This is my mongodb document
{
"_id" : "a3s6HzG9swNB3bQ78",
"comments" : [
{
"cmt_text" : "opp",
"vCount" : 2,
},
{
"cmt_text" : "o2",
"vCount" : 5,
},
{
"cmt_text" : "o3",
"vCount" : 3,
}
],
"question" : "test ques 3"
}
i want to sort the result using the vCount field how to achieve this
i tried the following but seems to be not working
Coll.findOne( {_id:this._id},{sort:{ "comments.vCount" : 1 }});
Coll.findOne( {_id:this._id},{sort:{ "comments.$.vCount" : 1 }});
anyone have idea about this???
EDIT
we are returning only one document and i want to display that document comment array values according to the vCount. my code
{{#each all_comments.comments}}
<br>{{cmt_text}}</p>
{{/each}}
i want to display like below
o2
o3
opp
EDIT
this is working fine in shell
db.testCol.aggregate([
{ $unwind: "$comments" },
{ $group: { _id: { id:"$_id", vcount:"$comments.vCount"} } },
{ $sort: { "_id.vcount":1 }}
])
why is it not working in my meteor app it says
error:object has no method aggregate
This is correct:
Coll.findOne( { _id: this._id }, { sort: { 'comments.vCount' : 1 } } );
No $ in front of sort.
EDIT:
If you want to sort the nested array, look here.
aggregate isn't currently available on the client. You can just do a findOne, extract the comments array, and return a sorted version to the template. For example:
Template.allComments.helpers({
comments: function() {
var coll = Coll.findOne(this._id);
return _.sortBy(coll.comments, function(comment) {
return -comment.vcount;
});
}
});
<template name="allComments">
{{#each comments}}
<br>{{cmt_text}}</p>
{{/each}}
</template>
You should learn aggregation for dealing with subdocuments, arrays and also with subdocuments in arrays. For your sort question this could work.
db.testCol.aggregate([
{ $unwind: "$comments" },
{ $group: { _id: { id:"$_id", vcount:"$comments.vCount"} } },
{ $sort: { "_id.vcount":1 }}
])
EDIT: In according to your edit you could add $project operator like;
db.testCol.aggregate([
{ $unwind: "$comments" },
{ $group: { _id: { id:"$_id", vcount:"$comments.vCount", text:"$comments.cmt_text"} } },
{ $sort: { "_id.vcount":1 }},
{ $project: { text: "$_id.text", _id:0}}
])
Related
I am using mongoDB, but I am a complete beginner. I have two different queries where I want to combine them both into one output (I'm hoping the answer is a single query)
Query 1:
db.fin.aggregate([
{ "$match": { "misc.incident_characteristics": { "$not": /Officer Involved Incident/ } } },
{ $group: {
_id: "NonOfficerInvolved",
nInjured: { $avg: "$casualties.n_injured" },
nKilled: { $avg: "$casualties.n_killed" }
}
}
])
Which returns
{ "_id" : "NonOfficerInvolved", "nInjured" : 0.5048153043227224, "nKilled" : 0.24339953718948618 }
Query 2:
db.fin.aggregate([
{ $match: { "misc.incident_characteristics": "Officer Involved Incident" } },
{ $group: {
_id: "OfficerInvolved",
nInjured: { $avg: "$casualties.n_injured" },
nKilled: { $avg: "$casualties.n_killed" }
}
}
])
Which returns
{ "_id" : "OfficerInvolved", "nInjured" : 0.3599233845980508, "nKilled" : 0.358965692073686 }
I would like to get the result of both into one table as seen below. Is it possible to do this in one query?
{ "_id" : "NonOfficerInvolved", "nInjured" : 0.5048153043227224, "nKilled" : 0.24339953718948618 }
{ "_id" : "OfficerInvolved", "nInjured" : 0.3599233845980508, "nKilled" : 0.358965692073686 }
Yes, you can definitely do that in one query.
