This question already has answers here:
count array occurrences across all documents with mongo
(2 answers)
Closed 7 years ago.
I have the following subdocument:
{
"id":1,
"url":"mysite.com",
"views":
[
{"ip":"1.1.1.1","date":"01-01-2015"},
{"ip":"2.2.2.2","date":"01-01-2015"},
{"ip":"1.1.1.1","date":"01-01-2015"},
{"ip":"1.1.1.1","date":"01-01-2015"}
]
}
and the following query so far:
db.collection.aggregate([
{
"$unwind": "$views"
},
{
"$group": {
"_id": "$views.ip",
"count": {
"$sum": 1
}
}
}
])
that returns the following result:
[
{
"_id":"2.2.2.2",
"count":1
},
{
"_id":"1.1.1.1",
"count":3
}
]
I would like to achieve the following output result, or quite similar:
[
{
"2.2.2.2": 1,
"1.1.1.1": 3
}
]
where each key would be every _id, and it's value their associated count. Is this possible to achieve?
I've tried so far adding at the end of the aggregate pipe a $project document like this one:
{
"$project": {
"_id": 0,
"$_id": "$count"
}
}
but it seems this doesn't work. Is there a way to achieve my desired query result or should I just stick to what I did so far and forget about it because it is not possible to achieve so?
The aggregation framework cannot output a key name from a value. The best place to do this in our application layer. As suggested in the comment, you could use either use the map reduce framework or forEach to achieve this, however, their performance is generally poorer, in particular in sharded environments, see the documentation.
Related
This question already has answers here:
mongo in() clause sort by most matches
(2 answers)
Closed 9 months ago.
I've been stuck with this for a quite while.
I want to filter specials.name field based on two inputs from my form - select and input field. The problem is, whatever I tried, it always chooses on of it.
Is there any way to put if, else block inside $or to make it work?
This is the last thing that I tried, and now it works always on searchInput, which is input field, it does not react on changes in select menu.
{
$match: {
$or: [
{ 'specials.name':{$in :[searchSpecials, searchInput]}},
]
}
}
const searchSpecials = req.body.searchSpecials
const searchInput = req.body.search
If the two inputs are strings, this works as expected:
db.collection.aggregate([
{
$match: {"specials.name": {$in: ["Malinda", "Joseph"]}}
}
])
and returns:
[
{
"_id": ObjectId("5a934e000102030405000000"),
"specials": {
"name": "Malinda"
}
},
{
"_id": ObjectId("5a934e000102030405000002"),
"specials": {
"name": "Joseph"
}
}
]
As you can see here, on the playground.
No need for the $or, since the $in will return all documents that have specials.name which is included in the list of options.
I am fairly new to MongoDB and cant seem to find a solution to this problem.
I have a database of documents that has this structure:
{
id: 1
elements: [ {elementId: 1, nr1: 1, nr2: 3}, {elementId:2, nr1:5, nr2: 10} ]
}
I am looking for a query that can add a value nr3 which is for example nr2/nr1 to all the objects in the elements array, so that the resulting document would look like this:
{
id: 1
elements: [ {elementId: 1, nr1: 1, nr2: 3, nr3:3}, {elementId:2, nr1:5, nr2: 10, nr3: 2} ]
}
So I imagine a query along the lines of this:
db.collection.updateOne({id:1}, {$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", "$$this.nr1"] } }
]
}
}
}
}
}]
)
Playground
db.collection.update(
{ id:1},
{ "$set": { "elements.$[elem].nr3":elements.$[elem].nr2/elements.$[elem].nr1} },
{ "multi": true }
);
I guess this should work
This question already has answers here:
Retrieve only the queried element in an object array in MongoDB collection
(18 answers)
Closed 3 years ago.
In my situation I just need my result but without my objectID in my array.
This is my method :
return Room.findOne(
{
_id: idRoom,
participants: {$elemMatch: {$ne: this.currentUser.profile}},
},
{
'participants.$': 1,
}
)
With elementMatch, the problem is when you found the object, only the first object is returned.
This is my result :
"result": {
"_id": "5da5e77f51e08708b79565e8",
"participants": [
"5da4d5b40cc94f04a7aaad40"
],
And this is the real result I need
"result": {
"_id": "5da5e77f51e08708b79565e8",
"participants": [
"5da4d5b40cc94f04a7aaad40"
"fwnert9248yrhnqwid13982r" // I have another participants
],
And my model is just like this :
const RoomSchema = new Schema({
participants: [{type: Schema.Types.ObjectId,ref: 'Profile'}],
...
