Related
Need help to sort these documents:
const docs = Docs.find(
{
'publishedOn.profileId': groupProfile._id,
},
{ sort: { ??? }}
);
I need to find documents which has defined 'publishedOn.profileId' and
sort by 'awards.type' = 'challengeWinner' and by its 'awards.score'
Not all document has awards.type = 'challengeWinner'. I need to
take on the top 'awards.score' = 1, then 2, then 3 and then the rest by 'writtenDate'.
I have no idea how to fix it. Is it possible?
[
{
"_id" : "5FW9EDW8gi3M8R7XK",
"createdAt" : ISODate("2021-06-13T00:11:48.638Z"),
"title" : "My solution",
"writtenDateType" : 4,
"writtenDate" : ISODate("2021-06-13T00:00:00.000Z"),
"userId" : "dC35hwe6XMRhvqWBv",
"publishedOn" : [
{
"profileId" : "36oPw2zxYCpKxfiu2",
"publishedDate" : ISODate("2021-06-13T00:11:48.787Z"),
"userId" : "dC35hwe6XMRhvqWBv"
},
{
"profileId" : "9y2RwJpzzyk29ApiC",
"userId" : "dC35hwe6XMRhvqWBv",
"publishedDate" : ISODate("2021-06-13T00:16:01.529Z")
}
],
"awards" : [
{
"type" : "topPoem",
"score" : 5,
"addedAt" : ISODate("2021-06-24T23:04:10.454Z"),
"updatedAt" : ISODate("2021-06-25T23:30:00.069Z")
},
{
"type" : "challengeWinner",
"score" : 2,
"challengeId" : "9y2RwJpzzyk29ApiC",
"addedAt" : ISODate("2021-06-24T23:04:10.454Z"),
"updatedAt" : ISODate("2021-06-25T23:30:00.069Z")
}
]
},
{
"_id" : "upzvo8BeHyQ9r9Yfv",
"createdAt" : ISODate("2021-06-19T15:35:13.716Z"),
"title" : "Briches",
"writtenDateType" : 2,
"writtenDate" : ISODate("2003-01-01T00:00:00.000Z"),
"userId" : "A32228XMuZqxFe4Kz",
"publishedOn" : [
{
"profileId" : "MLGkCtNyZ64bGKedG",
"publishedDate" : ISODate("2021-06-19T15:35:13.861Z"),
"userId" : "A32228XMuZqxFe4Kz"
},
{
"profileId" : "9y2RwJpzzyk29ApiC",
"userId" : "A32228XMuZqxFe4Kz",
"publishedDate" : ISODate("2021-06-19T15:35:36.280Z")
}
],
"awards" : [
{
"type" : "challengeWinner",
"score" : 1,
"challengeId" : "9y2RwJpzzyk29ApiC",
"addedAt" : ISODate("2021-06-24T22:59:00.948Z"),
"updatedAt" : ISODate("2021-06-25T23:30:00.067Z"),
"claps" : 19,
"clapsUsers" : 4
},
{
"type" : "suggestedHomepage",
"score" : 1,
"addedAt" : ISODate("2021-06-24T22:59:59.981Z"),
"updatedAt" : ISODate("2021-06-24T22:59:59.981Z")
}
]
}
]
I just learned and tried to solve your problem. I used aggregate to do the filter in your data.
First I selected all the items which $match the `publishedOn.profileId".
Then, I $project(ed) the items that are needed. In this case, I took the writtenDate and the matching awards.
In order to choose the needed value from awards, I $filter (ed) the award type.
Last, I did $sort for the award score first and then writtenDate,
db.collection.aggregate([
{
"$match": {
"publishedOn.profileId": "9y2RwJpzzyk29ApiC"
}
},
{
"$project": {
"writtenDate": 1,
"awards": {
"$filter": {
"input": "$awards",
"as": "award",
"cond": {
"$eq": [
"$$award.type",
"challengeWinner"
]
}
}
}
}
},
{
"$sort": {
"awards.score": 1,
"writtenDate": 1,
}
}
])
Working of above query: https://mongoplayground.net/p/MzWQCR2Gshg
Happy Coding !!!