Instead of just matching against the magic string, store the match result in a new field as a boolean or string, then group on that new field.
db.fin.aggregate([
{ "$addFields": { type:{
$cond:[
{$eq:["$misc.incident_characteristics","Officer Involved Incident"]},
"OfficerInvolved",
"NonOfficerInvolved"
]
}}},
{ $group: {
_id: "$type",
nInjured: { $avg: "$casualties.n_injured" },
nKilled: { $avg: "$casualties.n_killed" }
}
}
])
To match using a regular expression, replace the line
{$eq:["$misc.incident_characteristics","Officer Involved Incident"]},
with
{$regexMatch:{
input:"$misc.incident_characteristics",
regex:"Officer Involved Incident"
options:"i"
}},
I want to return Object as a field in my Aggregation result similar to the solution in this question. However in the solution mentioned above, the Aggregation results in an Array of Objects with just one item in that array, not a standalone Object. For example, a query like the following with a $push operation
$group:{
_id: "$publisherId",
'values' : { $push:{
newCount: { $sum: "$newField" },
oldCount: { $sum: "$oldField" } }
}
}
returns a result like this
{
"_id" : 2,
"values" : [
{
"newCount" : 100,
"oldCount" : 200
}
]
}
}
not one like this
{
"_id" : 2,
"values" : {
"newCount" : 100,
"oldCount" : 200
}
}
}
The latter is the result that I require. So how do I rewrite the query to get a result like that? Is it possible or is the former result the best I can get?
You don't need the $push operator, just add a final $project pipeline that will create the embedded document. Follow this guideline:
var pipeline = [
{
"$group": {
"_id": "$publisherId",
"newCount": { "$sum": "$newField" },
"oldCount": { "$sum": "$oldField" }
}
},
{
"$project" {
"values": {
"newCount": "$newCount",
"oldCount": "$oldCount"
}
}
}
];
db.collection.aggregate(pipeline);
I have the following collection
{
"_id" : ObjectId("57315ba4846dd82425ca2408"),
"myarray" : [
{
userId : ObjectId("570ca5e48dbe673802c2d035"),
point : 5
},
{
userId : ObjectId("613ca5e48dbe673802c2d521"),
point : 2
},
]
}
These are my questions
I want to push into myarray if userId doesn't exist, it should be appended to myarray. If userId exists, it should be updated to point.
I found this
db.collection.update({
_id : ObjectId("57315ba4846dd82425ca2408"),
"myarray.userId" : ObjectId("570ca5e48dbe673802c2d035")
}, {
$set: { "myarray.$.point": 10 }
})
But if userId doesn't exist, nothing happens.
and
db.collection.update({
_id : ObjectId("57315ba4846dd82425ca2408")
}, {
$push: {
"myarray": {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
})
But if userId object already exists, it will push again.
What is the best way to do this in MongoDB?
Try this
db.collection.update(
{ _id : ObjectId("57315ba4846dd82425ca2408")},
{ $pull: {"myarray.userId": ObjectId("570ca5e48dbe673802c2d035")}}
)
db.collection.update(
{ _id : ObjectId("57315ba4846dd82425ca2408")},
{ $push: {"myarray": {
userId:ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}}
)
Explination:
in the first statment $pull removes the element with userId= ObjectId("570ca5e48dbe673802c2d035") from the array on the document where _id = ObjectId("57315ba4846dd82425ca2408")
In the second one $push inserts
this object { userId:ObjectId("570ca5e48dbe673802c2d035"), point: 10 } in the same array.
The accepted answer by Flying Fisher is that the existing record will first be deleted, and then it will be pushed again.
A safer approach (common sense) would be to try to update the record first, and if that did not find a match, insert it, like so:
// first try to overwrite existing value
var result = db.collection.update(
{
_id : ObjectId("57315ba4846dd82425ca2408"),
"myarray.userId": ObjectId("570ca5e48dbe673802c2d035")
},
{
$set: {"myarray.$.point": {point: 10}}
}
);
// you probably need to modify the following if-statement to some async callback
// checking depending on your server-side code and mongodb-driver
if(!result.nMatched)
{
// record not found, so create a new entry
// this can be done using $addToSet:
db.collection.update(
{
_id: ObjectId("57315ba4846dd82425ca2408")
},
{
$addToSet: {
myarray: {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
}
);
// OR (the equivalent) using $push:
db.collection.update(
{
_id: ObjectId("57315ba4846dd82425ca2408"),
"myarray.userId": {$ne: ObjectId("570ca5e48dbe673802c2d035"}}
},
{
$push: {
myarray: {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
}
);
}
This should also give (common sense, untested) an increase in performance, if in most cases the record already exists, only the first query will be executed.
