray": ["$price", "$$item"] }
]}
]
}
}
},
}}])
can anybody please tell me what am i doing wrong?
db document structure:
{
"_id" : "module_settings",
"moduleChildren" : [
{
"_id" : "module_settings_general",
"name" : "General",
},
{
"_id" : "module_settings_users",
"name" : "Users",
},
{
"_id" : "module_settings_emails",
"name" : "Emails",
}
],
“permissions” : [
"module_settings_general",
"module_settings_emails"
]
}
pipeline stage:
{ $project: {
filteredChildren: {
$filter: {
input: "$moduleChildren",
as: "moduleChild",
cond: { $in : ["$$moduleChild._id", "$permissions"] }
}
},
}}
I need to filter "moduleChildren" array to show only modules which ids are in "permissions" array. Ive tried "$$ROOT.permissions" and "$$CURRENT.permissions" but none of them is working. I always get an error that $in is missing array as argument. It works when i hardcode the array like this: cond: { $in : ["$$moduleChild._id", [“module_settings_general", "module_settings_emails”]] } so it seems the problem is in passing of the array.
Thanks for any advices!
First option --> Use aggregation
Because your some of the documents in your collection may or may not contain permissions field or is type not equal to array that's why you are getting this error.
You can find the $type of the field and if it is not an array or not exists in your document than you can add it as an array with $addFields and $cond aggregation
db.collection.aggregate([
{ "$addFields": {
"permissions": {
"$cond": {
"if": {
"$ne": [ { "$type": "$permissions" }, "array" ]
},
"then": [],
"else": "$permissions"
}
}
}},
{ "$project": {
"filteredChildren": {
"$filter": {
"input": "$moduleChildren",
"as": "moduleChild",
"cond": {
"$in": [ "$$moduleChild._id", "$permissions" ]
}
}
}
}}
])
Second option -->
Go to your mongo shell or robomongo on any GUI you are using and run
this command
db.collection.update(
{ "permissions": { "$ne": { "$type": "array" } } },
{ "$set": { "permissions": [] } },
{ "multi": true }
)
i want fetch a unitHouse from my document which is an sub array of sub array
Here is the data
{
"_id" : ObjectId("5a17d305c438324308bffb19"),
"floorRow" : [
{
"floorRowNo" : "F1",
"floorRowInfo" : "Best Floor Ever that i have ever seen",
"_id" : ObjectId("5a17d333c438324308bffb1a"),
"unitHouse" : [
]
},
{
"floorRowNo" : "F2",
"floorRowInfo" : "view",
"_id" : ObjectId("5a1bdfbb4d63841c3cb6fc89"),
"unitHouse" : [
{
"unitHouseNo" : "Unit001",
"unitHouseType" : "OFFICE",
"unitHouseStatus" : "SELL",
"_id" : ObjectId("5a1d212bed3a552f0421fd6b"),
},
{
"unitHouseNo" : "Unit002",
"unitHouseType" : "CAT003",
"unitHouseStatus" : "SELL",
"_id" : ObjectId("5a1e3691af12544ff05690e3"),
}
]
}
],
}
Here is what I have queried so far, which i can get floor F2 that i wanted, but it came with both unit. I want only unitHouse with id : 5a1e3691af12544ff05690e3.
propertyDevModel.aggregate([
{
$match: {
_id: mongoose.Types.ObjectId("5a17d305c438324308bffb19"),
}
},
{
$project: {
floorRow: {
$filter: {
input: '$floorRow',
as: 'floorRow',
cond: {
$eq: ['$$floorRow._id', mongoose.Types.ObjectId("5a1bdfbb4d63841c3cb6fc89")],
}
}
}
}
},
])
I have answered this q by myself, but i will keep this post for others who have the same problem.
db.aggregate([
{
$match: {
_id: projectId,
'floorRow._id': floorRowId
}
},
{$unwind: '$floorRow'},
{
$match: {
'floorRow._id': floorRowId
}
},
{$unwind: '$floorRow.unitHouse'},
{
$match: {
'floorRow.unitHouse._id': unitId
}
},
{
$project:{
'floorRow.unitHouse': 1
}
}
])