I have the following document and want to update state
Document ID: ObjectId("5a4e5a448b70d50e34d204a5")
Target ID: ObjectId("5a4e5a438b70d50e34d203ea")
I have no idea how to update the state to e.g. 4
{
"_id" : ObjectId("5a4e5a448b70d50e34d204a5"),
"name" : "Wirtschaftsdienst",
"date" : ISODate("2012-10-07T00:00:00.000Z"),
"comment" : null,
"tasks" : [
{
"name" : "Speisen und Getränke",
"sections" : [
{
"start" : 46800,
"end" : 72000,
"entirely" : true,
"assistants" : [
{
"assistant" : {
"_id" : ObjectId("5a4e5a438b70d50e34d203ea")
},
"state" : 3
},
{
"assistant" : {
"_id" : ObjectId("5a4e5a438b70d50e34d203f4")
},
"state" : 3
}
]
}
]
}
]
}
Use positional operator $[] along with arrayFilters to get your job done!
Try this query:
db.collection.update(
{"_id" : ObjectId("5a4e5a448b70d50e34d204a5")},
{$set: {"tasks.$[].sections.$[].assistants.$[element].state":4}},
{arrayFilters: [ {"element.assistant":{"_id" :
ObjectId("5a4e5a438b70d50e34d203ea")} }
], multi:true}
)
And the output is:
/* 1 */
{
"_id" : ObjectId("5a4e5a448b70d50e34d204a5"),
"name" : "Wirtschaftsdienst",
"date" : ISODate("2012-10-07T00:00:00.000Z"),
"comment" : null,
"tasks" : [
{
"name" : "Speisen und Getränke",
"sections" : [
{
"start" : 46800,
"end" : 72000,
"entirely" : true,
"assistants" : [
{
"assistant" : {
"_id" : ObjectId("5a4e5a438b70d50e34d203ea")
},
"state" : 4.0
},
{
"assistant" : {
"_id" : ObjectId("5a4e5a438b70d50e34d203f4")
},
"state" : 3.0
}
]
}
]
}
]
}
I have datas of following format collection(projects) inside my database:
{ "_id" : ObjectId("5981a80f223e491a58230e5d"), "id" : 2, "name" : "gbqplhlqxzwl", "managerId" : 65151, "startDate" : "03.11.1999", "finishDate" : "02.01.2003", "projectStatus" : "POSTPONED", "participants" : [ ], "estimatedBudget" : 6017891.811079914 }
{ "_id" : ObjectId("5981a80f223e491a58230e5e"), "id" : 3, "name" : "erfekfsdgryu", "managerId" : 83749, "startDate" : "07.07.2007", "finishDate" : "26.12.2027", "projectStatus" : "POSTPONED", "participants" : [ 19229, 81856, 79270, 5509, 70344, 39424 ], "estimatedBudget" : 3086213.8981674756 }
{ "_id" : ObjectId("5981a80f223e491a58230e5f"), "id" : 1, "name" : "jvbzobhppntd", "managerId" : 18925, "startDate" : "29.04.1999", "finishDate" : "13.10.2008", "projectStatus" : "OPEN", "participants" : [ 46100, 96968, 6676, 56121, 4716, 68901, 43990, 48587, 62547, 30292, 65153, 17551, 27083, 20261, 27097, 50036, 86585, 69890, 18790, 22592, 60774, 93709, 78471, 27157, 4328, 36501, 47296, 16831 ], "estimatedBudget" : 3581496.7068344904 }
{ "_id" : ObjectId("5981a80f223e491a58230e60"), "id" : 4, "name" : "cdspkkqwvwld", "managerId" : 62042, "startDate" : "13.03.1998", "finishDate" : "20.06.2007", "projectStatus" : "OPEN", "participants" : [ 53480, 60897, 23677, 22064, 60807, 66637, 84609, 28378, 87143, 27675, 79283, 94992, 20429, 48769, 91671, 41747, 21651, 91134, 41684, 57228, 51949, 18756, 45679, 87781, 67287, 6902, 27526 ], "estimatedBudget" : 2126283.953787842 }
....