There is a option called update documents with aggregation pipeline starting from MongoDB v4.2,
check condition $cond if userId in myarray.userId or not
if yes then $map to iterate loop of myarray array and check condition if userId match then merge with new document using $mergeObjects
if no then $concatArrays to concat new object and myarray
let _id = ObjectId("57315ba4846dd82425ca2408");
let updateDoc = {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
};
db.collection.update(
{ _id: _id },
[{
$set: {
myarray: {
$cond: [
{ $in: [updateDoc.userId, "$myarray.userId"] },
{
$map: {
input: "$myarray",
in: {
$mergeObjects: [
"$$this",
{
$cond: [
{ $eq: ["$$this.userId", updateDoc.userId] },
updateDoc,
{}
]
}
]
}
}
},
{ $concatArrays: ["$myarray", [updateDoc]] }
]
}
}
}]
)
Playground
Unfortunately "upsert" operation is not possible on embedded array. Operators simply do not exist so that this is not possible in a single statement.Hence you must perform two update operations in order to do what you want. Also the order of application for these two updates is important to get desired result.
I haven't found any solutions based on a one atomic query. Instead there are 3 ways based on a sequence of two queries:
always $pull (to remove the item from array), then $push (to add the updated item to array)
db.collection.update(
{ _id : ObjectId("57315ba4846dd82425ca2408")},
{ $pull: {"myarray.userId": ObjectId("570ca5e48dbe673802c2d035")}}
)
db.collection.update(
{ _id : ObjectId("57315ba4846dd82425ca2408")},
{
$push: {
"myarray": {
userId:ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
}
)
try to $set (to update the item in array if exists), then get the result and check if the updating operation successed or if a $push needs (to insert the item)
var result = db.collection.update(
{
_id : ObjectId("57315ba4846dd82425ca2408"),
"myarray.userId": ObjectId("570ca5e48dbe673802c2d035")
},
{
$set: {"myarray.$.point": {point: 10}}
}
);
if(!result.nMatched){
db.collection.update({_id: ObjectId("57315ba4846dd82425ca2408")},
{
$addToSet: {
myarray: {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
);
always $addToSet (to add the item if not exists), then always $set to update the item in array
db.collection.update({_id: ObjectId("57315ba4846dd82425ca2408")},
myarray: { $not: { $elemMatch: {userId: ObjectId("570ca5e48dbe673802c2d035")} } } },
{
$addToSet : {
myarray: {
userId: ObjectId("570ca5e48dbe673802c2d035"),
point: 10
}
}
},
{ multi: false, upsert: false});
db.collection.update({
_id: ObjectId("57315ba4846dd82425ca2408"),
"myArray.userId": ObjectId("570ca5e48dbe673802c2d035")
},
{ $set : { myArray.$.point: 10 } },
{ multi: false, upsert: false});
1st and 2nd way are unsafe, so transaction must be established to avoid two concurrent requests could push the same item generating a duplicate.
3rd way is safer. the $addToSet adds only if the item doesn't exist, otherwise nothing happens. In case of two concurrent requests, only one of them adds the missing item to the array.
Possible solution with aggregation pipeline:
db.collection.update(
{ _id },
[
{
$set: {
myarray: { $filter: {
input: '$myarray',
as: 'myarray',
cond: { $ne: ['$$myarray.userId', ObjectId('570ca5e48dbe673802c2d035')] },
} },
},
},
{
$set: {
myarray: {
$concatArrays: [
'$myarray',
[{ userId: ObjectId('570ca5e48dbe673802c2d035'), point: 10 },
],
],
},
},
},
],
);
We use 2 stages:
filter myarray (= remove element if userId exist)
concat filtered myarray with new element;
When you want update or insert value in array try it
Object in db
key:name,
key1:name1,
arr:[
{
val:1,
val2:1
}
]
Query
var query = {
$inc:{
"arr.0.val": 2,
"arr.0.val2": 2
}
}
.updateOne( { "key": name }, query, { upsert: true }
key:name,
key1:name1,
arr:[
{
val:3,
val2:3
}
]
In MongoDB 3.6 it is now possible to upsert elements in an array.
array update and create don't mix in under one query, if you care much about atomicity then there's this solution:
normalise your schema to,
{
"_id" : ObjectId("57315ba4846dd82425ca2408"),
userId : ObjectId("570ca5e48dbe673802c2d035"),
point : 5
}
You could use a variation of the .forEach/.updateOne method I currently use in mongosh CLI to do things like that. In the .forEach, you might be able to set all of your if/then conditions that you mentioned.
Example of .forEach/.updateOne:
let medications = db.medications.aggregate([
{$match: {patient_id: {$exists: true}}}
]).toArray();
medications.forEach(med => {
try {
db.patients.updateOne({patient_id: med.patient_id},
{$push: {medications: med}}
)
} catch {
console.log("Didn't find match for patient_id. Could not add this med to a patient.")
}
})
This may not be the most "MongoDB way" to do it, but it definitely works and gives you the freedom of javascript to do things within the .forEach.
