I have the following data in MongoDB (simplified for what is necessary to my question).
{
_id: 0,
actions: [
{
type: "insert",
data: "abc, quite possibly very very large"
}
]
}
{
_id: 1,
actions: [
{
type: "update",
data: "def"
},{
type: "delete",
data: "ghi"
}
]
}
What I would like is to find the first action type for each document, e.g.
{_id:0, first_action_type:"insert"}
{_id:1, first_action_type:"update"}
(It's fine if the data structured differently, but I need those values present, somehow.)
EDIT: I've tried db.collection.find({}, {'actions.action_type':1}), but obviously that returns all elements of the actions array.
NoSQL is quite new to me. Before, I would have stored all this in two tables in a relational database and done something like SELECT id, (SELECT type FROM action WHERE document_id = d.id ORDER BY seq LIMIT 1) action_type FROM document d.
You can use $slice operator in projection. (but for what you do i am not sure that the order of the array remain the same when you update it. Just to keep in mind))
db.collection.find({},{'actions':{$slice:1},'actions.type':1})
You can also use the Aggregation Pipeline introduced in version 2.2:
db.collection.aggregate([
{ $unwind: '$actions' },
{ $group: { _id: "$_id", first_action_type: { $first: "$actions.type" } } }
])
Using the $arrayElemAt operator is actually the most elegant way, although the syntax may be unintuitive:
db.collection.aggregate([
{ $project: {first_action_type: {$arrayElemAt: ["$actions.type", 0]}
])
Related
Hello guys I'm writing a Message Application with Node.js and Mongoose. I keep datas in mongodb like that:
I want to list users who messaged before so I need to filter my 'Messages' collection but I can't do what exactly I want. If he sent a message to a person I need to take persons name but, if he take a message from a person I need to take persons name however in first situation person name in reciever, in second situation person name in sender. I made a table for explain more easily. I have left table and I need 3 name like second table.(Need to eliminate one John's name)
Sorry, if this problem asked before but I don't know how can I search this problem.
I tried this but it take user name who logged in and duplicate some names.
Message.find({$or: [{sender: req.user.username}, {reciever: req.user.username}]})
One option is to use an aggregation pipeline to create two sets and simply union them:
db.collection.aggregate([
{$match: {$or: [{sender: req.user.username}, {reciever: req.user.username}]}},
{$group: {
_id: 0,
recievers: {$addToSet: "$reciever"},
senders: {$addToSet: "$sender"}
}},
{$project: {
_id: req.user.username,
previousChats: {"$setDifference":
[
{$setUnion: ["$recievers", "$senders"]},
[req.user.username]
]
}
}}
])
See how it works on the playground example
This is a tricky one, but can be solved with a fairly simple aggregation pipeline.
Explanation
On our first stage of the pipeline, we will want to get all the messages sent or received by the user (in our case David), for that we will use a $match stage:
{
$match: {
$or: [
{sender: 'David'},
{receiver: 'David'}
]
}
}
After we found all the messages from or to David, we can start collecting the people he talks to, for that we will use a $group stage and use 2 operations that will help us to achieve this:
$addToSet - This will add all the names to a set. Sets only contain one instance of the same value and ignore any other instance trying to be added to the set of the same value.
$cond - This will be used to add either the receiver or the sender, depending which one of them is David.
The stage will look like this:
{
$group: {
_id: null,
chats: {$addToSet: {$cond: {
if: {$eq: ['$sender', 'David']},
then: '$receiver',
else: '$sender'
}}}
}
}
Combining these 2 stages together will give us the expected result, one document looking like this:
{
"_id": null, // We don't care about this
"chats": [
"John",
"James",
"Daniel"
]
}
Final Solution
Message.aggregate([{
$match: {
$or: [
{
sender: req.user.username
},
{
receiver: req.user.username
}
]
}
}, {
$group: {
_id: null,
chats: {
$addToSet: {
$cond: {
'if': {
$eq: [
'$sender',
req.user.username
]
},
then: '$receiver',
'else': '$sender'
}
}
}
}
}])
Sources
Aggregation
$match aggregation stage
$group aggregation stage
$addToSet operation
$cond operation
I am fairly new to MongoDB and cant seem to find a solution to this problem.