I need to find the busiest employee and list all his projects.
participants array contains employee ids who participate in the project.
I use the following query to find the busiest employee:
db.projects.aggregate(
{
$unwind: '$participants'
},
{
$addFields: {
count: 1
}
},
{
$group: {
_id : '$participants',
participation_count : {
'$sum':'$count'
}
}
},
{
$sort:{participation_count:-1}
},
{
$limit:1
}
)
and this work correctly. But I have no ideas how to list all his projects.
any ideas?
db.projects.aggregate(
[
{
$unwind: '$participants'
},
{
$addFields: {
count: 1
}
},
{
$group: {
_id : '$participants',
participation_count : {'$sum':'$count'},
projectId : {$push: '$id'}
}
},
{
$sort:{participation_count:-1}
},
{
$limit:1
}
],
{
allowDiskUse:true
}
)
I am trying to port an existing SQL schema into Mongo.
We have document tables, with sometimes several times the same document, with a different revision but the same reference. I want to get only the latest revisions of the documents.
A sample input data:
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC305",
"code" : "305-D",
"title" : "Document 305",
"creationdate" : ISODate("2011-11-24T15:13:28.887Z"),
"creator" : "X"
},
{
"Uid" : "xxx",
"status" : "COMMENTED",
"reference" : "DOC306",
"code" : "306-A",
"title" : "Document 306",
"creationdate" : ISODate("2011-11-28T07:23:18.807Z"),
"creator" : "X"
},
{
"Uid" : "xxx",
"status" : "COMMENTED",
"reference" : "DOC306",
"code" : "306-B",
"title" : "Document 306",
"creationdate" : ISODate("2011-11-28T07:26:49.447Z"),
"creator" : "X"
},
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC501",
"code" : "501-A",
"title" : "Document 501",
"creationdate" : ISODate("2011-11-19T06:30:35.757Z"),
"creator" : "X"
},
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC501",
"code" : "501-B",
"title" : "Document 501",
"creationdate" : ISODate("2011-11-19T06:40:32.957Z"),
"creator" : "X"
}
Given this data, I want this result set (sometimes I want only the last revision, sometimes I want all revisions with an attribute telling me whether it's the latest):
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC305",
"code" : "305-D",
"title" : "Document 305",
"creationdate" : ISODate("2011-11-24T15:13:28.887Z"),
"creator" : "X",
"lastrev" : true
},
{
"Uid" : "xxx",
"status" : "COMMENTED",
"reference" : "DOC306",
"code" : "306-B",
"title" : "Document 306",
"creationdate" : ISODate("2011-11-28T07:26:49.447Z"),
"creator" : "X",
"lastrev" : true
},
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC501",
"code" : "501-B",
"title" : "Document 501",
"creationdate" : ISODate("2011-11-19T06:40:32.957Z"),
"creator" : "X",
"lastrev" : true
}
I already have a bunch of filters, sorting, and skip/limit (for pagination of data), so the final result set should be mindful of these constraints.