}, options)
For others reasons, I can't use aggregate, thank you if you have the solution
So I think you are trying to shape a resultset in mongo with the findOne() method, and any use of the aggregation pipeline framework is out of the question and unavailable for other reasons.
I am not sure this is possible. I believe you will need to perform multiple steps to achieve your desired results. If you can use aggregation pipeline framework here is a pipeline to suit the desired results (I believe)...
db.Room.aggregate(
[
{
"$match": { _id: ObjectId(idRoom)}
},
{
$project: {
"participants": {
$filter: {
input: "$participants",
as: "participant",
cond: {$ne: ["$$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;
I wan't to run a query to get all Articles that have more than 6 com and then sort according length of com list,
for this i doing it:
ArticleModel.objects.filter(com__6__exists=True).order_by('-com.length')[:50]
suppose com is a ListField, but ordering not work, how can i fix it? thanks
Standard queries cannot do this as the "sort" needs to be done on a physical field present in the document. The best way to do this is to actually keep a count of your "list" as another field in the document. That also makes your query more efficient as well as that "counter" field can be indexed, so the basic query becomes ( Raw MongoDB sytax ) :
{ "comLength": { "$gt": 6 } }
If you cannot or do not want to change the document structure then the only way to otherwise sort on the length of your list is to $project it via .aggregate():
ArticleModel._get_collection().aggregate([
{ "$match": { "com.6": { "$exists": true } }},
{ "$project": {
"com": 1,
"otherField": 1,
"comLength": { "$size": "$com" }
}},
{ "$sort": { "comLength": -1 } }
])
And that considers that you have MongoDB 2.6 at least for the use of the $size aggregation operator. If you don't then you have to $unwind and $group in order to calculate the length of arrays:
ArticleModel._get_collection().aggregate([
{ "$match": { "com.6": { "$exists": true } }},
{ "$unwind": "$com" },
{ "$group": {
"_id": "$_id",
"otherField": { "$first": "$otherField" }
"com": { "$push": "$com" },
"comLength": { "$sum": 1 }
}},
{ "$sort": { "comLength": -1 } }
])
So if you are going to go down that route then take a good look at the documentation since you are possibly not used to the raw MongoDB syntax and have been using the query DSL that MongoEngine provdides.
Overall, only the aggregation providers in .aggregate() or .mapReduce() can actually "create a field" that is not present within the document. There is also not test for the "current" length that is available to standard projection or sorting of documents either.
Your best option to to add another field and keep it in sync with the actual array length. But failing that the above shows you the general approach.
If you're creating the database and you know such request will mostly be requested a lot it's recommended to add "com_length" field in A ArticleModel and make it automatically inserted on every save using save() method
add inside of your ArticleModel in models.py
def save(self, *args, **kwargs):
self.com_length = len(self.com)
return super(ArticleModel, self).save(*args, **kwargs)
then for requesting the asked question:
ArticleModel.objects.filter(com__6__exists=True).order_by('-com_length')[:50]
Mongo DB: I'm looking to make one query to return both the first and last element of an array. I realize that I can do this multiple queries, but I would really like to do it with one.
Assume a collection "test" where each objects has an array "arr" of numbers:
db.test.find({},{arr:{$slice: -1},arr:{$slice: 1}});
This will result in the following:
{ "_id" : ObjectId("xxx"), "arr" : [ 1 ] } <-- 1 is the first element
Is there a way to maybe alias the results? Similar to what the mysql AS keyword would allow in a query?
This is not possible at the moment but will be with the Aggregation Framework that's in development now if I understand your functional requirement correctly.
You have to wonder about your schema if you have this requirement in the first place though. Are you sure there isn't a more elegant way to get this to work by changing your schema accordingly?
This can be done with the aggregation framework using the operators $first and $last as follows:
db.test.aggregate([
{ '$addFields': {
'firstElem': { '$first': '$arr' },
'lastElem': { '$last': '$arr' }
} }
])
or using $slice as
db.test.aggregate([
{ '$addFields': {
'firstElem': { '$slice': [ '$arr', 1 ] },
'lastElem': { '$slice': [ '$arr', -1 ] }
} }
])