I tried searching on here but couldn't really find what I need. I have documents like this:
{
appletype:Granny,
color:Green,
datePicked:2015-01-26,
dateRipe:2015-01-24,
numPicked:3
},
{
appletype:Granny,
color:Green,
datePicked:2015-01-01,
dateRipe:2014-12-28,
numPicked:6
}
I would like to return only those apples picked latest, will all fields. I want my query to return me the first document only essentially. When I try to do:
db.collection.aggregate([
{ $match : { "appletype" : "Granny" } },
{ $sort : { "datePicked" : 1 } },
{ $group : { "_id" : { "appletype" : "$appletype" },
"datePicked" : { $max : "$datePicked" } },
])
It does return me all the apples picked latest, however with only appletype:Granny and datePicked:2015-01-26. I need the remaining fields. I tries using $project and adding all the fields, but it didn't get me what I needed. Also, when I added the other fields to the group, since datePicked is unique, it returned both records.
How can I go about returning all fields, for only the latest datePicked?
Thanks!
From your description, it sounds like you want one document for each of the types of apple in your collection and showing the document with the most recent datePicked value.
Here is an aggregate query for that:
db.collection.aggregate([
{ $sort: { "datePicked": -1 },
{ $group: { _id: "$appletype", color: { $first: "$color" }, datePicked: { $first: "$datePicked" }, dateRipe: { $first: "$dateRipe" }, numPicked: { $first: "$numPicked" } } },
{ $project: { _id: 0, color: 1, datePicked: 1, dateRipe: 1, numPicked: 1, appletype: "$_id" } }
])
But then based on the aggregate query you've written, it looks like you're trying to get this:
db.collection.find({appletype: "Granny"}).sort({datePicked: -1}).limit(1);
I am trying to write an aggregation to identify accounts that use multiple payment sources. Typical data would be.
{
account:"abc",
vendor:"amazon",
}
...
{
account:"abc",
vendor:"overstock",
}
Now, I'd like to produce a list of accounts similar to this
{
account:"abc",
vendorCount:2
}
How would I write this in Mongo's aggregation framework
I figured this out by using the $addToSet and $unwind operators.
Mongodb Aggregation count array/set size
db.collection.aggregate([
{
$group: { _id: { account: '$account' }, vendors: { $addToSet: '$vendor'} }
},
{
$unwind:"$vendors"
},
{
$group: { _id: "$_id", vendorCount: { $sum:1} }
}
]);
Hope it helps someone
I think its better if you execute query like following which will avoid unwind
db.t2.insert({_id:1,account:"abc",vendor:"amazon"});
db.t2.insert({_id:2,account:"abc",vendor:"overstock"});
db.t2.aggregate([
{ $group : { _id : { "account" : "$account", "vendor" : "$vendor" }, number : { $sum : 1 } } },
{ $group : { _id : "$_id.account", number : { $sum : 1 } } }
]);
Which will show you following result which is expected.
{ "_id" : "abc", "number" : 2 }
You can use sets
db.test.aggregate([
{$group: {
_id: "$account",
uniqueVendors: {$addToSet: "$vendor"}
}},
{$project: {
_id: 1,
vendorsCount: {$size: "$uniqueVendors"}
}}
]);
I do not see why somebody would have to use $group twice
db.t2.aggregate([ { $group: {"_id":"$account" , "number":{$sum:1}} } ])
This will work perfectly fine.
This approach doesn't make use of $unwind and other extra operations. Plus, this won't affect anything if new things are added into the aggregation. There's a flaw in the accepted answer. If you have other accumulated fields in the $group, it would cause issues in the $unwind stage of the accepted answer.
db.collection.aggregate([{
"$group": {
"_id": "$account",
"vendors": {"$addToSet": "$vendor"}
}
},
{
"$addFields": {
"vendorCount": {
"$size": "$vendors"
}
}
}])
To identify accounts that use multiple payment sources:
Use grouping to count data from multiple account records and group the result by account with count
Use a match case is to filter only such accounts having more than one payment method
db.payment_collection.aggregate([ { $group: {"_id":"$account" ,
"number":{$sum:1}} }, {
"$match": {
"number": { "$gt": 1 }
}
} ])
This will work perfectly fine,
db.UserModule.aggregate(
{ $group : { _id : { "companyauthemail" : "$companyauthemail", "email" : "$email" }, number : { $sum : 1 } } },
{ $group : { _id : "$_id.companyauthemail", number : { $sum : 1 } } }
);
An example
db.collection.distinct("example.item").forEach( function(docs) {
print(docs + "==>>" + db.collection.count({"example.item":docs}))
});