The current "find" query (built with the .Net driver), which filters fine but gives me all revisions of each document:
coll.find(
{ "$and" : [
{ "$or" : [
{ "deletedid" : { "$exists" : false } },
{ "deletedid" : null }
] },
{ "$or" : [
{ "taskid" : { "$exists" : false } },
{ "taskid" : null }
] },
{ "objecttypeuid" : { "$in" : ["xxxxx"] } }
] },
{ "_id" : 0, "Uid" : 1, "lastrev" : 1, "title" : 1, "code" : 1, "creator" : 1, "owner" : 1, "modificator" : 1, "status" : 1, "reference": 1, "creationdate": 1 }
).sort({ "creationdate" : 1 }).skip(0).limit(10);
Using another question, I have been able to build this aggregation, which gives me the latest revision of each document, but with not enough attributes in the result:
coll.aggregate([
{ $sort: { "creationdate": 1 } },
{
$group: {
"_id": "$reference",
result: { $last: "$creationdate" },
creationdate: { $last: "$creationdate" }
}
}
]);
I would like to integrating the aggregate with the find query.
I have found the way to mix aggregation and filtering:
coll.aggregate(
[
{ $match: {
"$and" : [
{ "$or" : [
{ "deletedid" : { "$exists" : false } },
{ "deletedid" : null }
] },
{ "$or" : [
{ "taskid" : { "$exists" : false } },
{ "taskid" : null }
] },
{ "objecttypeuid" : { "$in" : ["xxx"] } }
]
}
},
{ $sort: { "creationdate": 1 } },
{ $group: {
"_id": "$reference",
"doc": { "$last": "$$ROOT" }
}
},
{ $sort: { "doc.creationdate": 1 } },
{ $skip: skip },
{ $limit: limit }
],
{ allowDiskUse: true }
);
For each result node, this gives me a "doc" node with the document data. It has too much data still (it's missing projections), but it's a start.
Translated in .Net:
FilterDefinitionBuilder<BsonDocument> filterBuilder = Builders<BsonDocument>.Filter;
FilterDefinition<BsonDocument> filters = filterBuilder.Empty;
filters = filters & (filterBuilder.Not(filterBuilder.Exists("deletedid")) | filterBuilder.Eq("deletedid", BsonNull.Value));
filters = filters & (filterBuilder.Not(filterBuilder.Exists("taskid")) | filterBuilder.Eq("taskid", BsonNull.Value));
foreach (var f in fieldFilters) {
filters = filters & filterBuilder.In(f.Key, f.Value);
}
var sort = Builders<BsonDocument>.Sort.Ascending(orderby);
var group = new BsonDocument {
{ "_id", "$reference" },
{ "doc", new BsonDocument("$last", "$$ROOT") }
};
var aggregate = coll.Aggregate(new AggregateOptions { AllowDiskUse = true })
.Match(filters)
.Sort(sort)
.Group(group)
.Sort(sort)
.Skip(skip)
.Limit(rows);
return aggregate.ToList();
I'm pretty sure there are better ways to do this, though.
You answer is pretty close. Instead of $last, $max is better.
About $last operator:
Returns the value that results from applying an expression to the last document in a group of documents that share the same group by a field. Only meaningful when documents are in a defined order.
Get the last revision in each group, see code below in mongo shell:
db.collection.aggregate([
{
$group: {
_id: '$reference',
doc: {
$max: {
"creationdate" : "$creationdate",
"code" : "$code",
"Uid" : "$Uid",
"status" : "$status",
"title" : "$title",
"creator" : "$creator"
}
}
}
},
{
$project: {
_id: 0,
Uid: "$doc.Uid",
status: "$doc.status",
reference: "$_id",
code: "$doc.code",
title: "$doc.title",
creationdate: "$doc.creationdate",
creator: "$doc.creator"
}
}
]).pretty()
The output as your expect:
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC501",
"code" : "501-B",
"title" : "Document 501",
"creationdate" : ISODate("2011-11-19T06:40:32.957Z"),
"creator" : "X"
}
{
"Uid" : "xxx",
"status" : "COMMENTED",
"reference" : "DOC306",
"code" : "306-B",
"title" : "Document 306",
"creationdate" : ISODate("2011-11-28T07:26:49.447Z"),
"creator" : "X"
}
{
"Uid" : "xxx",
"status" : "ACCEPTED",
"reference" : "DOC305",
"code" : "305-D",
"title" : "Document 305",
"creationdate" : ISODate("2011-11-24T15:13:28.887Z"),
"creator" : "X"
}
Is it possible to use the aggregation framework to group by a specific element of an array?