I have a database of documents that has this structure:
{
id: 1
elements: [ {elementId: 1, nr1: 1, nr2: 3}, {elementId:2, nr1:5, nr2: 10} ]
}
I am looking for a query that can add a value nr3 which is for example nr2/nr1 to all the objects in the elements array, so that the resulting document would look like this:
{
id: 1
elements: [ {elementId: 1, nr1: 1, nr2: 3, nr3:3}, {elementId:2, nr1:5, nr2: 10, nr3: 2} ]
}
So I imagine a query along the lines of this:
db.collection.updateOne({id:1}, {$set:{"elements.$[].nr3": nr2/nr1}})
But I cant find how to get the value of nr2 and nr1 of the same object in the array.
I found some similar questions on stackoverflow stating this is not possible, but they were 5+ years old, so I thought maybe they have added support for something like this.
I realize I can achieve this with first querying the document and iterate over the elements-array doing updates along the way, but for the purpose of learning I would love to see if its possible to do this in one query.
You can use update with aggregation pipeline starting from MongoDB v4.2,
$map to iterate loop of elements
divide nr2 with nr1 using $divide
merge current object and new field nr3 using $mergeObjects
db.collection.updateOne(
{ id: 1 },
[{
$set: {
elements: {
$map: {
input: "$elements",
in: {
$mergeObjects: [
"$$this",
{ nr3: { $divide: ["$$this.nr2", "$$this.nr1"] } }
]
}
}
}
}
}]
)
Playground
db.collection.update(
{ id:1},
{ "$set": { "elements.$[elem].nr3":elements.$[elem].nr2/elements.$[elem].nr1} },
{ "multi": true }
);
I guess this should work
I have a schema which has one field named ownerId and a field which is an array named participantIds. In the frontend users can select participants. I'm using these ids to filter documents by querying the participantIds with the $all operator and the list of participantsIds from the frontend. This is perfect except that the participantsIds in the document don't include the ownerId. I thought about using aggregate to add a new field which consists of a list like this one: [participantIds, ownerId] and then querying against this new field with $all and after that delete the field again since it isn't need in the frontend.
How would such a query look like or is there any better way to achieve this behavior? I'm really lost right now since I'm trying to implement this with mongo_dart for the last 3 hours.
This is how the schema looks like:
{
_id: ObjectId(),
title: 'Title of the Event',
startDate: '2020-09-09T00:00:00.000',
endDate: '2020-09-09T00:00:00.000',
startHour: 1,
durationHours: 1,
ownerId: '5f57ff55202b0e00065fbd10',
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13'],
classesIds: [],
categoriesIds: [],
roomsIds: [],
creationTime: '2020-09-10T16:42:14.966',
description: 'Some Desc'
}
Tl;dr I want to query documents with the $all operator on the participantsIds field but the ownerId should be included in this query.
What I want is instead of querying against:
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13']
I want to query against:
participantsIds: ['5f57ff55202b0e00065fbd14', '5f57ff55202b0e00065fbd15', '5f57ff55202b0e00065fbd13', '5f57ff55202b0e00065fbd10']
Having fun here, by the way, it's better to use Joe answer if you are doing the query frequently, or even better a "All" field on insertion.
Additional Notes: Use projection at the start/end, to get what you need
https://mongoplayground.net/p/UP_-IUGenGp
db.collection.aggregate([
{
"$addFields": {
"all": {
$setUnion: [
"$participantsIds",
[
"$ownerId"
]
]
}
}
},
{
$match: {
all: {
$all: [
"5f57ff55202b0e00065fbd14",
"5f57ff55202b0e00065fbd15",
"5f57ff55202b0e00065fbd13",
"5f57ff55202b0e00065fbd10"
]
}
}
}
])
Didn't fully understand what you want to do but maybe this helps:
db.collection.find({
ownerId: "5f57ff55202b0e00065fbd10",
participantsIds: {
$all: ['5f57ff55202b0e00065fbd14',
'5f57ff55202b0e00065fbd15',
'5f57ff55202b0e00065fbd13']
})
You could use the pipeline form of update to either add the owner to the participant list or add a new consolidated field:
db.collection.update({},[{$set:{
allParticipantsIds: {$setUnion: [
"$participantsIds",
["$ownerId"]
]}
}}])
This question already has answers here:
Retrieve only the queried element in an object array in MongoDB collection
(18 answers)
Closed 3 years ago.
In my situation I just need my result but without my objectID in my array.
This is my method :
return Room.findOne(
{
_id: idRoom,
participants: {$elemMatch: {$ne: this.currentUser.profile}},
},
{
'participants.$': 1,
}
)
With elementMatch, the problem is when you found the object, only the first object is returned.