Such that with documents like this:
{
name: 'Russell',
favourite_foods: [
{ name: 'Pizza', type: 'Four Cheeses' },
{ name: 'Burger', type: 'Veggie'}
],
height: 6
}
I could get a distinct list of top favourite foods (ie. foods at index 0) along with the height of the tallest person who's top favourite food that is?
Something like this (although it doesn't work as the array index access dot notation doesn't seem to work in the aggregation framework):
db.people.aggregate([
{ $group : { _id: "$favourite_foods.0.name", max_height: { $max : "$height" } } }
])
Seems like you are relying on the favorite food for each person being first in the array. If so, there is an aggregation framework operator you can take advantage of.
Here is the pipeline you can use:
db.people.aggregate(
[
{
"$unwind" : "$favourite_foods"
},
{
"$group" : {
"_id" : {
"name" : "$name",
"height" : "$height"
},
"faveFood" : {
"$first" : "$favourite_foods"
}
}
},
{
"$group" : {
"_id" : "$faveFood.name",
"height" : {
"$max" : "$_id.height"
}
}
}
])
On this sample dataset:
> db.people.find().pretty()
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 6
}
{
"_id" : ObjectId("5088950bd4197aa2b9490742"),
"name" : "Lucy",
"favourite_foods" : [
{
"name" : "Pasta",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 5.5
}
{
"_id" : ObjectId("5088951dd4197aa2b9490743"),
"name" : "Landy",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 5
}
{
"_id" : ObjectId("50889541d4197aa2b9490744"),
"name" : "Augie",
"favourite_foods" : [
{
"name" : "Sushi",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 6.2
}
You get these results:
{
"result" : [
{
"_id" : "Pasta",
"height" : 5.5
},
{
"_id" : "Pizza",
"height" : 6
},
{
"_id" : "Sushi",
"height" : 6.2
}
],
"ok" : 1
}
Looks like it isn't currently possible to extract a specific element from an array in aggregation:
https://jira.mongodb.org/browse/SERVER-4589
JUST add more information about the result after using "$wind":
DOCUMENT :
> db.people.find().pretty()
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 6
},
...
AGGREAGATION :
db.people.aggregate([{
$unwind: "$favourite_foods"
}]);
RESULT :
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" :{
"name" : "Pizza",
"type" : "Four Cheeses"
},
"height" : 6
},
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : {
"name" : "Burger",
"type" : "Veggie"
},
"height" : 6
}
In Addition:
If there are more than two array fields in one collection record,
we can use "$project" stage to specify the array field.
db.people.aggregate([
{
$project:{
"favourite_foods": 1
}
},
{
$unwind: "$favourite_foods"
}
]);
I think you can make use of the $project and $unwind operators (let me know if this isn't what you're trying to accomplish):
> db.people.aggregate(
{$unwind: "$favourite_foods"},
{$project: {food : "$favourite_foods", height: 1}},
{$group : { _id: "$food", max_height: { $max : "$height" } } })
{
"result" : [
{
"_id" : {
"name" : "Burger",
"type" : "Veggie"
},
"max_height" : 6
},
{
"_id" : {
"name" : "Pizza",
"type" : "Four Cheeses"
},
"max_height" : 6
}
],
"ok" : 1
}
http://docs.mongodb.org/manual/applications/aggregation/
Since mongoDB version 3.2 You can simply use $arrayElemAt and $max:
db.collection.aggregate([
{
$set: {favourite_foods: {$arrayElemAt: ["$favourite_foods", 0]}}
},
{
$group: {
_id: "$favourite_foods.name",
maxHeight: {$max: "$height"}
}
}
])
Playground example