This is my result :
"result": {
"_id": "5da5e77f51e08708b79565e8",
"participants": [
"5da4d5b40cc94f04a7aaad40"
],
And this is the real result I need
"result": {
"_id": "5da5e77f51e08708b79565e8",
"participants": [
"5da4d5b40cc94f04a7aaad40"
"fwnert9248yrhnqwid13982r" // I have another participants
],
And my model is just like this :
const RoomSchema = new Schema({
participants: [{type: Schema.Types.ObjectId,ref: 'Profile'}],
...
}, options)
For others reasons, I can't use aggregate, thank you if you have the solution
So I think you are trying to shape a resultset in mongo with the findOne() method, and any use of the aggregation pipeline framework is out of the question and unavailable for other reasons.
I am not sure this is possible. I believe you will need to perform multiple steps to achieve your desired results. If you can use aggregation pipeline framework here is a pipeline to suit the desired results (I believe)...
db.Room.aggregate(
[
{
"$match": { _id: ObjectId(idRoom)}
},
{
$project: {
"participants": {
$filter: {
input: "$participants",
as: "participant",
cond: {$ne: ["$$participant", this.currentUser.profile] }
}
}
}
}
]
)
...but if you cannot use aggregation pipeline then here is a mongoshell script that accomplishes the task in several steps. The strategy is to capture the whole document by _id then remove the data element from the array then echo the results...
var document = db.Room.findOne( { _id: ObjectId("5da64a62cd63abf99d11f210") } );
document.participants.splice(document.participants.indexOf("5da4d5b40cc94f04a7aaad40"), 1);
document;
Is it possible to rename the name of fields returned in a find query? I would like to use something like $rename, however I wouldn't like to change the documents I'm accessing. I want just to retrieve them differently, something that works like SELECT COORINATES AS COORDS in SQL.
What I do now:
db.tweets.findOne({}, {'level1.level2.coordinates': 1, _id:0})
{'level1': {'level2': {'coordinates': [10, 20]}}}
What I would like to be returned is:
{'coords': [10, 20]}
So basically using .aggregate() instead of .find():
db.tweets.aggregate([
{ "$project": {
"_id": 0,
"coords": "$level1.level2.coordinates"
}}
])
And that gives you the result that you want.
MongoDB 2.6 and above versions return a "cursor" just like find does.
See $project and other aggregation framework operators for more details.
For most cases you should simply rename the fields as returned from .find() when processing the cursor. For JavaScript as an example, you can use .map() to do this.
From the shell:
db.tweets.find({},{'level1.level2.coordinates': 1, _id:0}).map( doc => {
doc.coords = doc['level1']['level2'].coordinates;
delete doc['level1'];
return doc;
})
Or more inline:
db.tweets.find({},{'level1.level2.coordinates': 1, _id:0}).map( doc =>
({ coords: doc['level1']['level2'].coordinates })
)
This avoids any additional overhead on the server and should be used in such cases where the additional processing overhead would outweigh the gain of actual reduction in size of the data retrieved. In this case ( and most ) it would be minimal and therefore better to re-process the cursor result to restructure.
As mentioned by #Neil Lunn this can be achieved with an aggregation pipeline:
And starting Mongo 4.2, the $replaceWith aggregation operator can be used to replace a document by a sub-document:
// { level1: { level2: { coordinates: [10, 20] }, b: 4 }, a: 3 }
db.collection.aggregate(
{ $replaceWith: { coords: "$level1.level2.coordinates" } }
)
// { "coords" : [ 10, 20 ] }
Since you mention findOne, you can also limit the number of resulting documents to 1 as such:
db.collection.aggregate([
{ $replaceWith: { coords: "$level1.level2.coordinates" } },
{ $limit: 1 }
])
Prior to Mongo 4.2 and starting Mongo 3.4, $replaceRoot can be used in place of $replaceWith:
db.collection.aggregate(
{ $replaceRoot: { newRoot: { coords: "$level1.level2.coordinates" } } }
)
As we know, in general, $project stage takes the field names and specifies 1 or 0/true or false to include the fields in the output or not, we also can specify the value against a field instead of true or false to rename the field. Below is the syntax
db.test_collection.aggregate([
{$group: {
_id: '$field_to_group',
totalCount: {$sum: 1}
}},
{$project: {
_id: false,
renamed_field: '$_id', // here assigning a value instead of 0 or 1 / true or false effectively renames the field.
totalCount: true
}}
])
Stages (>= 4.2)
$addFields : {"New": "$Old"}
$unset : {"$Old": 1}